php page run another script in background process not working

Question

I've read this How can I run another PHP script using PHP? and some other sources.

so i have 2 files. testexec.php and testfile.php, the purpose of the testfile.php is to insert some query into table. the result running testexec.php is a success but no data inserted. i try accessing testfile.php directly and the data is inserted.

in testexec.php:

<?php
    ignore_user_abort(true);
    $pathFile = dirname(__FILE__) . "/textfile.php";
    $cmd = "usr/bin/php " . $pathFile;
    exec( $cmd , $output, $return);

    if(!$return) {
        echo "Fail";
    } else {
        echo "Success" . "<br/>";
    }
?>

in testfile.php

<?php
    include("database.php");
    $myDB = new Database();

    $dateTime = new DateTime();
    $query = "INSERT INTO marking (description) VALUES ('Test Exec');";
    $result = $myDB->insertMysqli($query);
    if($result) {
        echo "success";
    } else {
        echo "fail";
    }
?>

I think that's because you use relative path here "usr/bin/php" Try to use absolute - "/usr/bin/php" — Ihor Skliar
– Ihor Skliar, Commented Mar 3, 2017 at 10:30
Definitely. And a general remark: a "background script" always should log its activity somewhere. Preferably in a separate log file. And another thing: Most likely you would have been able to spot the cause if the issue yourself by taking a look into your http servers error log file... — arkascha
– arkascha, Commented Mar 3, 2017 at 10:34
try change the path to "/usr/bin/php". page return success but the data is not inserted. thanks for the quick reply. — Alex
– Alex, Commented Mar 3, 2017 at 10:52
i found solution to the problem. just wrong name file in $pathFile. should be testfile not textfile. :D the logfile really help arkascha. thanks very much. — Alex
– Alex, Commented Mar 3, 2017 at 11:27

Guillaume Sainthillier · Accepted Answer · 2017-03-03 10:35:55Z

0

Try to replace the lines

$cmd = "usr/bin/php " . $pathFile;

By

$cmd = "/usr/bin/php " . $pathFile;

And

include("database.php");

By

include(dirname(__FILE__) ."/database.php");

answered Mar 3, 2017 at 10:35

Guillaume Sainthillier

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