Shell Programming and Scripting

Trying to do something like this

Code:

ls -lrt | awk '$9=="test5"'
-rw-r--r-- 1 lrmq db2iadm1  381 Sep 20 21:56 test5

But now, I need to give a variable in place of test5. For example let's define x as test5

Code:

x=test5
ls -lrt | awk '$9=="$x"'

This doesn't seem to be working. It doesn't take the value of x if we input it as "$x".

Any help as to how use shell variable in awk command is greatly appreciated.

I also tried double quotes to $x, which makes two double quotes after ==. Even this didn't work.

Code:

ls -lrt | awk '$9==""$x""'

Code:

 
ls -l | awk -v x=test5 '$9==x{print x}'

Do I need to declare it again inside the awk. That variable gets changed time to time. So I would need a way to avoid declaring the variable inside the awk.

Code:

ls -l | awk -v y=$X '$9==y{print y}

Works great, thanks !