how to execute unix command in php

Question

i am trying to execute unix command in php script like this.

<?php

echo shell_exec('head -n 1 log_list_23072014|awk  -F ',' '{print $2}'');

?>

This is the file , trying to get the first column of the first row.

NODE,CGR,TERMID,VMGW,ET

but the error message i am getting

Parse error: syntax error, unexpected 'shell_exec' (T_STRING), expecting ',' or ';'.

cant able to find please help.

epects 1 parameter means, i dint get and the parameters which i passed is correct or wrong or else please share the code to get 1st column of the 1st row — John
– John, Commented Jul 24, 2014 at 5:33
You're doing this: shell_exec(FIRSTITEM , SECOND ITEM); and the function can only do this shell_exec(FIRSTITEM); — Darren
– Darren, Commented Jul 24, 2014 at 5:35

denisvm · Accepted Answer · 2014-07-24 05:32:39Z

1

The string you've used is not valid, you have to escape single quotes inside your string:

<?php echo shell_exec('head -n 1 log_list_23072014|awk  -F \',\' \'{print $2}\'');

You can also use exec()

answered Jul 24, 2014 at 5:32

denisvm

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1 Comment

dave_thompson_085 Over a year ago

Slight tweak: you need not quote the comma, and then you need not escape those quotes. In addition, awk can do the equivalent of head -n1 by itself: ' awk -F, \'NR==1{print $2;exit}\' '

l'L'l · Accepted Answer · 2014-07-24 06:02:17Z

1

There was an extra ' within your command. Using your command in a variable can help with identifying errors, and when using the standard exec it's required.

$cmd = 'head -n 1 log_list_23072014 | awk -F , \'{print $2}\'';
echo shell_exec($cmd);

changing it to the format above should work.

edited Jul 24, 2014 at 6:02

answered Jul 24, 2014 at 5:34

l'L'l

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1 Comment

Darren Over a year ago

Wrong. read the manual. Only one $cmd can be passed. EDIT: +1 better for referencing exec

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