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up vote 1 down vote favorite

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);
    ?>
  1. $output returns no value
  2. $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);
  1. $output returns no value
  2. $code returns 1

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

share|improve this question
    
Any reason for not using mysql_create_db()‌​? – manatwork Nov 4 '11 at 9:58
    
Unfortunately it doesn't suit me. I write my own t_exec() function which executes and captures errors for shell commands. – Mooncat No Nov 4 '11 at 10:31

2 Answers 2

up vote 1 down vote

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);
share|improve this answer
up vote 0 down vote

Use system()

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

share|improve this answer
    
I write my own exec() function which executes and captures errors for shell commands. – Mooncat No Nov 4 '11 at 10:04
    
system returns an empty result too – Mooncat No Nov 4 '11 at 10:26

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