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在modelSlikeVrijednost我有一个对model-s 主键的引用。ModelSlikeVrijednost可以包含很多图像（取决于用户）。我需要删除基于modelID.

路径示例：/home/mainSite/public_html/site/img/1/1/.

是否有可能做到这一点 ？

代码：

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();

}

我收到此错误：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

4

2 回答 2

3

不，您不能……您需要遍历所有结果，关闭光标或使用单独的连接。

但是，无论如何，您要尝试做的事情最好通过加入来完成...

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

这将为您提供每行所需的所有信息。

但是，您也错误地使用了准备好的语句。您不应该直接传递 php 变量，您应该将它们作为参数绑定到查询：

$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);
        }
   }
}
于 2013-02-20T14:12:30.617 回答
2

不要使用裸 mysqli API。
给自己一个帮助类，比如safemysql
然后你的代码是

$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，最好在一个查询中完成，就像 prodigitalson 说的：

$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);
        }
    }
}

主要思想是从查询中获取数据，然后使用它。

于 2013-02-20T14:23:13.430 回答