[SOLVED] awk(ward) question: How does awk handle expansions with bash?

[deuteros]

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Howdy folks.

I'm trying to iteratively narrow down a list of values held in the variable $toprint. Each value is seperated by a newline.

I want to remove lines whose fields $3 and $4, together, contain the same value as a variable I defined previously in bash called $seccord.

The best way to do this is obviously to tell awk to print only lines for which the string $seccord does not equal fields $3 and $4 in each line:

toprint=`printf '%b\n' $toprint | awk 'BEGIN { FS="," } { if($3$4!='$seccord') print }'`

This doesn't work, however. It prints EVERY line, even though I know for sure that the fields $3 and $4 are equal to $seccord in at least some of them.

I figure it's because of some weirdness in how bash vs. awk expand variables. Does anyone have an idea?

-Deut


EDIT:

I apologize. I got really desperate and experimented with all sorts of quote combinations. This one works:

" ' $seccord ' "

That is, single quotes inside to expand the variable in bash;
Double quotes outside the single quotes to tell awk that it's looking for a string equal to the value of $seccords

You can delete this thread if you want, or if it's a commmon newbie problem you can keep it.

[japettyjohn]

You're on the right track, there is no shell expansion in single quotes in bash. Another approach instead of the bash concentration approach is to use awk's variable setting, e.g. -v testCase="${seccord}" ' awk script ...'.

PS: If the issue is solved, please marked is solved.