Question

以下代码不会引发任何错误，但数据库不会备份新数据;

<?php
include('conn.php');
$dbhost="localhost";
$dbuser="landshop_anitgop";
$dbpass="password";

$source=array("landshop_formdata.sql","landshop_clientdata.sql","landshop_blogs.sql","landshop_counter.sql","landshop_hitems.sql");

$path = '"C:/Program Files/EasyPHP-5.3.8.1/www/Landshoppe/Downloaded Dbs/"';

        $mysqldump = '"pathToExe".mysqldump.exe"';
       foreach($source as $db) {
            $dbsource= $path.$db;
            $command = "mysql -u{$dbuser} -p{$dbpass} {$db} < ".$path."backup_{$db}.sql";
       system($command,$output);
       echo "Backed Up : ".$db." Output [".print_r($output,1)."]<br>";

     //system('mysql -u <user> -p<password> dbname < filename.sql');  



}

?>

不知道什么是错的..

Answer 1

两点：

我认为你必须引用你的路径

 //$path = "C:\Program Files\EasyPHP-5.3.8.1\www\Landshoppe/Downloaded Dbs/";
 // change this into
 $path = '"C:/Program Files/EasyPHP-5.3.8.1/www/Landshoppe/Downloaded Dbs/"';

为什么不使用 mysqldump ？

本准则应该有效

       <?php 
           $path = '"C:/Program Files/EasyPHP-5.3.8.1/www/Landshoppe/Downloaded Dbs/"';
           $mysqldump = '"pathToExe".mysqldump.exe"';
           foreach($source as $db) {
                $dbsource= $path.$db;
                $command = $mysqldump." -u{$dbuser} -p{$dbpass} {$db} > ".$path."backup_{$db}.sql";
           system($command,$output);
           echo "Backed Up : ".$db." Output [".print_r($output,1)."]<br>";
            }

sql文件没有备份

1 个答案: