Using variable in bash command and collecting output in a variable is not working

Question

I have the following command which, if fired with a hardcoded IP works fine -

ad_request_output="$(/usr/local/nagios/libexec/check_http -w 0.18 -c 0.25 -H <Some private IP> -u 'http://<Some private IP>/fam/postGetAd.php?site_id=76986&partner_id=27de34b6f8b03d81&banner_id=183517&timeout=5000&version=1.5.1&language=jsp&format=wap&phone_headers=REMOTE_ADDR=>166.137.8.134||REMOTE_HOST=>http://localhost||HTTP_USER_AGENT=>Mozilla/5.0')"

echo $ad_request_output gives expected output -

HTTP OK: HTTP/1.1 200 OK - 217 bytes in 0.055 second response time |time=0.054961s;0.180000;0.250000;0.000000 size=217B;;;0

But, using a variable IP gives a different output -

private_ip=<Some private IP>
ad_request_output=$(/usr/local/nagios/libexec/check_http -w 0.18 -c 0.25 -H $private_ip -u 'http://$private_ip/fam/postGetAd.php?site_id=76986&partner_id=27de34b6f8b03d81&banner_id=183517&timeout=5000&version=1.5.1&language=jsp&format=wap&phone_headers=REMOTE_ADDR=>166.137.8.134||REMOTE_HOST=>http://localhost||HTTP_USER_AGENT=>Mozilla/5.0')

echo $ad_request_output gives -

HTTP WARNING: HTTP/1.1 400 Bad Request - 311 bytes in 0.001 second response time |time=0.000703s;0.180000;0.250000;0.000000 size=311B;;;0

Tried with this format of putting variable ${private_ip} as well, but got same output -

 ad_request_output=`/usr/local/nagios/libexec/check_http -w 0.18 -c 0.25 -H ${private_ip} -u 'http://${private_ip}/fam/postGetAd.php?site_id=76986&partner_id=27de34b6f8b03d81&banner_id=183517&timeout=5000&version=1.5.1&language=jsp&format=wap&phone_headers=REMOTE_ADDR=>166.137.8.134||REMOTE_HOST=>http://localhost||HTTP_USER_AGENT=>Mozilla/5.0'`

I checked some related questions Bash - Using variable inside command not working but got no clue what am I doing wrong.

I have working code to use variable in a command -

php /var/cake_1.2.0.6311-beta/beforeInstall.php ${OUTPUT}

But, not sure how to do it when the output needs to be collected in a variable.

Arkadiusz Drabczyk · Accepted Answer · 2016-10-14 07:37:50Z

1

In bash all characters inside ' are preserved. That means that $<variable> is not expanded. See this:

$ something=value
$ echo $something
value
$ echo "$something"
value
$ echo '$something'
$something

In your specific case $private_ip would not be expanded to the value of private_ip. Relevant section from man bash:

Enclosing characters in single quotes preserves the literal value of each character within the quotes. A single quote may not occur between single quotes, even when preceded by a backslash.

answered Oct 14, 2016 at 7:37

Arkadiusz Drabczyk

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Comments

oliv · Accepted Answer · 2016-10-14 07:35:02Z

1

The private_ip variable doesn't get expanded if you use single quote. You should use double quotes:

ad_request_output=$(/usr/local/nagios/libexec/check_http -w 0.18 -c 0.25 -H ${private_ip} -u "http://${private_ip}/fam/postGetAd.php?site_id=76986&partner_id=27de34b6f8b03d81&banner_id=183517&timeout=5000&version=1.5.1&language=jsp&format=wap&phone_headers=REMOTE_ADDR=>166.137.8.134||REMOTE_HOST=>http://localhost||HTTP_USER_AGENT=>Mozilla/5.0")

answered Oct 14, 2016 at 7:35

oliv

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Comments

adiesner · Accepted Answer · 2016-10-14 07:55:06Z

1

Oliv already answered your question correctly. But I would like to add some help so that you can find the answer yourself in the future.

Given your script to begin with (I simplified the command):

#!/bin/bash
private_ip=127.0.0.1
ad_request_output=$(check_http -H $private_ip -u 'http://$private_ip/fam/postGetAd.php')
echo $ad_request_output;

You could simply call bash with the parameter -x, which prints every command before it is executed

bash -x test.sh

Output:

adiesner@local /tmp> bash -x test.sh 
+ private_ip=127.0.0.1
++ check_http -H 127.0.0.1 -u 'http://$private_ip/fam/postGetAd.php'
t.sh: Line 4: check_http: Command not found
[...]

You can clearly see that $private_ip was replaced once, but not the second time.

Another way is to simply output the command by placing "echo" in front of it.

#!/bin/bash
private_ip=127.0.0.1
ad_request_output=$(echo check_http -H $private_ip -u 'http://$private_ip/fam/postGetAd.php')
echo $ad_request_output;

Output:

adiesner@local /tmp> ./test.sh
check_http -H 127.0.0.1 -u http://$private_ip/fam/postGetAd.php

As soon as you know what is going on it should be easy to enter the correct search words.

answered Oct 14, 2016 at 7:55

adiesner

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1 Comment

Sandeepan Nath Over a year ago

Thanks for the debugging tip.

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