PHP MySQL準備好的語句中的語句

Question

在modelSlikeVrijednost我參考了model -s主鍵。 ModelSlikeVrijednost可以包含很多圖片（取決於用戶）。我需要刪除基於modelID的文件夾。

路徑示例：/home/mainSite/public_html/site/img/1/1/。

可以做到這一點嗎？

代碼：

if ($stmt = $mysqli->prepare("SELECT modelID FROM model WHERE proizvodacID='$id'")) {  
    $stmt->execute(); 

    $stmt->bind_result($modelID); 

    while ($stmt->fetch()) { 
     $path="/home/mainSite/public_html/site/img/".$id."/".$modelID."/"; 

     if ($stmt1 = $mysqli->prepare("SELECT modelSlikeVrijednost FROM modelSlike WHERE modelID='$modelID'")) {  
      $stmt1->execute(); 

      $stmt1->bind_result($slike); 

      while ($stmt1->fetch()) { 
       if(is_null($slike)){ 
        rmdir($path); 
       } 
       else{ 
        $slikePath="/home/mainSite/public_html/site/".$slike; 
        if($slikePath!=$path){ 
         unlink($slikePath); 
        } 
        rmdir($path); 
       } 
      } 

      $stmt1->close(); 

     } 
     else { 
      printf("Prepared Statement Error: %s\n", $mysqli->error); 
     } 
    } 

    $stmt->close(); 

} 

我得到這個錯誤：Prepared Statement Error: Commands out of sync; you can't run this command now Prepared Statement Error: Commands out of sync; you can't run this command now

Answer 1

不，你不能....您完成所有的結果需要循環，關閉遊標，或使用一個單獨的連接。

但是你正在嘗試做的是更好地完成了反正加入...

SELECT ms.modelSlikeVrijednost, m.modelID FROM model m, modelSlike ms 
WHERE ms.modelID= m.modelID 
AND m.proizvodacID ='$id' 

這會給你你每一行中需要的所有信息。

但是你也不正確地使用準備好的語句。你不應該直接傳遞php變量，你應該將它們作爲參數綁定到查詢中：

$sql = 'SELECT ms.modelSlikeVrijednost, m.modelID FROM model m, modelSlike ms' 
     .' WHERE ms.modelID= m.modelID' 
     .' AND m.proizvodacID = ?'; 

if($stmt = $mysqli->prepare($sql)) { 

    // bind the $id to the parameter as an integer 
    $stmt->bind_param('i', $id); 

    $stmt->execute(); 

    // bind the fields of the result to the same variables you had before 
    $stmt->bind_result($slike, $modelID); 

    // less prone to error if we only type this manually once :-) 
    $basePath = "/home/mainSite/public_html/site"; 

    while($stmt->fetch()) { 

     $path= $basePath . "/img/".$id."/".$modelID."/"; 
     $slikePath = $basePath . "/" . $slike; 

     if(is_null($slike)){ 
      rmdir($path); 
     } else { 
      if($slikePath!=$path) { 
       unlink($slikePath); 
      } 

      rmdir($path); 
     } 
    } 
} 

Answer 2

不要使用裸mysqli API。
讓自己一個輔助類，如safemysql
那麼你的代碼將

$models = $db->getCol("SELECT modelID FROM model WHERE proizvodacID=?i",$id); 
foreach($models as $modelID) { 
    $path = "/home/mainSite/public_html/site/img/$id/$modelID/"; 
    $sql = "SELECT modelSlikeVrijednost FROM modelSlike WHERE modelID=?i"; 
    $sarr = $db->getCol($sql, $modelID)); 

    foreach($sarr as $silke) { 
     if(!$slike)) { 
      rmdir($path); 
     } else { 
      $slikePath="/home/mainSite/public_html/site/".$slike; 
      if($slikePath!=$path){ 
       unlink($slikePath); 
      } 
       rmdir($path); 
      } 
     } 
    } 
} 

但是，是的，這是更好地做一個查詢，如prodigitalson說：

$sql = "SELECT ms.modelSlikeVrijednost, m.modelID FROM model m, modelSlike ms 
     WHERE ms.modelID= m.modelID AND m.proizvodacID=?i"; 
$sarr = $db->getCol($sql, $id); 
foreach($sarr as $silke) { 
    if(!$slike)) { 
     rmdir($path); 
    } else { 
     $slikePath="/home/mainSite/public_html/site/".$slike; 
     if($slikePath!=$path){ 
      unlink($slikePath); 
     } 
      rmdir($path); 
     } 
    } 
} 

主要思路是從查詢中獲取數據，然後使用它。