PhP shell_exec output convert to variable, and variable to image

Question

In the days I started developing a VPS contol panel that communicates with the OpenVZ kernel using PhP commands.

I would like to display the status of the VPS, so it works so far.

<?php
$VMstatus = shell_exec("sudo vzctl status 17 | cut -d ' ' -f 5");
echo $VMstatus;
?>

It prints "running".

However, if the term "running" word is in the output, I would like to display an image, but the following code does not display any images.

<?php
$VMstatus = shell_exec("sudo vzctl status 17 | cut -d ' ' -f 5");
echo $VMstatus;

if ($VMstatus == 'running') {
  echo "<img src='/assets/img/vps-running.png'>";
} else {
  echo "<img src='/assets/img/vps-down.png'>";
}
?>

What might be the problem?

Thank you in advance for your help!
Have a nice day!

How does the html looks like in your browser? also do you see any 404's in your browsers devtool? — Sander Visser, Jul 16 '19 at 16:30
Could it be that the images are corrupt? can you open them directly in your browser? — Sander Visser, Jul 16 '19 at 16:35
Also if you do `var_dump($VMstatus)` what do you get? it could have whitespaces. — Sander Visser, Jul 16 '19 at 16:37

score 1 · Accepted Answer · answered Jul 16 '19 at 16:43

1

Your shell outputs whitespaces, you can remove those with the trim method. https://www.php.net/manual/en/function.trim.php

<?php
$VMstatus = trim(shell_exec("sudo vzctl status 17 | cut -d ' ' -f 5"));
echo $VMstatus;

if ($VMstatus == 'running') {
  echo "<img src='/assets/img/vps-running.png'>";
} else {
  echo "<img src='/assets/img/vps-down.png'>";
}
?>

answered Jul 16 '19 at 16:43

Sander Visser

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Thank you very much for your answer! :) It works! High five. – Erik Jul 16 '19 at 16:45

PhP shell_exec output convert to variable, and variable to image

1 Answers1