2016-01-08 33 views
0

我有一個導出腳本將多個表格導入.sql文件作爲備份。該腳本很好,但我想在服務器上保存一份備份!我可以將它作爲記錄插入,但不能像我以前用於照片上傳一樣使用方法move_uploaded_file保存/上傳導出的文件到服務器

這裏是我的腳本:

<?php 
    //ENTER THE RELEVANT INFO BELOW 
    //http://stackoverflow.com/a/31531996/1620626 
    $mysqlUserName  = "root"; 
    $mysqlPassword  = "xxx"; 
    $mysqlHostName  = "localhost"; 
    $DbName    = "xxx_db"; 
    $backup_name  ="mybackup.sql"; 
    $tables    = array("table1","table2"); 

    //or add 5th parameter(array) of specific tables: array("mytable1","mytable2","mytable3") for multiple tables 

    Export_Database($mysqlHostName,$mysqlUserName,$mysqlPassword,$DbName, $tables=false, $backup_name=false); 

    function Export_Database($host,$user,$pass,$name, $tables=false, $backup_name=false) 
    { 
     $mysqli = new mysqli($host,$user,$pass,$name); 
     $mysqli->select_db($name); 
     $mysqli->query("SET NAMES 'utf8'"); 

     $queryTables = $mysqli->query('SHOW TABLES'); 
     while($row = $queryTables->fetch_row()) 
     { 
      $target_tables[] = $row[0]; 
     } 
     if($tables !== false) 
     { 
      $target_tables = array_intersect($target_tables, $tables); 
     } 
     foreach($target_tables as $table) 
     { 
      $result   = $mysqli->query('SELECT * FROM '.$table); 
      $fields_amount = $result->field_count; 
      $rows_num=$mysqli->affected_rows;  
      $res   = $mysqli->query('SHOW CREATE TABLE '.$table); 
      $TableMLine  = $res->fetch_row(); 
      $content  = (!isset($content) ? '' : $content) . "\n\n".$TableMLine[1].";\n\n"; 

      for ($i = 0, $st_counter = 0; $i < $fields_amount; $i++, $st_counter=0) 
      { 
       while($row = $result->fetch_row()) 
       { //when started (and every after 100 command cycle): 
        if ($st_counter%100 == 0 || $st_counter == 0) 
        { 
          $content .= "\nINSERT INTO ".$table." VALUES"; 
        } 
        $content .= "\n("; 
        for($j=0; $j<$fields_amount; $j++) 
        { 
         $row[$j] = str_replace("\n","\\n", addslashes($row[$j])); 
         if (isset($row[$j])) 
         { 
          $content .= '"'.$row[$j].'"' ; 
         } 
         else 
         { 
          $content .= '""'; 
         }  
         if ($j<($fields_amount-1)) 
         { 
           $content.= ','; 
         }  
        } 
        $content .=")"; 
        //every after 100 command cycle [or at last line] ....p.s. but should be inserted 1 cycle eariler 
        if ((($st_counter+1)%100==0 && $st_counter!=0) || $st_counter+1==$rows_num) 
        { 
         $content .= ";"; 
        } 
        else 
        { 
         $content .= ","; 
        } 
        $st_counter=$st_counter+1; 
       } 
      } $content .="\n\n\n"; 
     } 
     $backup_name = $backup_name ? $backup_name : $name."_(".date('H-i-s')."_".date('d-m-Y').").sql"; 
     //$backup_name = $backup_name ? $backup_name : $name.".sql"; 
     header('Content-Type: application/octet-stream'); 
     header("Content-Transfer-Encoding: Binary"); 
     header("Content-disposition: attachment; filename=\"".$backup_name."\""); 

     move_uploaded_file($backup_name, 'files/bak/'.$backup_name); 
     echo $content; exit; 
    } 
?> 

所以,我要導出的文件保存到files/bak作爲用戶下載.sql文件。

+0

這豈不是更容易使用phpMyAdmin,那可幾乎無處不 – RiggsFolly

+0

我需要從PHP執行。 – Wilf

+0

你能告訴我如何使用phpMyadmin導出到服務器嗎? – Wilf

回答

1

因此,這一點的代碼獲取備份文件下載到您的瀏覽器,因此您的客戶端機器,備份是安全和健全的!

header('Content-Type: application/octet-stream'); 
    header("Content-Transfer-Encoding: Binary"); 
    header("Content-disposition: attachment; filename=\"".$backup_name."\""); 

    echo $content; exit; 

move_uploaded_file();有着密切的聯繫通過$ _FILES機制上傳文件,所以不要使用。

所有你需要做的就是寫$content到一個文件,它可以像做簡單的

file_put_contents('files/bak/'.$backup_name, $content); 

Link to manual - file-put-contents

因此,對於你的腳本尾部的完整代碼將

file_put_contents('files/bak/'.$backup_name, $content); 

header('Content-Type: application/octet-stream'); 
header("Content-Transfer-Encoding: Binary"); 
header("Content-disposition: attachment; filename=\"".$backup_name."\""); 

echo $content; 
exit; 
+0

謝謝,但該文件沒有保存到路徑中。我怎樣才能看到錯誤? – Wilf

+0

刷新頁面,我忘記把文件路徑放在原來的 – RiggsFolly

+0

是的,我已經在使用你編輯的最新版本。 – Wilf

0

感謝@RiggsFolly的幫助。我拿出我自己的解決方案:

$bakDir='files/bak/'; 
if (!is_dir($bakDir) or !is_writable($bakDir)) { 
    // Error if directory doesn't exist or isn't writable. 
    exit($bakDir." is unwritable"); 
} elseif (!empty($backup_name)) { 
    // Error if the file exists and isn't writable. 
    if(file_put_contents($bakDir.$backup_name, $content)){ 
     echo "<script>alert('Backup successfully');</script>"; 
    }else{ 
     echo "<script>alert('Backup failed');</script>"; 
    } 
} 
+0

我在我的答案中說過'它可能是儘可能簡單地完成,但這是一個更好的解決方案。 – RiggsFolly

+0

非常感謝。你教會我在哪裏學習。 :d – Wilf