Re: Using variables in variables names
Oh Mike,
thanks a lot for such detailled and well structured clarification! That
did it, now I can use it (and it gave my script a speedup of nearly 70 %
less running time).
Many thanks!
Dirk
Mike Stroyan schrieb:
On 3/13/06, Paul Jarc [EMAIL PROTECTED] wrote:
Dirk H. Schulz [EMAIL PROTECTED] wrote:
Paul Jarc schrieb:
ac=12 eval dings$ac=wasannersder
And how do I reference it then?
ac=12 eval value=\$dings$ac
echo $value
Or:
ac=12 name=dings$ac echo ${!name}
It seems that you need to use the eval form instead of the ${!var} form
to handle array variables. Here are some examples I played with. The
pattern is to use a backslash to quote the $ for the array name. The $i
in the array examples could be done as \$i because it works out the same
if it is expanded in either the first pass or the second pass.
$ suffix=one
$ eval pre_${suffix}=simple1
$ suffix=two
$ eval pre_${suffix}=simple2
$ suffix=one
$ eval echo \$pre_${suffix}
simple1
$ suffix=two
$ eval echo \$pre_${suffix}
simple2
$ suffix=one
$ i=1
$ eval pre_A_${suffix}[$i]=array1_1
$ i=2
$ eval pre_A_${suffix}[$i]=array1_2
$ suffix=two
$ i=1
$ eval pre_A_${suffix}[$i]=array2_1
$ i=3
$ eval pre_A_${suffix}[$i]=array2_3
$ set | grep pre_
_='pre_A_two[3]=array2_3'
pre_A_one=([1]=array1_1 [2]=array1_1)
pre_A_two=([1]=array2_1 [3]=array2_3)
pre_one=simple1
pre_two=simple2
$ i=1
$ eval echo \${pre_A_${suffix}[$i]}
array2_1
$ eval echo \${pre_A_${suffix}[$i]}
array2_1
$ i=3
$ eval echo \${pre_A_${suffix}[$i]}
array2_3
$ i=2
$ suffix=one
$ eval echo \${pre_A_${suffix}[$i]}
array1_2
--
Mike Stroyan
[EMAIL PROTECTED]
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Using variables in variables names
Hi folks, I am sure this has been asked quite some times, but I did not find anything inspiring or helpful - in fact not too much at all. For accelerating a script I need the possibility to set up an unknown number of arrays and to name them (at least partly) with values of a variable. It is like defining arrays with names that contain an increasing number: array$x. While poking around I found out: This does not work at all, even with simple variables it does not. dings=bums echo $dings bums ac=12 dings$ac=wasannersder -bash: dings12=wasannersder: command not found I looked very deeply into man bash and any manual and howto I found, but did not find out any reason why this does not work. It must be one of the basic principles of expansion and assignment, but I would like to understand it. And then the question of multi-dimensional arrays (see above). The Advanced Bash Scripting Guide mentions that it is possible to have them using indirect referencing - but I did not understand how this could be done. Any idea, hint or help? It would be a great relief after two days of search and research. Dirk ___ Bug-bash mailing list Bug-bash@gnu.org http://lists.gnu.org/mailman/listinfo/bug-bash
Re: Using variables in variables names
Dirk H. Schulz [EMAIL PROTECTED] wrote: ac=12 dings$ac=wasannersder -bash: dings12=wasannersder: command not found Variable names in assignments are not subject to expansion. So since dings$ac, as-is, does not fit the syntax for variable names, it isn't treated as an assignment. This will work: ac=12 eval dings$ac=wasannersder paul ___ Bug-bash mailing list Bug-bash@gnu.org http://lists.gnu.org/mailman/listinfo/bug-bash
Re: Using variables in variables names
Paul Jarc schrieb:
Dirk H. Schulz [EMAIL PROTECTED] wrote:
ac=12 dings$ac=wasannersder
-bash: dings12=wasannersder: command not found
Variable names in assignments are not subject to expansion. So since
dings$ac, as-is, does not fit the syntax for variable names, it
isn't treated as an assignment. This will work:
ac=12 eval dings$ac=wasannersder
And how do I reference it then?
ac=12
eval dings$ac=wasannersder
echo $dings12
wasannersder # that works
echo $('$dings'$ac) # trying to
substitute $ac before $dings...
-bash: $dings12: command not found
Is there any way to reference it without anticipating the result of
indirect referencing?
Dirk
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Re: Using variables in variables names
Dirk H. Schulz [EMAIL PROTECTED] wrote:
Paul Jarc schrieb:
ac=12 eval dings$ac=wasannersder
And how do I reference it then?
ac=12 eval value=\$dings$ac
echo $value
Or:
ac=12 name=dings$ac echo ${!name}
paul
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Re: Using variables in variables names
On 3/13/06, Paul Jarc [EMAIL PROTECTED] wrote:
Dirk H. Schulz [EMAIL PROTECTED] wrote:
Paul Jarc schrieb:
ac=12 eval dings$ac=wasannersder
And how do I reference it then?
ac=12 eval value=\$dings$ac
echo $value
Or:
ac=12 name=dings$ac echo ${!name}
It seems that you need to use the eval form instead of the ${!var} form
to handle array variables. Here are some examples I played with. The
pattern is to use a backslash to quote the $ for the array name. The $i
in the array examples could be done as \$i because it works out the same
if it is expanded in either the first pass or the second pass.
$ suffix=one
$ eval pre_${suffix}=simple1
$ suffix=two
$ eval pre_${suffix}=simple2
$ suffix=one
$ eval echo \$pre_${suffix}
simple1
$ suffix=two
$ eval echo \$pre_${suffix}
simple2
$ suffix=one
$ i=1
$ eval pre_A_${suffix}[$i]=array1_1
$ i=2
$ eval pre_A_${suffix}[$i]=array1_2
$ suffix=two
$ i=1
$ eval pre_A_${suffix}[$i]=array2_1
$ i=3
$ eval pre_A_${suffix}[$i]=array2_3
$ set | grep pre_
_='pre_A_two[3]=array2_3'
pre_A_one=([1]=array1_1 [2]=array1_1)
pre_A_two=([1]=array2_1 [3]=array2_3)
pre_one=simple1
pre_two=simple2
$ i=1
$ eval echo \${pre_A_${suffix}[$i]}
array2_1
$ eval echo \${pre_A_${suffix}[$i]}
array2_1
$ i=3
$ eval echo \${pre_A_${suffix}[$i]}
array2_3
$ i=2
$ suffix=one
$ eval echo \${pre_A_${suffix}[$i]}
array1_2
--
Mike Stroyan
[EMAIL PROTECTED]
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