Question

我初始化变量，使用move_uploaded_file函数然后我写一个基本表单用户将用于上传他们的照片。我试图打印一个error_msg或sucess_msg，显示他们在加载文件时是否成功。没有什么东西出现了。救命！

           $error_msg = "";
           $success_msg = "";
           $name2 = "";


       if ($_POST['parse_var'] == "pic"){

        if (!$_FILES['fileField']['tmp_name']) { 

            $error_msg = '<font color="#FF0000">ERROR: Please browse for an image before you press submit.</font>';

        } else {

            $maxfilesize = 51200; // 51200 bytes equals 50kb
            if($_FILES['fileField']['size'] > $maxfilesize ) { 

                        $error_msg = '<font color="#FF0000">ERROR: Your image was too large, please try again.</font>';
                        unlink($_FILES['fileField']['tmp_name']); 

            } else if (!preg_match("/\.(gif|jpg|png)$/i", $_FILES['fileField']['name'] ) ) {

                        $error_msg = '<font color="#FF0000">ERROR: Your image was not one of the accepted formats, please try again.</font>';
                        unlink($_FILES['fileField']['tmp_name']); 

            } else { 

                        $newname = $id'.jpg';
                        $place_file = move_uploaded_file( $_FILES['fileField']['tmp_name'], "images/$id/".$newname);
                        $success_msg = '<font color="#009900">Your image has been updated, it may take a few minutes for the changes to show... please                                                    be patient.</font>';
            }

        } // close else that checks file exists

}


    <table width="709" align="center" cellpadding="5">
            <form action="edit_profile.php" enctype="multipart/form-data" method="post" name="pic1_form" id="pic1_form">
          <!--    <tr>
                <td width="125" class="style7"><div align="center"><strong>Please Do First &rarr;</strong></div></td>

              </tr>-->
              <tr>
            <td width="16%"><?php print "$user_pic"; ?></td>
            <td width="74%">
              <input name="fileField" type="file" class="formFields" id="fileField" size="42" />
              50 kb max
           </td>
            <td width="10%">
              <input name="parse_var" type="hidden" value="pic" />
              <input  type="submit" name="button" id="button" value="Submit" />
            </td>
           </tr>
           </form></table>

Answer 1

您正在检查上传是否以错误的方式执行：

if ($_SERVER['REQUEST_METHOD'] == 'POST') {
   ... we're in a post situation, so check if a file was uploaded ...
   if ($_FILES['fileField']['error'] !== UPLOAD_ERR_OK) {
      $error_msg = "File upload failed with error code " . $_FILES['fileField']['error'];
   } else {
      ... a file was successfully uploaded ...
      ... process it ...
      if (!move_uploaded_file( ... )) {
        $error_msg = "Failed to move file";
      }
   }
}

检查是否存在表单字段是检查POST是否执行的一种不好的方法。您可能忘记将该字段放入表单，名称可能会更改等等...检查REQUEST_METHOD保证可以正常工作，因为无论正在执行何种请求，都会设置，并且始终是请求的类型（get，post，head等）。

同样，请勿使用用户提供的文件名来验证他们是否上传了图片。只需将其他类型的文件重命名为＆＃34; whatever.jpg＆＃34;即可简单地伪造这个名称。使用服务器端方法检查它是否是图像，例如FileInfo或getimagesize()

Answer 2

我认为您正在寻找的错误就是这个错误：

$newname = $id'.jpg';

你忘记了一个点，整个剧本崩溃了：

$newname = $id.'.jpg';

Answer 3

我没有在该代码中看到您回显错误或成功消息的任何地方。

将图片上传到文件夹时出现问题

3 个答案: