How to overwrite variable value in linux

Question

0.00/5 (No votes)

See more:

i want to overwrite value of variable s1 from "abc" to ---> [more /home/mohsin/Desktop/system_usage_bash_file | grep TIME | awk 'BEGIN {FS=","} {print $3}']

but when i echo i get "abc" on both echos,means it is not overwritten

What I have tried:

#!/bin/bash
s1="abc"
s2="USED"
echo "before"
echo $s1
s1= more /home/mohsin/Desktop/system_usage_bash_file | grep TIME | awk 'BEGIN {FS=","} {print $3}'
echo "after"
echo $s1

Posted 25-Apr-16 22:13pm

Member 12485314

Updated 16-May-16 7:53am

Add a Solution

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Валентин Ким · Answer 1 · 2016-05-16T07:53:00

Solution 2

You need to write it that way:

root@server:~# s1="$(echo 5)"
root@vpn-1gb-nyc3-01:~# echo $s1
5

So you'll need to rewrite your code like this:

C#

s1= "$(more /home/mohsin/Desktop/system_usage_bash_file | grep TIME | awk 'BEGIN {FS=","} {print $3}')"

Posted 16-May-16 7:53am

Валентин Ким

Richard MacCutchan · Answer 2 · 2016-04-25T22:19:00

Solution 1

s1= more /home/mohsin/Desktop/system_usage_bash_file | grep TIME | awk 'BEGIN {FS=","} {print $3}'

As far as I remember shell scripts, you need to put the command in back quotes thus:

s1= `more /home/mohsin/Desktop/system_usage_bash_file | grep TIME | awk 'BEGIN {FS=","} {print $3}'`

See the section titled Command Substitution at bash(1): GNU Bourne-Again SHell - Linux man page[^].

Posted 25-Apr-16 22:19pm

Richard MacCutchan

Comments

Member 12485314 26-Apr-16 4:35am

On putting back quotes i got error that command not found

Richard MacCutchan 26-Apr-16 4:42am

Well I don't expect it was more, grep or awk.

Member 12485314 26-Apr-16 4:54am

actually i am moving to file directory [more /home/mohsin/Desktop/system_usage_bash_file] than i am greping time and after that printing some value out of it

Richard MacCutchan 26-Apr-16 5:10am

That makes no sense. The more command types the contents of a file on the console. Type the pipeline at the prompt and see what happens.

Jochen Arndt 26-Apr-16 4:58am

Start entering the first command on the shell and continue by appending the additional commands:

more /home/mohsin/Desktop/system_usage_bash_file
more /home/mohsin/Desktop/system_usage_bash_file | grep TIME

and so on.

Once it is working, copy it to your shell script and use the backticks as suggested by Richard or use the $(command) format.