exec和只有变量应该通过引用传递[关闭]

Question

I'm using the PHP export script from 1 and 1:

<?php
//ENTER THE RELEVANT INFO BELOW
$mysqlDatabaseName ='db123456789';
$mysqlUserName ='dbo123456789';
$mysqlPassword ='myPassword';
$mysqlHostName ='db1234.perfora.net';
$mysqlExportPath ='chooseFilenameForBackup.sql';

//DONT EDIT BELOW THIS LINE
//Export the database and output the status to the page
$command='mysqldump --opt -h' .$mysqlHostName .' -u' .$mysqlUserName .' -p' .$mysqlPassword .' ' .$mysqlDatabaseName .' > ~/' .$mysqlExportPath;
exec($command,$output=array(),$worked);
switch($worked){
    case 0:
        echo 'Database <b>' .$mysqlDatabaseName .'</b> successfully exported to <b>~/' .$mysqlExportPath .'</b>';
        break;
    case 1:
        echo 'There was a warning during the export of <b>' .$mysqlDatabaseName .'</b> to <b>~/' .$mysqlExportPath .'</b>';
        break;
    case 2:
        echo 'There was an error during export. Please check your values:<br/><br/><table><tr><td>MySQL Database Name:</td><td><b>' .$mysqlDatabaseName .'</b></td></tr><tr><td>MySQL User Name:</td><td><b>' .$mysqlUserName .'</b></td></tr><tr><td>MySQL Password:</td><td><b>NOTSHOWN</b></td></tr><tr><td>MySQL Host Name:</td><td><b>' .$mysqlHostName .'</b></td></tr></table>';
        break;
}
?>   

If I'm executing it I get Strict Standards: Only variables should be passed by reference in and it points to this line

exec($command,$output=array(),$worked);

How should I adapt the script?

Answer 1

You should define $output before calling exec:

$output = array();
exec($command, $output, $worked);

The error occurs because PHP does not like being passed values created on the spot as a reference (even though in this case it would be perfectly obvious to a more advanced compiler how this should not be an issue).