Saving awk output to variable [duplicate]

Answer 1

#!/bin/bash

variable=`ps -ef | grep "port 10 -" | grep -v "grep port 10 -" | awk '{printf $12}'`
echo $variable

Notice that there's no space after the equal sign.

You can also use $() which allows nesting and is readable.

Answer 2

I think the $() syntax is easier to read...

variable=$(ps -ef | grep "port 10 -" | grep -v "grep port 10 -"| awk '{printf "%s", $12}')

But the real issue is probably that $12 should not be qouted with ""

Edited since the question was changed, This returns valid data, but it is not clear what the expected output of ps -ef is and what is expected in variable.

Answer 3

as noted earlier, setting bash variables does not allow whitespace between the variable name on the LHS, and the variable value on the RHS, of the '=' sign.

awk can do everything and avoid the "awk"ward extra 'grep'. The use of awk's printf is to not add an unnecessary "\n" in the string which would give perl-ish matcher programs conniptions. The variable/parameter expansion for your case in bash doesn't have that issue, so either of these work:

variable=$(ps -ef | awk '/port 10 \-/ {print $12}')

variable=`ps -ef | awk '/port 10 \-/ {print $12}'`

The '-' int the awk record matching pattern removes the need to remove awk itself from the search results.

Answer 4

variable=$(ps -ef | awk '/[p]ort 10/ {print $12}')

The [p] is a neat trick to remove the search from showing from ps

@Jeremy If you post the output of ps -ef | grep "port 10", and what you need from the line, it would be more easy to help you getting correct syntax