[SOLVED] Cannot pass bash variable to awk (in order to cut out text)

readmore · 12-31-2013, 07:14 PM

Hi, I am trying to pass a variable to awk so awk can then cut out text to the left. Attempting this in a bash script.

Here's the script....

Code:

#!/bin/bash

var1=five

# this works ... static config 
echo "--manual"
grep "line2" file.txt | awk -F'five' '{print $2}'
echo "end manual"

echo

# this fails ... trying to pass variable from shell to awk
echo "--pass shell var to awk"
grep "line2" file.txt | awk -v awkvar=$var1 -F'awkvar' '{print $2}'
echo "end pass var"

This is file.txt

Code:

[win@lotto ~]$ cat file.txt 
line1     one     two     three
line2     four    five    six
line3     seven   eight   nine

This is my result....

Code:

win@lotto ~]$ ./script.sh 
--manual
    six
end manual

--pass shell var to awk

end pass var
[win@lotto ~]$

It works if I manually pass a value. But not if I try to pass a variable from the script.

I've googled'd a whole lot & tried different syntax. No luck. Hope someone can help.

grail · 12-31-2013, 10:27 PM

The answer to your question is you are doing it wrong. In your manual method you are assigning a value to FS (using -F) but then in second process you create an awk variable using -v and
assign the string 'awkvar' as FS.

The reason this did not work in the way you think is because awk variables are only visible to awk and before entering the single quotes of the awk script you are not in awk (hope that made sense)

So to do what you wanted:

Code:

grep "line2" file.txt | awk -F"$var1" '{print $2}'

That being said, this would seem to be the long way to do this.
Try:

Code:

awk -vawkvar="$var1" '$0 ~ awkvar && /line2/{print $NF}' file.txt

readmore · 01-01-2014, 03:35 AM

Thanks for the info. I ended up going a different direction but your reply got me thinking.

After doing some grep & sed stuff to end up with one line, I was able to just do an awk for a random column. No idea why I was trying that other stuff (trying to cut left/right portions etc...

Anyway...I did this:

Code:

echo $passtoawk1 | awk -vawkvar1="$randomcol1" '{print $awkvar1}'

Thanks again.

grail · 01-01-2014, 04:45 AM

Glad it worked

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