上传后如何让swfupload返回图片网址？ [英] How do i get swfupload to return image URL after upload?

问题描述

我面临一个非常奇怪的问题。我使用swfupload将多个图像上传到ima​​geshack服务器。我还将返回的URL存储在数据库中。我收到后如何显示从imageshack返回给用户的URL？上传完成后有没有办法让swfupload返回URL？以下是我用于上传图片的代码：

I am facing a really weird problem. I am using swfupload to upload multiple images to the imageshack server. I am also storing the returned URLs in the database. How do i show the URL returned from imageshack to the user as soon as i receive it? Is there any way to make swfupload return the URL once upload finishes? Here is the code that i use for uploading images:

   $('#swfupload-control').swfupload({  
      upload_url: "<%= upload_url-%>",  
      file_post_name: 'photo',  
      post_params: {"authenticity_token" : "<%= form_authenticity_token %>"},
      file_size_limit : "100024",  
      file_types : "*.jpg;*.png;*.gif",  
      file_types_description : "Image files",  
      file_upload_limit : 5,  
      flash_url : "/flash/swfupload.swf",  
      button_image_url : '/javascripts/swfupload/XPButtonUploadText_61x22.png',  
      button_width : 62,  
      button_height : 22,  
      button_placeholder : $('#button')[0],  
      debug: false  
  })  
      .bind('fileQueued', function(event, file){  
          var listitem='<li id="'+file.id+'" >'+  
              'File: <em>'+file.name+'</em> ('+Math.round(file.size/1024)+' KB) <span class="progressvalue" ></span>'+  
            '<div class="progressbar" ><div class="progress" ></div></div>'+  
            '<p class="status" >Pending</p>'+  
            '<span class="cancel" >&nbsp;</span>'+  
            '</li>';  
        $('#log').append(listitem);  
        $('li#'+file.id+' .cancel').bind('click', function(){ //Remove from queue on cancel click  
             var swfu = $.swfupload.getInstance('#swfupload-control');  
             swfu.cancelUpload(file.id);  
             $('li#'+file.id).slideUp('fast');  
         });  
         // start the upload since it's queued  
         $(this).swfupload('startUpload'); 
     }) 
     .bind('fileQueueError', function(event, file, errorCode, message){ 
         alert('Size of the file '+file.name+' is greater than limit'); 
     }) 
     .bind('fileDialogComplete', function(event, numFilesSelected, numFilesQueued){ 
         $('#queuestatus').text('Files Selected: '+numFilesSelected+' / Queued Files: '+numFilesQueued); 
     }) 
     .bind('uploadStart', function(event, file){ 
         $('#log li#'+file.id).find('p.status').text('Uploading...'); 
         $('#log li#'+file.id).find('span.progressvalue').text('0%'); 
         $('#log li#'+file.id).find('span.cancel').hide(); 
     }) 
     .bind('uploadProgress', function(event, file, bytesLoaded){ 
         //Show Progress 
         var percentage=Math.round((bytesLoaded/file.size)*100); 
         $('#log li#'+file.id).find('div.progress').css('width', percentage+'%'); 
         $('#log li#'+file.id).find('span.progressvalue').text(percentage+'%');
     }) 
     .bind('uploadSuccess', function(event, file, serverData){ 
         var item=$('#log li#'+file.id), fetchedfile;  
         item.find('div.progress').css('width', '100%'); 
         item.find('span.progressvalue').text('100%'); 

         var pathtofile='<a href="'+fetchedfile+'" target="_blank" >view &raquo;</a>'; 
         item.addClass('success').find('p.status').html('Done!!! | '+pathtofile); 
     }) 
     .bind('uploadComplete', function(event, file){ 
         // upload has completed, try the next one in the queue 

         $(this).swfupload('startUpload');  
     })  
});

在swfupload文档中没有内置的回调函数来获取上传的图像文件的URL。 / p>

There is no inbuilt callback function for fetching the URL of the uploaded image file in the swfupload documentation.

推荐答案

哦，这很简单。 uploadSuccess返回serverData！

Oh it was simple. uploadSuccess returns serverData!

.bind('uploadSuccess', function(event, file, serverData){ 
         var item=$('#log li#'+file.id), fetchedfile;  
         item.find('div.progress').css('width', '100%'); 
         item.find('span.progressvalue').text('100%'); 
          fetchedfile = serverData;
         var pathtofile='<a href="'+fetchedfile+'" target="_blank" >view &raquo;</a>'; 
         item.addClass('success').find('p.status').html('Done!!! | '+pathtofile); 
     }) 

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