Executing bash script from php

Question

I am trying to execute a bash script from within a php script.

The php script is:

    <?php

    $clientName = $_POST['clientName'];
    $startDate = $_POST['startDate'];
    $endDate = $_POST['endDate'];
    $mode = $_POST['mode'];

    echo("Current working directory = " . getcwd());
    echo("Client Name = " . $clientName . "<br/>\n");
    echo("Start date = " . $startDate . "<br/>\n");
    echo("End date = " . $endDate . "<br/>\n");
    echo("Mode = " . $mode . "<br/>\n");

    $cmd = "/webroot/argRepeater.bash escapeshellarg($clientName) escapeshellarg($startDate) escapeshellarg($endDate) escapeshellarg($mode)";

echo("Command = " . $cmd . "<br/>\n");
var_dump($cmd);

    exec("/bin/bash ./argRepeater.bash escapeshellarg($clientName) escapeshellarg($startDate) escapeshellarg($endDate) escapeshellarg($mode)", $output, $output2);

    echo("Output array = " . print_r($output) . "<br/>\n");
    echo("Output = " . $output2 . "<br/>\n");

    ?>

The above php script takes arguments from an html form. The bash script argRepeater.bash only repeats whatever arguments were given to it. The output is as follows:

 Current working directory = /home/content/31/10199331/htmlClient Name = yum
    Start date = 2013-05-14
    End date = 2013-05-24
    Mode = fir
   Command = ./argRepeater.bash escapeshellarg(yum) escapeshellarg(2013-05-14) escapeshellarg(2013-05-24) escapeshellarg(fir)
string(128) "./argRepeater.bash escapeshellarg(yum) escapeshellarg(2013-05-14) escapeshellarg(2013-05-24) escapeshellarg(fir)" Array ( ) Output array = 1
    Output = 1

My questions:
1. What more needs to be done to ensure argRepeater gets executed?
2. How do I display argRepeater's outputs on the webpage?

hakre · Accepted Answer · 2013-05-04 20:08:29Z

1

1. What more needs to be done to ensure argRepeater gets executed?

Well it needs all that it needs. Troubleshoot your needs if you are not so sure if it worked out or not. Output = 1 signals it did work in the sense that something was executed.

2. How do I display argRepeater's outputs on the webpage?

You already do it:

echo("Output array = " . print_r($output) . "<br/>\n");

so as well, you ask about something that already is solved by your own code.

So probably the problem is just that you are uncertain if you understand your code or not? If so, just double check every statement you write with the manual.

edited May 4, 2013 at 20:08

answered May 4, 2013 at 19:58

hakre

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6 Comments

Sriram Over a year ago

Correct me if I am wrong, but output2 will contain the exit status of the command. since bash returns non-zero on unsuccessful completion, I think the bash script did not execute properly. Am I right?

hakre Over a year ago

That depends on the bash script. each bash script can use it's own return codes, and 1 must not signal failure per-se. Which means you might be right, yes. However it also can show that something did work (also I didn't say you should not further trouble-shoot this). tldp.org/LDP/abs/html/exit-status.html

hakre Over a year ago

I did have written some good troubleshooting guide for shell commands from PHP, I will dig them up, they might be helpful here as well if I see that right. Edit: Please see my answer on "php shell exec acting different to running command directly" for an effective debugging technique.

Sriram Over a year ago

To the above bash script, I added echo "Hello World" >> hello.txt. there is no hello.txt in the directory. These files are hosted on godaddy web hosting. could that be a problem?

hakre Over a year ago

Put the command into a variable first. Output the variable (without executing the command). Does it contain your expected command? (this is independent to your hoster)

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