Saving awk output to variable

Question

Can anyone help me out with this problem?

I'm trying to save the awk output into a variable.

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

EDIT: $12 corresponds to a parameter running on that process.

Thanks!

The problem is not really clear. What is the error you get? What is the line from `ps -ef` and what do you expect in $variable ? (if you can not provide real data, then generate a sanitized example that shows the problem instead.) — NiKiZe, Sep 06 '13 at 01:35
Problem: variable does not show up in the printf statement. End goal: $12 corresponds to a serial number that is running with that process. I want to save that parameter into a variable so I can work with it. — Jeremy, Sep 06 '13 at 02:39

score 67 · Accepted Answer · answered Sep 06 '13 at 01:25

67

#!/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.

answered Sep 06 '13 at 01:25

vvns

3,548
3
41
57

thanks RUI & NiKiZe, it was the spacing before and after the equal sign. – Jeremy Sep 06 '13 at 02:54
@wens , how to save multiple variables, from a single command? – user2713461 Jun 26 '20 at 09:06

NiKiZe · Answer 2 · 2017-01-28T23:05:33.860

24

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.

edited Jan 28 '17 at 23:05

answered Sep 06 '13 at 01:16

NiKiZe

1,256
10
26

2

There's supposed to be an `awk` after the last pipe and before the `'{printf`. It was probably accidentally removed when the answer was edited. – aDroid Jan 27 '17 at 22:02

score 3 · Answer 3 · answered Sep 06 '13 at 03:40

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.

score 0 · Answer 4 · answered Sep 06 '13 at 04:48

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

Saving awk output to variable

4 Answers4

Linked