awk computed variable names - Shell Programming and Scripting

craig06y · #1 (**permalink**) 02-04-2008

Is there a way to do make-style computed variable names in awk? e.g. in make

Code:

foo = bar
bar = wocket

I can get "wocket" with

Code:

$($(foo))

Alternatively can you list all defined variables in awk?

thanks

Klashxx · #2 (**permalink**) 02-04-2008

Maybe this helps:

Code:

> foo=var
> awk 'BEGIN{var="wocket";print '${foo}'}'
wocket

fpmurphy · #3 (**permalink**) 02-04-2008

The following code snippet outputs "wooket"

Code:

#!/usr/bin/awk -f

BEGIN {
    bar="wooket";
    foo=bar;
}

END { print foo }

craig06y · #4 (**permalink**) 02-05-2008

Thanks for the reply, but that's not exactly what I'm after. That just assigns "wocket" to bar, copies the value of that to foo, then prints the value of foo.

I was wondering if it's possible to say (as in a makefile)....
- assign the string "bar" to the variable foo.
- assign some string value to the variable bar.
- retrieve the value of a variable (in this case bar), where the name of the variable to retrieve from is specified in another variable (in this case foo).

So writing $($(foo)) accesses the value of a variable whose name is stored in foo.

Klashxx's one is closer to what I need, but it doesn't work if i move the foo=var from the shell into the awk script.

I guess this can't be done in awk. I'll workaround it a different way.
Thanks

vgersh99 · #5 (**permalink**) 02-05-2008

there's no 'eval' in awk: FYI

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