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HomeBackend DevelopmentPHP Tutorialjavascript - webuploader 怎么验证上传返回的信息呢

webuploader 给的图片上传demo里面的js没有给验证上传是否成功的回调信息
而且那个js里好多代码都是没有经过验证的信息
我现在已经能上传成功 而且保存到数据库里了 但是 上传失败的时候 上传页面没法验证上传成功几个图片 都是返回成功的
我是这样验证的

<code>  uploader.on('uploadSuccess', function (file,response) {
        if(response.status==1){
            $('#' + file.id).addClass('upload-state-done');
            alert('response.url')
        }
        
    });</code>

我现在用的是examples里的image-upload文件夹里的那个例子 是有个按钮点击上传 而且要缩略图的那个 它下面的upload.js里的updateStatus这些都没法检测

有没有一个简单的示例 给发一个

回复内容:

webuploader 给的图片上传demo里面的js没有给验证上传是否成功的回调信息
而且那个js里好多代码都是没有经过验证的信息
我现在已经能上传成功 而且保存到数据库里了 但是 上传失败的时候 上传页面没法验证上传成功几个图片 都是返回成功的
我是这样验证的

<code>  uploader.on('uploadSuccess', function (file,response) {
        if(response.status==1){
            $('#' + file.id).addClass('upload-state-done');
            alert('response.url')
        }
        
    });</code>

我现在用的是examples里的image-upload文件夹里的那个例子 是有个按钮点击上传 而且要缩略图的那个 它下面的upload.js里的updateStatus这些都没法检测

有没有一个简单的示例 给发一个

你没有仔细看文档。。。其实是有的,成功回调里除了file,还有一个response参数,这个就是用来接收服务器返回的数据,你只要返回json,它就能收到。你可以仔细看文档。

// 文件上传成功,给item添加成功class, 用样式标记上传成功。

<code>   uploader.on( 'uploadSuccess', function( file,response) {
        var imgurl=response.url;    //上传图片路径
       $('#images').val(imgurl);
       $( '#'+file.id ).addClass('upload-state-success').find(".state").text("已上传");
   });

response.url 中的url  返回的是我的上传地址,如果它不能判断是否上传成功,那你就是后台返回一个status, if(status==1)  上传成功就可以了</code>
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