通过JSON响应将状态数据发送到POST - 通过JavaScript解析

时间:2016-09-19 18:00:38

标签: javascript php jquery json dropzone.js

我有一个用于上传文件的PHP代码,工作正常,作为最终结果,我发送了JSON响应,上传状态是我无法检索的。

上传文件(POST)后,我的回复如下:

{"html_content":"<p>File was uploaded<\/p>"}   

PHP上传代码如下:

if (!is_file($targetFile)) {
    move_uploaded_file($tempFile,$targetFile);
    $html_content="<p>File was uploaded</p>";
}
else {
    $html_content="<p>You have uploaded duplicate.</p>";
    move_uploaded_file($tempFile,$targetFile);  
}

$json_array=array('html_content'=>$html_content);

header('Content-type: text/json');
header('Content-type: application/json');
echo json_encode($json_array);

和用于显示消息的JavaScript主代码:

this.on("success", function(file,responseText) {
    $.ajax({
        dataType: 'json',
        success: function (response) {

             var htmlElement = document.createElement("div");
             htmlElement.setAttribute("class","success-message");
             var responseText = response.html_content;
             var messageText = document.createTextNode(responseText);
             htmlElement.appendChild(messageText);
             file.previewTemplate.appendChild(htmlElement);

             console.log(response.html_content);
        }
    });
});

当我从AJAX部分解开上面的JS并将变量responseText设置为静态时,一切正常。

此外,当我不使用AJAX并输出console.log(responseText);时,我会在控制台中看到这个:

Object {html_content: "<p>File was uploaded</p>"}

在我的案例中,我有什么想法?

2 个答案:

答案 0 :(得分:0)

您没有向PHP服务器发送请求。指定URL参数。

 $.ajax({url: "your_php_file",
         dataType: 'json',
         success: function (response) {
                   ...
                }
            });

甚至更好:

$.getJSON( "your_php_file", function( data ) { ... });

答案 1 :(得分:0)

好的,解决了。 JavaScript代码看起来像

        this.on("success", function(file,responseText) {

            str = JSON.stringify(responseText);
            responseMessage = $.parseJSON(str);

            var htmlElement = document.createElement("div");
            htmlElement.setAttribute("class","success-message");
            htmlElement.innerHTML = responseMessage.html_content;
            file.previewTemplate.appendChild(htmlElement);
        });