shell即使安全模式关闭，MySQL也无法正常工作

Question

I have a string of script which working in terminal but does not work when I use it in PHP with shell_exec().I know a lot of questions similar to this question has been asked already but in my case the problem I am facing is that I have already tried the proposed solutions I found. Below is my simple code.

<?php

 $output = shell_exec('mysql -u root -pmypass -h 127.0.0.1 mydatabase< db.sql');
 echo "<pre>$output</pre>";

?>

So far this is the best solution I have found.

Does anyone knows what might be wrong?

Answer 1

Your shell_exec is this:

$output = shell_exec('mysql -u root -pmypass -h 127.0.0.1 mydatabase< db.sql');

And your command is this:

mysql -u root -pmypass -h 127.0.0.1 mydatabase< db.sql

The reason that command works when you are in the shell is the binary path to mysql is part of your user login profile.

To see what I mean, login to the shell as yourself and then type echo $PATH and what you will see is a list of search paths the shell uses to figure out where binaries you are attempting to run are located.

But when you attempt to run a script via shell_exec() the Apache server user running PHP is making the sell call. And that user typically does not have binary paths set. So you need to provide the full path to mysql which might be:

/usr/bin/mysql

Or:

/usr/local/bin/mysql

The best solution is from the shell use the which command like so:

which mysql

And then take the full path provided and adjust your shell_exec() command as follows; using /usr/bin/ for example:

$output = shell_exec('/usr/bin/mysql -u root -pmypass -h 127.0.0.1 mydatabase < db.sql');

Also, where is db.sql actually located? You would have to prepend the full path to that MySQL script like this as well; using /full/path/to/this/ for example:

$output = shell_exec('/usr/bin/mysql -u root -pmypass -h 127.0.0.1 my database < /full/path/to/this/db.sql');