php - 使用 php 导出数据库时出现严格标准错误

Question

这是我的 php 代码

//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;
}
?>

当我尝试执行这个文件时。显示此错误。 “严格标准:第 12 行 C:\wamp\www\dbest.php 中只能通过引用传递变量”

为什么会出现这个错误？请帮助我...谢谢

Answer 1

The value of an assignment expression is the value assigned

Assignment Operators

$output=array() 是返回值(空数组)但不返回变量的表达式。因此它不能用作通过引用获取此参数的函数的参数。

但是你可以尝试使用this bug .