Create arrays with brace expansion in loop
I would like to generate a number of arrays that differ only by year. In the loop, I create the arrays with brace expansion and a variable.
I tried the following code without success:
LIST={JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}
do
declare -a 'y$year=('"$LIST"'-$year)'
echo "${y$year[@]}"
done
The outcome should be the list of the following entries:
y1998: JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
y1999: JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
...
y2011: JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
I don't need to print them, just use them in a loop as well as pass them as arguments. Because of the second, eval
in the loop is not sufficient.
shell-script array brace-expansion
add a comment |
I would like to generate a number of arrays that differ only by year. In the loop, I create the arrays with brace expansion and a variable.
I tried the following code without success:
LIST={JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}
do
declare -a 'y$year=('"$LIST"'-$year)'
echo "${y$year[@]}"
done
The outcome should be the list of the following entries:
y1998: JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
y1999: JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
...
y2011: JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
I don't need to print them, just use them in a loop as well as pass them as arguments. Because of the second, eval
in the loop is not sufficient.
shell-script array brace-expansion
Why the hardquotes there? I think you might be trying too hard with the brace expansion...
– mikeserv
Dec 29 '14 at 0:39
The hardquotes, because I want to use the array in a loop while passing the entries as arguments.
– MERose
Dec 29 '14 at 0:43
Please explain in detail (i.e. with example values) what you want to happen.
– Hauke Laging
Dec 29 '14 at 0:56
add a comment |
I would like to generate a number of arrays that differ only by year. In the loop, I create the arrays with brace expansion and a variable.
I tried the following code without success:
LIST={JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}
do
declare -a 'y$year=('"$LIST"'-$year)'
echo "${y$year[@]}"
done
The outcome should be the list of the following entries:
y1998: JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
y1999: JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
...
y2011: JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
I don't need to print them, just use them in a loop as well as pass them as arguments. Because of the second, eval
in the loop is not sufficient.
shell-script array brace-expansion
I would like to generate a number of arrays that differ only by year. In the loop, I create the arrays with brace expansion and a variable.
I tried the following code without success:
LIST={JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}
do
declare -a 'y$year=('"$LIST"'-$year)'
echo "${y$year[@]}"
done
The outcome should be the list of the following entries:
y1998: JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
y1999: JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
...
y2011: JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
I don't need to print them, just use them in a loop as well as pass them as arguments. Because of the second, eval
in the loop is not sufficient.
shell-script array brace-expansion
shell-script array brace-expansion
edited Dec 29 '14 at 1:13
MERose
asked Dec 29 '14 at 0:35
MERoseMERose
299418
299418
Why the hardquotes there? I think you might be trying too hard with the brace expansion...
– mikeserv
Dec 29 '14 at 0:39
The hardquotes, because I want to use the array in a loop while passing the entries as arguments.
– MERose
Dec 29 '14 at 0:43
Please explain in detail (i.e. with example values) what you want to happen.
– Hauke Laging
Dec 29 '14 at 0:56
add a comment |
Why the hardquotes there? I think you might be trying too hard with the brace expansion...
– mikeserv
Dec 29 '14 at 0:39
The hardquotes, because I want to use the array in a loop while passing the entries as arguments.
– MERose
Dec 29 '14 at 0:43
Please explain in detail (i.e. with example values) what you want to happen.
– Hauke Laging
Dec 29 '14 at 0:56
Why the hardquotes there? I think you might be trying too hard with the brace expansion...
– mikeserv
Dec 29 '14 at 0:39
Why the hardquotes there? I think you might be trying too hard with the brace expansion...
– mikeserv
Dec 29 '14 at 0:39
The hardquotes, because I want to use the array in a loop while passing the entries as arguments.
– MERose
Dec 29 '14 at 0:43
The hardquotes, because I want to use the array in a loop while passing the entries as arguments.
– MERose
Dec 29 '14 at 0:43
Please explain in detail (i.e. with example values) what you want to happen.
– Hauke Laging
Dec 29 '14 at 0:56
Please explain in detail (i.e. with example values) what you want to happen.
– Hauke Laging
Dec 29 '14 at 0:56
add a comment |
3 Answers
3
active
oldest
votes
Deferring brace expansion is really a case for eval
, particularly if you want to stringify things — ordinary parameter expansion doesn't do the right thing at the right time.
This should do what it seems you wanted:
LIST={JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}
do
eval "y$year=($LIST-$year)"
tmp="y$year[@]"
echo "${!tmp}"
done
You can't indirect into an array, so it's necessary to stringify the array inside an eval
too if you want to print it out. If you don't, you can take out everything after the ;
.tmp
is used for indirect expansion: with tmp
set to "y$year[@]"
, where $year
is replaced with its value, the expansion ${!tmp}
gives the contents of this iteration's array (what ${y1998[@]}
, etc, would have expanded to).
The above will output:
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
and also create arrays y1998
...y2011
. The declare
s aren't strictly necessary, although they do let you skip an eval
if you're aiming for that and don't need to stringify.
I suggest that this probably isn't the way you really want to achieve your underlying goal, though, whatever that is. Nested loops aren't evil, and if any bit of it's hardcoded you can abstract that out.
why do you not justeval "y$year=...; echo "${y$year[@]}""
? As written your command is susceptible to all kinds of unintended effects depending on the values of$IFS
and filename expansion.
– mikeserv
Dec 29 '14 at 1:16
Minimising escaped characters. All of it is definitely not something you should generally want to do for those reasons and others, yes. If you have an odd enoughIFS
value you're on your own.
– Michael Homer
Dec 29 '14 at 1:19
Yes, my intention is maybe not healthy, but everything works in my little script. What do you mean by$IFS
?
– MERose
Dec 29 '14 at 1:28
If you have, say,-
inIFS
(the input field separator) everything gets cut up in the wrong place. The same happens if you have the defaultIFS
and a space in one of your words.
– Michael Homer
Dec 29 '14 at 1:30
1
I meant, originally, that${!foo[@]}
didn't give the values of the array named infoo
in the way that indirection to a scalar does, and I discovered thatfoo='bar[@]'
means that${!foo}
does what I wanted.
– Michael Homer
Dec 29 '14 at 2:18
|
show 9 more comments
Here's one way (with brace expansion):
unset y _y
for y in {JF,JFE,RFS,JBF,JFI,JMCB,}-{{1998..2000},{2009..2011}}
do case "${_y=y${y#*-}[@]}" in
(y${y#-}*) echo "${!_y}" ;;
(*) declare -a "${_y%???}+=($y)";;
esac; unset y _y; done
That will expand your parameter set for the for
loop out to all of your desired values plus an extra last set which will consist only of the years. For each that does not begin with a digit declare
adds an array member to whatever y${y##*[!0-9]}
comes to, and for each that does, echo
prints it.
So for the first 36 iterations it builds each array, and for the last 6 it prints each. The output is:
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
Here is an alternative, perhaps...
for year in 1998 1999 2000 2009 2010 2011
do printf "%s-$year " JF JFE RFS JBF JFI JMCB
echo; done
That would at least get you the same output...
To also store that in an array as well, probably you could...
for year in 1998 1999 2000 2009 2010 2011
do declare -a "y$year=($(printf "%s-$year " JF JFE RFS JBF JFI JMCB |
tee /dev/fd/2 ))"
year=y$year[@]; year=(${!year})
echo "${#year[@]}"
done 2>&1; unset year
This is the output from the second command, but the first prints the same sans the 6 - which just indicates the array member count.
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998 6
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999 6
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000 6
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009 6
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010 6
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011 6
Remember that declare
is a command and the var$expand=(something $expand)
argument is just that - an arg. declare
's args are expanded in the same way as any other - which is not true of a var$expand=$expand
statement. So you can indirectly declare
in the same way you can indirectly export
or whatever - the hardquotes you use are not necessary I think.
add a comment |
Convert the brace expansion to an array via set --
.
Then you have a simple two dimensional loop to construct your output:
set -- {JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}; do
printf "y%s: " $year
for code do
printf "%s-%s " $code $year
done
echo
done
Yields:
y1998: JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
y1999: JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
y2000: JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
y2009: JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
y2010: JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
y2011: JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Deferring brace expansion is really a case for eval
, particularly if you want to stringify things — ordinary parameter expansion doesn't do the right thing at the right time.
This should do what it seems you wanted:
LIST={JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}
do
eval "y$year=($LIST-$year)"
tmp="y$year[@]"
echo "${!tmp}"
done
You can't indirect into an array, so it's necessary to stringify the array inside an eval
too if you want to print it out. If you don't, you can take out everything after the ;
.tmp
is used for indirect expansion: with tmp
set to "y$year[@]"
, where $year
is replaced with its value, the expansion ${!tmp}
gives the contents of this iteration's array (what ${y1998[@]}
, etc, would have expanded to).
The above will output:
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
and also create arrays y1998
...y2011
. The declare
s aren't strictly necessary, although they do let you skip an eval
if you're aiming for that and don't need to stringify.
I suggest that this probably isn't the way you really want to achieve your underlying goal, though, whatever that is. Nested loops aren't evil, and if any bit of it's hardcoded you can abstract that out.
why do you not justeval "y$year=...; echo "${y$year[@]}""
? As written your command is susceptible to all kinds of unintended effects depending on the values of$IFS
and filename expansion.
– mikeserv
Dec 29 '14 at 1:16
Minimising escaped characters. All of it is definitely not something you should generally want to do for those reasons and others, yes. If you have an odd enoughIFS
value you're on your own.
– Michael Homer
Dec 29 '14 at 1:19
Yes, my intention is maybe not healthy, but everything works in my little script. What do you mean by$IFS
?
– MERose
Dec 29 '14 at 1:28
If you have, say,-
inIFS
(the input field separator) everything gets cut up in the wrong place. The same happens if you have the defaultIFS
and a space in one of your words.
– Michael Homer
Dec 29 '14 at 1:30
1
I meant, originally, that${!foo[@]}
didn't give the values of the array named infoo
in the way that indirection to a scalar does, and I discovered thatfoo='bar[@]'
means that${!foo}
does what I wanted.
– Michael Homer
Dec 29 '14 at 2:18
|
show 9 more comments
Deferring brace expansion is really a case for eval
, particularly if you want to stringify things — ordinary parameter expansion doesn't do the right thing at the right time.
This should do what it seems you wanted:
LIST={JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}
do
eval "y$year=($LIST-$year)"
tmp="y$year[@]"
echo "${!tmp}"
done
You can't indirect into an array, so it's necessary to stringify the array inside an eval
too if you want to print it out. If you don't, you can take out everything after the ;
.tmp
is used for indirect expansion: with tmp
set to "y$year[@]"
, where $year
is replaced with its value, the expansion ${!tmp}
gives the contents of this iteration's array (what ${y1998[@]}
, etc, would have expanded to).
The above will output:
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
and also create arrays y1998
...y2011
. The declare
s aren't strictly necessary, although they do let you skip an eval
if you're aiming for that and don't need to stringify.
I suggest that this probably isn't the way you really want to achieve your underlying goal, though, whatever that is. Nested loops aren't evil, and if any bit of it's hardcoded you can abstract that out.
why do you not justeval "y$year=...; echo "${y$year[@]}""
? As written your command is susceptible to all kinds of unintended effects depending on the values of$IFS
and filename expansion.
– mikeserv
Dec 29 '14 at 1:16
Minimising escaped characters. All of it is definitely not something you should generally want to do for those reasons and others, yes. If you have an odd enoughIFS
value you're on your own.
– Michael Homer
Dec 29 '14 at 1:19
Yes, my intention is maybe not healthy, but everything works in my little script. What do you mean by$IFS
?
– MERose
Dec 29 '14 at 1:28
If you have, say,-
inIFS
(the input field separator) everything gets cut up in the wrong place. The same happens if you have the defaultIFS
and a space in one of your words.
– Michael Homer
Dec 29 '14 at 1:30
1
I meant, originally, that${!foo[@]}
didn't give the values of the array named infoo
in the way that indirection to a scalar does, and I discovered thatfoo='bar[@]'
means that${!foo}
does what I wanted.
– Michael Homer
Dec 29 '14 at 2:18
|
show 9 more comments
Deferring brace expansion is really a case for eval
, particularly if you want to stringify things — ordinary parameter expansion doesn't do the right thing at the right time.
This should do what it seems you wanted:
LIST={JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}
do
eval "y$year=($LIST-$year)"
tmp="y$year[@]"
echo "${!tmp}"
done
You can't indirect into an array, so it's necessary to stringify the array inside an eval
too if you want to print it out. If you don't, you can take out everything after the ;
.tmp
is used for indirect expansion: with tmp
set to "y$year[@]"
, where $year
is replaced with its value, the expansion ${!tmp}
gives the contents of this iteration's array (what ${y1998[@]}
, etc, would have expanded to).
The above will output:
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
and also create arrays y1998
...y2011
. The declare
s aren't strictly necessary, although they do let you skip an eval
if you're aiming for that and don't need to stringify.
I suggest that this probably isn't the way you really want to achieve your underlying goal, though, whatever that is. Nested loops aren't evil, and if any bit of it's hardcoded you can abstract that out.
Deferring brace expansion is really a case for eval
, particularly if you want to stringify things — ordinary parameter expansion doesn't do the right thing at the right time.
This should do what it seems you wanted:
LIST={JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}
do
eval "y$year=($LIST-$year)"
tmp="y$year[@]"
echo "${!tmp}"
done
You can't indirect into an array, so it's necessary to stringify the array inside an eval
too if you want to print it out. If you don't, you can take out everything after the ;
.tmp
is used for indirect expansion: with tmp
set to "y$year[@]"
, where $year
is replaced with its value, the expansion ${!tmp}
gives the contents of this iteration's array (what ${y1998[@]}
, etc, would have expanded to).
The above will output:
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
and also create arrays y1998
...y2011
. The declare
s aren't strictly necessary, although they do let you skip an eval
if you're aiming for that and don't need to stringify.
I suggest that this probably isn't the way you really want to achieve your underlying goal, though, whatever that is. Nested loops aren't evil, and if any bit of it's hardcoded you can abstract that out.
edited Dec 29 '14 at 2:16
answered Dec 29 '14 at 1:10
Michael HomerMichael Homer
46.6k8122161
46.6k8122161
why do you not justeval "y$year=...; echo "${y$year[@]}""
? As written your command is susceptible to all kinds of unintended effects depending on the values of$IFS
and filename expansion.
– mikeserv
Dec 29 '14 at 1:16
Minimising escaped characters. All of it is definitely not something you should generally want to do for those reasons and others, yes. If you have an odd enoughIFS
value you're on your own.
– Michael Homer
Dec 29 '14 at 1:19
Yes, my intention is maybe not healthy, but everything works in my little script. What do you mean by$IFS
?
– MERose
Dec 29 '14 at 1:28
If you have, say,-
inIFS
(the input field separator) everything gets cut up in the wrong place. The same happens if you have the defaultIFS
and a space in one of your words.
– Michael Homer
Dec 29 '14 at 1:30
1
I meant, originally, that${!foo[@]}
didn't give the values of the array named infoo
in the way that indirection to a scalar does, and I discovered thatfoo='bar[@]'
means that${!foo}
does what I wanted.
– Michael Homer
Dec 29 '14 at 2:18
|
show 9 more comments
why do you not justeval "y$year=...; echo "${y$year[@]}""
? As written your command is susceptible to all kinds of unintended effects depending on the values of$IFS
and filename expansion.
– mikeserv
Dec 29 '14 at 1:16
Minimising escaped characters. All of it is definitely not something you should generally want to do for those reasons and others, yes. If you have an odd enoughIFS
value you're on your own.
– Michael Homer
Dec 29 '14 at 1:19
Yes, my intention is maybe not healthy, but everything works in my little script. What do you mean by$IFS
?
– MERose
Dec 29 '14 at 1:28
If you have, say,-
inIFS
(the input field separator) everything gets cut up in the wrong place. The same happens if you have the defaultIFS
and a space in one of your words.
– Michael Homer
Dec 29 '14 at 1:30
1
I meant, originally, that${!foo[@]}
didn't give the values of the array named infoo
in the way that indirection to a scalar does, and I discovered thatfoo='bar[@]'
means that${!foo}
does what I wanted.
– Michael Homer
Dec 29 '14 at 2:18
why do you not just
eval "y$year=...; echo "${y$year[@]}""
? As written your command is susceptible to all kinds of unintended effects depending on the values of $IFS
and filename expansion.– mikeserv
Dec 29 '14 at 1:16
why do you not just
eval "y$year=...; echo "${y$year[@]}""
? As written your command is susceptible to all kinds of unintended effects depending on the values of $IFS
and filename expansion.– mikeserv
Dec 29 '14 at 1:16
Minimising escaped characters. All of it is definitely not something you should generally want to do for those reasons and others, yes. If you have an odd enough
IFS
value you're on your own.– Michael Homer
Dec 29 '14 at 1:19
Minimising escaped characters. All of it is definitely not something you should generally want to do for those reasons and others, yes. If you have an odd enough
IFS
value you're on your own.– Michael Homer
Dec 29 '14 at 1:19
Yes, my intention is maybe not healthy, but everything works in my little script. What do you mean by
$IFS
?– MERose
Dec 29 '14 at 1:28
Yes, my intention is maybe not healthy, but everything works in my little script. What do you mean by
$IFS
?– MERose
Dec 29 '14 at 1:28
If you have, say,
-
in IFS
(the input field separator) everything gets cut up in the wrong place. The same happens if you have the default IFS
and a space in one of your words.– Michael Homer
Dec 29 '14 at 1:30
If you have, say,
-
in IFS
(the input field separator) everything gets cut up in the wrong place. The same happens if you have the default IFS
and a space in one of your words.– Michael Homer
Dec 29 '14 at 1:30
1
1
I meant, originally, that
${!foo[@]}
didn't give the values of the array named in foo
in the way that indirection to a scalar does, and I discovered that foo='bar[@]'
means that ${!foo}
does what I wanted.– Michael Homer
Dec 29 '14 at 2:18
I meant, originally, that
${!foo[@]}
didn't give the values of the array named in foo
in the way that indirection to a scalar does, and I discovered that foo='bar[@]'
means that ${!foo}
does what I wanted.– Michael Homer
Dec 29 '14 at 2:18
|
show 9 more comments
Here's one way (with brace expansion):
unset y _y
for y in {JF,JFE,RFS,JBF,JFI,JMCB,}-{{1998..2000},{2009..2011}}
do case "${_y=y${y#*-}[@]}" in
(y${y#-}*) echo "${!_y}" ;;
(*) declare -a "${_y%???}+=($y)";;
esac; unset y _y; done
That will expand your parameter set for the for
loop out to all of your desired values plus an extra last set which will consist only of the years. For each that does not begin with a digit declare
adds an array member to whatever y${y##*[!0-9]}
comes to, and for each that does, echo
prints it.
So for the first 36 iterations it builds each array, and for the last 6 it prints each. The output is:
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
Here is an alternative, perhaps...
for year in 1998 1999 2000 2009 2010 2011
do printf "%s-$year " JF JFE RFS JBF JFI JMCB
echo; done
That would at least get you the same output...
To also store that in an array as well, probably you could...
for year in 1998 1999 2000 2009 2010 2011
do declare -a "y$year=($(printf "%s-$year " JF JFE RFS JBF JFI JMCB |
tee /dev/fd/2 ))"
year=y$year[@]; year=(${!year})
echo "${#year[@]}"
done 2>&1; unset year
This is the output from the second command, but the first prints the same sans the 6 - which just indicates the array member count.
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998 6
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999 6
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000 6
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009 6
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010 6
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011 6
Remember that declare
is a command and the var$expand=(something $expand)
argument is just that - an arg. declare
's args are expanded in the same way as any other - which is not true of a var$expand=$expand
statement. So you can indirectly declare
in the same way you can indirectly export
or whatever - the hardquotes you use are not necessary I think.
add a comment |
Here's one way (with brace expansion):
unset y _y
for y in {JF,JFE,RFS,JBF,JFI,JMCB,}-{{1998..2000},{2009..2011}}
do case "${_y=y${y#*-}[@]}" in
(y${y#-}*) echo "${!_y}" ;;
(*) declare -a "${_y%???}+=($y)";;
esac; unset y _y; done
That will expand your parameter set for the for
loop out to all of your desired values plus an extra last set which will consist only of the years. For each that does not begin with a digit declare
adds an array member to whatever y${y##*[!0-9]}
comes to, and for each that does, echo
prints it.
So for the first 36 iterations it builds each array, and for the last 6 it prints each. The output is:
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
Here is an alternative, perhaps...
for year in 1998 1999 2000 2009 2010 2011
do printf "%s-$year " JF JFE RFS JBF JFI JMCB
echo; done
That would at least get you the same output...
To also store that in an array as well, probably you could...
for year in 1998 1999 2000 2009 2010 2011
do declare -a "y$year=($(printf "%s-$year " JF JFE RFS JBF JFI JMCB |
tee /dev/fd/2 ))"
year=y$year[@]; year=(${!year})
echo "${#year[@]}"
done 2>&1; unset year
This is the output from the second command, but the first prints the same sans the 6 - which just indicates the array member count.
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998 6
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999 6
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000 6
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009 6
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010 6
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011 6
Remember that declare
is a command and the var$expand=(something $expand)
argument is just that - an arg. declare
's args are expanded in the same way as any other - which is not true of a var$expand=$expand
statement. So you can indirectly declare
in the same way you can indirectly export
or whatever - the hardquotes you use are not necessary I think.
add a comment |
Here's one way (with brace expansion):
unset y _y
for y in {JF,JFE,RFS,JBF,JFI,JMCB,}-{{1998..2000},{2009..2011}}
do case "${_y=y${y#*-}[@]}" in
(y${y#-}*) echo "${!_y}" ;;
(*) declare -a "${_y%???}+=($y)";;
esac; unset y _y; done
That will expand your parameter set for the for
loop out to all of your desired values plus an extra last set which will consist only of the years. For each that does not begin with a digit declare
adds an array member to whatever y${y##*[!0-9]}
comes to, and for each that does, echo
prints it.
So for the first 36 iterations it builds each array, and for the last 6 it prints each. The output is:
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
Here is an alternative, perhaps...
for year in 1998 1999 2000 2009 2010 2011
do printf "%s-$year " JF JFE RFS JBF JFI JMCB
echo; done
That would at least get you the same output...
To also store that in an array as well, probably you could...
for year in 1998 1999 2000 2009 2010 2011
do declare -a "y$year=($(printf "%s-$year " JF JFE RFS JBF JFI JMCB |
tee /dev/fd/2 ))"
year=y$year[@]; year=(${!year})
echo "${#year[@]}"
done 2>&1; unset year
This is the output from the second command, but the first prints the same sans the 6 - which just indicates the array member count.
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998 6
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999 6
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000 6
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009 6
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010 6
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011 6
Remember that declare
is a command and the var$expand=(something $expand)
argument is just that - an arg. declare
's args are expanded in the same way as any other - which is not true of a var$expand=$expand
statement. So you can indirectly declare
in the same way you can indirectly export
or whatever - the hardquotes you use are not necessary I think.
Here's one way (with brace expansion):
unset y _y
for y in {JF,JFE,RFS,JBF,JFI,JMCB,}-{{1998..2000},{2009..2011}}
do case "${_y=y${y#*-}[@]}" in
(y${y#-}*) echo "${!_y}" ;;
(*) declare -a "${_y%???}+=($y)";;
esac; unset y _y; done
That will expand your parameter set for the for
loop out to all of your desired values plus an extra last set which will consist only of the years. For each that does not begin with a digit declare
adds an array member to whatever y${y##*[!0-9]}
comes to, and for each that does, echo
prints it.
So for the first 36 iterations it builds each array, and for the last 6 it prints each. The output is:
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
Here is an alternative, perhaps...
for year in 1998 1999 2000 2009 2010 2011
do printf "%s-$year " JF JFE RFS JBF JFI JMCB
echo; done
That would at least get you the same output...
To also store that in an array as well, probably you could...
for year in 1998 1999 2000 2009 2010 2011
do declare -a "y$year=($(printf "%s-$year " JF JFE RFS JBF JFI JMCB |
tee /dev/fd/2 ))"
year=y$year[@]; year=(${!year})
echo "${#year[@]}"
done 2>&1; unset year
This is the output from the second command, but the first prints the same sans the 6 - which just indicates the array member count.
JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998 6
JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999 6
JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000 6
JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009 6
JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010 6
JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011 6
Remember that declare
is a command and the var$expand=(something $expand)
argument is just that - an arg. declare
's args are expanded in the same way as any other - which is not true of a var$expand=$expand
statement. So you can indirectly declare
in the same way you can indirectly export
or whatever - the hardquotes you use are not necessary I think.
edited Dec 29 '14 at 4:52
answered Dec 29 '14 at 1:01
mikeservmikeserv
45.5k668154
45.5k668154
add a comment |
add a comment |
Convert the brace expansion to an array via set --
.
Then you have a simple two dimensional loop to construct your output:
set -- {JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}; do
printf "y%s: " $year
for code do
printf "%s-%s " $code $year
done
echo
done
Yields:
y1998: JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
y1999: JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
y2000: JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
y2009: JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
y2010: JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
y2011: JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
add a comment |
Convert the brace expansion to an array via set --
.
Then you have a simple two dimensional loop to construct your output:
set -- {JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}; do
printf "y%s: " $year
for code do
printf "%s-%s " $code $year
done
echo
done
Yields:
y1998: JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
y1999: JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
y2000: JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
y2009: JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
y2010: JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
y2011: JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
add a comment |
Convert the brace expansion to an array via set --
.
Then you have a simple two dimensional loop to construct your output:
set -- {JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}; do
printf "y%s: " $year
for code do
printf "%s-%s " $code $year
done
echo
done
Yields:
y1998: JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
y1999: JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
y2000: JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
y2009: JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
y2010: JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
y2011: JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
Convert the brace expansion to an array via set --
.
Then you have a simple two dimensional loop to construct your output:
set -- {JF,JFE,RFS,JBF,JFI,JMCB}
for year in {1998..2000} {2009..2011}; do
printf "y%s: " $year
for code do
printf "%s-%s " $code $year
done
echo
done
Yields:
y1998: JF-1998 JFE-1998 RFS-1998 JBF-1998 JFI-1998 JMCB-1998
y1999: JF-1999 JFE-1999 RFS-1999 JBF-1999 JFI-1999 JMCB-1999
y2000: JF-2000 JFE-2000 RFS-2000 JBF-2000 JFI-2000 JMCB-2000
y2009: JF-2009 JFE-2009 RFS-2009 JBF-2009 JFI-2009 JMCB-2009
y2010: JF-2010 JFE-2010 RFS-2010 JBF-2010 JFI-2010 JMCB-2010
y2011: JF-2011 JFE-2011 RFS-2011 JBF-2011 JFI-2011 JMCB-2011
edited 2 hours ago
mikeserv
45.5k668154
45.5k668154
answered Apr 21 '17 at 13:33
bishopbishop
2,0762822
2,0762822
add a comment |
add a comment |
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Why the hardquotes there? I think you might be trying too hard with the brace expansion...
– mikeserv
Dec 29 '14 at 0:39
The hardquotes, because I want to use the array in a loop while passing the entries as arguments.
– MERose
Dec 29 '14 at 0:43
Please explain in detail (i.e. with example values) what you want to happen.
– Hauke Laging
Dec 29 '14 at 0:56