Bash

BlueSpirit · 01-23-2012, 10:18 AM

Hi,

I'd like to do something like that

Code:

#!/bin/bash
hello=varname
$hello=kennedy

echo $varname

I expected the output to be "kennedy".
However it just won't work

Any thoughts?

BlueSpirit

Cedrik · 01-23-2012, 10:23 AM

Although powerfull, bash is just shell language

I mean for whatever variable value it sees at the start of line, it will try to execute it as a command

The indirect reference works like this

Code:

#!/bin/bash
varname=hello
hello=kennedy

echo ${!varname}

flamelord · 01-23-2012, 12:15 PM

You can do something like this

Code:

#!/bin/bash
hello=varname
eval $hello=kennedy
echo $varname

then in the second line the $hello is expanded by bash so that you are actually evaluating "varname=kennedy"

David the H. · 01-23-2012, 12:39 PM

Please don't use eval for things like this. It's a very insecure option.

There are much better ways to do this. Have a look here:

http://mywiki.wooledge.org/BashFAQ/006

I'd give good odds that an associative array, or possibly even a regular array, would be a better solution than an indirect reference, although we'd need more details about your actual goals to tell for sure.

devUnix · 01-24-2012, 11:34 PM

David ->

Thanks for sharing this knowledge-base:

http://mywiki.wooledge.org/BashFAQ/006

I have now got that "eval" is a really a misspelling for "evil" and would look for better approaches/substitutions of it.

David the H. · 01-25-2012, 05:56 AM

Glad to be of service.

To be fair, there are legitimate uses for eval. You just have to be very cautious about what you are doing, and only use it when you are very sure about what's happening every step of the way.

The main point I want to convey is that there are nearly always other, safer, better options available for doing this kind of thing.