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Php MKDIR不工作。不上传_Php_Image_Upload_Mkdir - Fatal编程技术网

Php MKDIR不工作。不上传

Php MKDIR不工作。不上传,php,image,upload,mkdir,Php,Image,Upload,Mkdir,我的代码如下: for($i=0; $i<count($_FILES['upload_obra']['name']); $i++) { //Get the temp file path $tmpFilePath = $_FILES['upload_obra']['tmp_name'][$i]; echo $tmpFilePath . " tmpFilePath"; //Mak

我的代码如下:

            for($i=0; $i<count($_FILES['upload_obra']['name']); $i++) {
          //Get the temp file path
          $tmpFilePath = $_FILES['upload_obra']['tmp_name'][$i];

           echo $tmpFilePath . "     tmpFilePath";
          //Make sure we have a filepath
          if ($tmpFilePath != ""){

            mkdir('images/fotos-geral/' . $ultimoID, 0700);
            mkdir('images/fotos-geral/' . $ultimoID . '/obra', 0700);

            $newFilePath = 'images/fotos-geral/' . $ultimoID . '/obra/' . $_FILES['upload_obra']['name'][$i];
            if(move_uploaded_file($tmpFilePath, $newFilePath)) {
                echo "rolou";

              //Handle other code here

            }
          }
        }

        for($i=0; $i<count($_FILES['upload_doc']['name']); $i++) {
          //Get the temp file path
          $tmpFilePath = $_FILES['upload_doc']['tmp_name'][$i];

           echo $tmpFilePath . "     tmpFilePath";
          //Make sure we have a filepath
          if ($tmpFilePath != ""){

            mkdir('images/fotos-geral/' . $ultimoID, 0700);
            mkdir('images/fotos-geral/' . $ultimoID . '/doc', 0700);

            $newFilePath = 'images/fotos-geral/' . $ultimoID . '/doc/' . $_FILES['upload_doc']['name'][$i];
            if(move_uploaded_file($tmpFilePath, $newFilePath)) {
                echo "rolou";

              //Handle other code here

            }
          }
        }
在我的表单中存在两次图像上载。每一个都是一个特定的上传。
但是,没有创建粘贴!上载未发生。我已经选择了十张图片,但它不起作用。

你确定要在html表单中使用吗

 <form enctype="multipart/form-data"> </form>

将error_reporting-1添加到代码顶部,查看返回的错误。您在哪里定义$ultimoID?如果它是未定义的,那么您尝试执行mkdir'images/fotos general'——这是毫无意义的。无法创建未指定的目录。$query=从obra中选择MAXcodObra作为ID;$ultimoID=mysql\u query$query或diemysql\u错误;$ultimoID=mysql\u fetch\u数组$ultimoID;$ultimoID=$ultimoID[ID];是的