Assignment to a variable variable.

nikefalcon · 03-17-2011, 04:11 AM

This loop is part of a bash script which takes multiple arguments.

Code:

for ((i=1;i<=$number;++i)) ; do
	offset=$(($i+5))
	url$i=${!offset}
	echo $url1	
done

I get the following error-"./test2: line 6: url1=URLf1: command not found".

Note: "URLf1" is one of the arguments passed to the script.

Any ideas as to how to work around this??

Thanks

Nominal Animal · 03-17-2011, 06:21 AM

You need to do variable expansion before the assignment -- a perfect job for the eval Bash built-in (see man bash-builtins for details). Try

Code:

for ((i=1;i<=$number;++i)) ; do
	offset=$(($i+5))
	eval url$i=${offset}
	eval echo "\\\$url\\\$i = \$url$i"
done

Note how the $ characters in the echo are encoded. eval is applied first; the resulting command (for i=1) is echo "\$url\$i = $url1". The escaping is sometimes pretty hairy, but it's easier if you work backwards: start from the expression you wish to execute, then add escaping, and finally prepend the eval to the command.

Hope this helps.

nikefalcon · 03-17-2011, 06:54 AM

Thank You, eval works perfectly.
So it now reads:

Code:

for ((i=1;i<=$number;++i)) ; do
    offset=$(($i+5))
    eval url$i=${!offset}
    ltemp=url$i
    URL=${!ltemp}
    echo CHECK$URL    
done

grail · 03-17-2011, 07:56 AM

Personally I would try and use arrays instead of the needless call to eval. In general you will find that its use is often not required and generally cause more headaches than it is worth.
Obviously in the simple case as demonstrated you may not run into an issue.