php exec mysql empty output

Question

I try to capture the error output.

   <?php
        $output = array();
        $command = <<<END
mysql -h$host -u$user --password='$pass' --execute="create database $name;" 2>&1
END;
        exec($command, $output, $code);
    ?>

$output returns no value
$code returns 0

But this query returns an error in the terminal: "database already exists".

When I remove 2>&1

     $command = <<<END
        mysql -h$host -u$user --password='$pass' --execute="create database $name;"
END;
     exec($command, $output, $code);

$output returns no value
$code returns 1

How can I get the correct $output and $code value?

Unfortunately it doesn't suit me. I write my own t_exec() function which executes and captures errors for shell commands. — Mooncat No
– Mooncat No, Commented Nov 4, 2011 at 10:31

Paul Salameh · Accepted Answer · 2012-01-20 11:35:09Z

2

Had the same problem, except that I was importing a .sql file instead of using --execute

It worked after I added 2>&1 and using double quotes instead of singles.

exec("mysql -h $host -u $user -p$pass my_db < \"faulty_sql.sql\" 2>&1",$output);

answered Jan 20, 2012 at 11:35

Paul Salameh

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Doctor Putricide · Accepted Answer · 2011-11-04 09:58:40Z

0

Use system()

p.s: why don't use msyql_query('your query'); ?

answered Nov 4, 2011 at 9:58

Doctor Putricide

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2 Comments

Mooncat No Over a year ago

I write my own exec() function which executes and captures errors for shell commands.

Mooncat No Over a year ago

system returns an empty result too

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