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Output
bar

does it work your end?

once again it works in my TEST directory, but not the real thing. Checked all variables this time...

Thanks for the help but that doesn't produce any output for me.

do you not have to -v and assign the shell variable as an awk variable before using it with an awk command?

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Thanks for the reply.

awk -F: -v var=${Username} '$0 ~ var {gsub(/foo/,"bar",$2); print}' FILENAME

Changes every instance of foo in the second field into bar, not just on lines starting with...

Would really appreciate it if someone could point out my mistake in this line of code, i've been staring blankly at it trying everything i can think of some time now and coming up with nothing.

...

Update:

$sort -t: -k1 AdditionalInfo > AdditionalInfo.srt
$sort -t: -k1 /etc/passwd > etcpasswd.srt
$join -t: AdditionalInfo.srt etcpasswd.srt

Joins the matching fields correctly thanks....

So i'm currently working on a project where I'm attempting to display information of users from the /etc/passwd file and also another information file holding addition information about users.
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