exec 并且只有变量应该通过引用传递

Question

我正在使用1 和 1的PHP 导出脚本：

<?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;
}
?>   

如果我正在执行它，我会得到Strict Standards: Only variables should be passed by reference in并且它指向这一行

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

我应该如何改编剧本？

Answer 1

您应该$output在调用之前定义exec：

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

发生错误是因为 PHP 不喜欢将当场创建的值作为引用传递（即使在这种情况下，对于更高级的编译器来说这应该不是问题是非常明显的）。