通過客戶端上傳（.sql文件）在PHP中恢復MySQL數據庫

Question

好吧，我已經成功地在PHPMyAdmin中創建了一個數據庫，並使用php備份到客戶端;現在我試圖將該數據庫恢復爲同一環境中的新數據庫。

我發現了在mysql中工作的命令，但我似乎無法讓它與客戶端從他們的計算機上傳的文件一起工作。

下面是代碼，我現在所擁有的：

<?php 
    // display form if user has not clicked submit 
    if (!isset($_POST["btn_submit"])) 
    { 
?> 

<!--This will be the form that will hold the information of the entire page.--> 
<form enctype="multipart/form-data"class="elegant-aero" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> 
    <h1>Restore Database</h1> 
    <p> 
     <label> 
      <span>Database File</span> 
      <input type="file" name="backupFile" value="Upload File"> 
     </label> 

     <!--Submit Button--> 
     <label> 
      <span>&nbsp;</span> 
      <input type="submit" name="btn_submit" class="button" value="Add Scenario"/> 
     </label> 
    </p> 
</form> 

<?php 

    } //end if 

    else 
    { 
     if(isset($_FILES['backupFile']['error']) && UPLOAD_ERR_OK == $_FILES['backupFile']['error']) 
     { 
      //Format sql file 
      $file = addslashes(file_get_contents($_FILES['backupFile']['tmp_name'])); 
     } //end if 

     //Set up the MySQL Server 
     $servername = "localhost"; 
     $username = "root"; 
     $password = "password";//hide your password 
     $dbname = "test"; 

     //Create connection to the MySQL server 
     $conn = new mysqli($servername, $username, $password); 

     $sql1 = "CREATE DATABASE test"; 

     if($conn->query($sql1) === TRUE) 
     { 
      echo "Database created successfully<br>"; 
     } else { 
      echo "Error creating database: " . $conn->error; 
     } //end else 

     $sql2 = "mysql --user={$username} --password={$password} --database={$dbname} < $file"; 

     if($conn->query($sql2) === TRUE) 
     { 
      echo "Information uploaded successfully<br>"; 
     } else { 
      echo "Error uploading information: " . $conn->error; 
     } //end else 

    } //end else 

?> 

Answer 1

有你的代碼（哪裏fileScenDetails來自什麼形式發生backupFile？），但根據您的問題一堆問題：

$sql2 = "mysql --user={$username} --password={$password} --database={$dbname} < $file"; 

    if($conn->query($sql2) === TRUE) 
    { 
     echo "Information uploaded successfully<br>"; 
    } else { 
     echo "Error uploading information: " . $conn->error; 
    } //end else 

您試圖運行* nix命令作爲數據庫查詢。相反的：

if($conn->query($sql2) === TRUE) 

你需要做這樣的事情：

if(exec($sql2) === TRUE) 

你真的需要在你想在這裏做什麼讀了，因爲它是真的危險的，不應該由初學者嘗試。