gawk - using variable in search pattern

[washad]

Nice - forum, first time user but hope to be more regular now.

So, yes, this is a homework question, but no - I'm not trying to get anyone to do it for me. I think that I am really close, but can't quite get one small aspect to work.

in gawk, I want to include a variable name in the search string, but the below code doesn't work.

read -p 'Login name please? ' uLogName
userID=$(gawk -F: '/"$uLogName"/{print $3}' /etc/passwd)
echo $userID

This, does, however:

read -p 'Login name please? ' uLogName
userID=$(gawk -F: '/stevej/{print $3}' /etc/passwd)
echo $userID

I've explored the -v option, but have had no luck. I've Googled myself blue in the face looking for a usable example.

Thanks,

[theNbomr]

Assuming you are doing this with Bash, it could be that your version of bash does not grok the $( command ) syntax. Try backticks instead:

userID=`gawk -F: '/stevej/{print $3}' /etc/passwd`

--- rod.

[washad]

Thank you for the prompt reply; I may not have written my question clearly enough, though.

The one that works is

userID=$(gawk -F: '/stevej/{print $3}' /etc/passwd)

The one that doesn't work is

userID=$(gawk -F: '/"$uLogName"/{print $3}' /etc/passwd)

If I use the code example that you gave me, but put in the variable name instead,it continues not to work.

userID=`gawk -F: '/"$userID"/{print $3}' /etc/passwd`

So my question still remains as to how to substitute a variable in gawk.

Incidentally, I finally went with the solution below. It works, so I was able to move on. I still would like to understand gawk better for the future.

read -p 'Login name please? ' uLogName
userID=$(grep "$uLogName" /etc/passwd | gawk -F: '{print $3}')
if (( ${#userID} != 0))
	then
		echo; echo "According to /etc/passwd,"
		echo "Your userID is: $userID"
	else
		echo "The userID was not found"
		exit 1
fi

Thanks again,

SteveJ

[theNbomr]

The shell variable in your original example was enclosed in single quotes, preventing it from being interpolated. Once it released from those quotes, it can be interpolated, rather than taking its value as the literal string '$uLogName'.

--- rod.