Calling MySQL exe using PHP exec doesn’t work

Question

I have a simple script:

<?php

  $dbhost = 'localhost';
  $dbuser = 'root';
  $dbpass = 'password';
  $db = 'BPOJob';
  $file =dirname(__FILE__).'\\'.'job_create.sql';
  $mySQLDir='"C:\Program Files\MySQL\MySQL Server 5.1\bin\mysql.exe"';     

  $cmd = $mySQLDir.' -h '.$dbhost.' --user='.$dbuser.' --password='.$dbpass.' < "'.$file.'"';

   exec($cmd,$out,$retval);
   echo "\n";
echo ($retval);

?>

The thing is when I run the above command in PHP.exe, it doesn't work; I got a 1 as a return value from exec.

But if I run the command line directly by calling

mysql.exe -h localhost --user=admin --password=password < job_create.sql

Then the .sql file can run.

Is there anything I miss?

Edit: Sorry! Wrong info. I have edited the question and it's still valid.

Answer 1

Hm, shouldn't you exec $cmd instead of $mySQLDir?

EDIT: Try to echo $cmd maybe you could see an error.

Answer 2

The spaces in the command line are probably being treated as separators, so it's actually trying to run:

c:\program

with arguments:

files ...

Try enclosing the whole of the executable path in double-quotes as it's passed to exec(), so that the windows shell knows that it's supposed to be one word.

Answer 3

I found the solution: you need to explicitly enclosed the $cmd in quote:

 exec('"'.$cmd.'"',$out ,$retval);

This is the full script, for your reference:

<?php

  $dbhost = 'localhost';
  $dbuser = 'root';
  $dbpass = 'password';
  $db = 'BPOJob';
  $file =dirname(__FILE__).'\\'.'job_create.sql';
  $mySQLDir='"C:\Program Files\MySQL\MySQL Server 5.1\bin\mysql.exe"';     

  $cmd = $mySQLDir.' -h '.$dbhost.' --user='.$dbuser.' --password='.$dbpass.' < "'.$file.'"';

   exec('"'.$cmd.'"',$out,$retval);
   echo "\n";
echo ($retval);

?>