Question

我正在处理用户同时发送图像和评论的表单。无论发生什么，ajax调用直接导致错误（即使文件和注释成功上传并写入DB）。

我尝试记录错误的反馈，这就是我收到的：

Object {readyState: 4, responseText: "<pre class='xdebug-var-dump' dir='ltr'>↵<b>array</…DstagramPost succesfully made.{"check":"success"}", status: 200, statusText: "OK"}

这是我的ajax电话：

<script type="text/javascript"> 
$(document).ready(function(){

    $("#postSubmit").on("click", function(e){
        $.ajax({
            url: "ajax/postUpload.php", 
            type: "POST",             
            data: new FormData($("#postForm")[0]),
            dataType: 'json',
            contentType: false,       
            cache: false,             
            processData:false,        
            success: function(data)  
            {
                console.log(data);
            },
            error: function (request, status, error) {
                console.log(request);
                alert('Something is wrong');
            }
        });
        e.preventDefault();
    });
});
</script>

我的PHP代码：

<?php
include_once("../classes/Post.class.php");
session_start();
$post = new Post();

var_dump($_FILES);
if(!empty($_POST)){

$file_tmp_name = $_FILES['postImage']['tmp_name'];

$post->checkIfImage();
$post->checkFileSize();
$post->checkFileFormat();
$post->checkAspectRatio();

if(!$post->checkIfImage()){
    $uploadStatus['image'] = "false";
}

if(!$post->checkFileSize()){
    $uploadStatus['size'] = "false";
}

if(!$post->checkFileFormat()){
    $uploadStatus['format'] = "false";
}

if(!$post->checkAspectRatio()){
    $uploadStatus['ratio'] = "false";
}

if($post->checkIfImage() && $post->checkFileSize() && $post-    >checkFileFormat() && $post->checkAspectRatio()){
    $result = $post->createPost($file_tmp_name);
    $uploadStatus['check'] = "success";
}else{
    $uploadStatus['check'] = "error";
}

header('Content-Type: application/json; charset=utf-8', true);
echo json_encode($uploadStatus);
}

?>

总而言之，如果图像通过了checkIfImage（）等所有要求，它就会上传到文件夹，图像路径和描述会在数据库中发布。如果它没有通过所有要求，它就不会在那里发布。但是我每次都会收到错误。

如果我删除了dataType：＆＃39; json＆＃39;部分它会出错，如果我将其更改为其他任何内容，我会得到以下输出：

<pre class='xdebug-var-dump' dir='ltr'>
<b>array</b> <i>(size=1)</i>
'postImage' <font color='#888a85'>=&gt;</font> 
<b>array</b> <i>(size=5)</i>
  'name' <font color='#888a85'>=&gt;</font> <small>string</small> <font        color='#cc0000'>'passfoto.png'</font> <i>(length=12)</i>
  'type' <font color='#888a85'>=&gt;</font> <small>string</small> <font color='#cc0000'>'image/png'</font> <i>(length=9)</i>
  'tmp_name' <font color='#888a85'>=&gt;</font> <small>string</small> <font color='#cc0000'>'D:\xampp\tmp\phpDFE9.tmp'</font> <i>(length=24)</i>
  'error' <font color='#888a85'>=&gt;</font> <small>int</small> <font color='#4e9a06'>0</font>
  'size' <font color='#888a85'>=&gt;</font> <small>int</small> <font color='#4e9a06'>112825</font>
</pre>D:\xampp\htdocs\IMDstagramPost succesfully made.{"check":"success"}

我需要的部分当然是检查，所以我可以例如清除表格如果检查=成功或如果不是则给出错误..

Answer 1

问题是因为您在private void CreateZipContentFolder(string zipsPath, string destinationPath) { Zips = Directory.GetFiles(zipsPath, "*.zip", SearchOption.TopDirectoryOnly).ToList(); if (Zips.Count != 0) { MyLog.WriteToLog("Creating Folder of ZipFiles... From: " + zipsPath + " To: " + destinationPath, MyLog.Messages.Info); foreach (string zip in Zips) { FileInfo fileInfo = new FileInfo(zip); string dirName = destinationPath + "\\" + fileInfo.Name.Substring(0, fileInfo.Name.Length - 4); using (ZipArchive archive = ZipFile.OpenRead(zip)) { foreach (ZipArchiveEntry entry in archive.Entries) { if (entry.FullName.EndsWith("/")) { try { ZipFile.ExtractToDirectory(zip, destinationPath); } catch (IOException e) { MyLog.WriteToLog(e.Message, MyLog.Messages.Error); } break; } else if (new FileInfo(dirName).Exists == false) { try { Directory.CreateDirectory(dirName); ZipFile.ExtractToDirectory(zip, dirName); } catch (IOException e) { MyLog.WriteToLog(e.Message, MyLog.Messages.Error); } break; } } } } MyLog.WriteToLog("Created Temporary Folders", MyLog.Messages.Info); } else { MyLog.WriteToLog("No Zips Found in: " + zipsPath, MyLog.Messages.Warning); } } JSON之前在PHP文件中输出了一些HTML。您可以在JSON echo参数中看到它：

response

删除该HTML。响应文本应仅此JSON格式的字符串：

<pre class='xdebug-var-dump' dir='ltr'>↵<b>array</…DstagramPost succesfully made.

Ajax调用返回错误，状态为200，状态文本为OK

1 个答案: