Process multiple PHP scripts in one array

Question

I have multiple php scripts to ping each of our locations, and I'm trying to list all results on one page.

Here's the ping script:

<?php
  $host  = "10.10.10.10"; //IP adress to ping
  $loc = ("HQ"); //Name of location
  $output = array();
  echo("<b>$loc</b> <i>(IP: $host)</i> is ");
        exec("ping -n 1 $host 2>&1", $output);
     //print_r($output);
      //you can use  print_r($output)  to view the output result
      if (count($output) > 7) {
        $output = null;
        die ("<font color='green'><b>up</b></font>");
        }
        else {
        $output = null;
        die ("<font color='red'><b>down</b></font>");
        }       
?>

So, I have many php files with this script, where the only difference is host and loc. I've tried to include each file in a new php file using include like this:

<?php
include "file1.php";
include "file2.php";
include "file3.php";
...and so on...
?>

But this only outputs the result of the first file.

How can I do this in any other way?

Thanks!

using die means you are exiting your code. you should use echo instead of die — Kamran
– Kamran, Commented Apr 22, 2015 at 10:43

ksbg · Accepted Answer · 2015-04-22 10:43:48Z

4

die(), or its equivalent exit() ends the script. Simply change die() to echo as in:

if (count($output) > 7) {
    $output = null;
    echo "<font color='green'><b>up</b></font>";
} else {    
    $output = null;
    echo "<font color='red'><b>down</b></font>";
}

answered Apr 22, 2015 at 10:43

ksbg

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