Linux : Get total cpu usage by httpd - php

Question

I need to display the total Percentage of CPU utilized by httpd processes on a server in a php report.
I am calling following from exec :
ps -e -o %mem,%cpu,cmd | grep httpd | awk ' {memory+=$1;cpu+=$2} END {printf("%05.2f ",memory);printf("%05.2f\n",cpu)}'
But the above command's reported CPU usage and the one reported by top command are not matching.
I need to report --> If CPU is busy at 40%, 10% of httpd processes, 20% of mysqld processes, 10% of perl processes, then I need to report the 10% of httpd. (Assuming that there are no other processes).
I saw this : get apache total cpu usage in (linux)
But I understand that ps command returns the percentage of CPU consumed by a process out of the total percentage of CPU consumed. I understand that it is getting messy, so the below example should help.
If httpd is consuming 10% of CPU which is busy at 60% then the actual contribution of httpd to make CPU busy was ((100/60)*10) = 16.66 %. Is this correct? What else are the best way to get cpu usage by a group of processes by the same name.

Answer 1

try this in ssh
ps aux | grep "httpd" | awk '{sum1 +=$3}; END {print sum1}'
output is:
10.5
and this for sum of memory
ps aux | grep "httpd" | awk '{sum1 +=$4}; END {print sum1}'

Answer 2

I'm not 100% sure on what you're asking, but if I'm right, this answer might help you:
<?php
exec('ps -aux', $processes);
foreach($processes as $process){
$cols = split(' ', ereg_replace(' +', ' ', $process));
if (strpos($cols[2], '.') > -1){
$cpuUsage += floatval($cols[2]);
}
}
print($cpuUsage);
?>
and after searching many forms also found the another way:
after searching on forums and trying many methods but I have not tried it:
$stat1 = file('/proc/stat');
sleep(1);
$stat2 = file('/proc/stat');
$info1 = explode(" ", preg_replace("!cpu +!", "", $stat1[0]));
$info2 = explode(" ", preg_replace("!cpu +!", "", $stat2[0]));
$dif = array();
$dif['user'] = $info2[0] - $info1[0];
$dif['nice'] = $info2[1] - $info1[1];
$dif['sys'] = $info2[2] - $info1[2];
$dif['idle'] = $info2[3] - $info1[3];
$total = array_sum($dif);
$cpu = array();
foreach($dif as $x=>$y) $cpu[$x] = round($y / $total * 100, 1);

Answer 3

This works for me on OSX:
<?php
exec('ps -e -o %mem,%cpu,command | grep httpd', $output);
$proc_data = [];
foreach($output as $key => $value) {
// Make sure it's only path httpd and not the grep included
if (strstr($value, '/httpd')) {
$info = explode(' ', trim($value), 5);
unset($info[1]);
unset($info[2]);
$proc_data[] = array_merge($info);
}
}
echo '<pre>';
print_r($proc_data);
echo '</pre>';
// Caclulate total CPU percentages
$total_cpu = 0;
foreach ($proc_data as $key => $value) {
$total_cpu += $value[1];
}
echo $total_cpu;
?>
This is the Terminal output for the bash:
MacBook-Pro:~ user$ ps -e -o %mem,%cpu,command | grep httpd
0,2 0,0 /Applications/MAMP/Library/bin/httpd -k start
0,0 0,0 /Applications/MAMP/Library/bin/httpd -k start
0,1 0,0 /Applications/MAMP/Library/bin/httpd -k start
0,0 0,0 /Applications/MAMP/Library/bin/httpd -k start
0,1 0,0 /Applications/MAMP/Library/bin/httpd -k start
0,0 0,0 /Applications/MAMP/Library/bin/httpd -k start
0,0 0,0 /Applications/MAMP/Library/bin/httpd -k start
0,0 0,0 /Applications/MAMP/Library/bin/httpd -k start
0,0 0,0 /Applications/MAMP/Library/bin/httpd -k start
0,0 0,0 /Applications/MAMP/Library/bin/httpd -k start
0,0 0,0 /Applications/MAMP/Library/bin/httpd -k start
0,0 9,0 /Applications/MAMP/Library/bin/httpd -k start
0,0 0,0 grep httpd
I noticed ps -e -o %mem,%cpu,cmd has to be command, but it might be an OSX-only thing tho. Hope you can work with this.
Good luck!