我是Jquery新手,在使用Uploadify上传文件时遇到问题。我试图将文件发送到Java Servlet,后者使用apache commons上传文件。
但是,在执行代码时,并不会全部使用servlet。实际上,我从网络上获取了此示例示例并进行了尝试。谁能告诉我我想念什么?我在下面粘贴客户端代码:

我的理解是uploadify函数中的“脚本”参数将调用servlet。因此,我在脚本参数中写了我的Servlet名称。

请告诉我如何继续进行。

<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">

<html>
  <head>
    <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
    <title>SimpleFileUpload</title>
    <script type="text/javascript" src="js/swfobject.js"></script>
    <script type="text/javascript" src="js/jquery-1.3.2.min.js"></script>
    <script type="text/javascript" src="js/jquery.uploadify.v2.1.0.min.js"></script>
    <link rel="stylesheet" href="css/uploadify.css" type="text/css" media="screen"/>

    <script type="text/javascript">
        $(function() {
            $('#file_upload').uploadify({
                'uploader' : 'swf/uploadify.swf',
                'script' : 'fileupload',
                'cancelImg' : 'img/cancel.png',
                'multi' : false
            });
        });
    </script>
</head>
<body>
    <h1>Simple File Upload</h1>

    <h3>Multiple file upload made easy</h3>

    <div id="file_upload"></div>
    <br/>
    <input type="button" value="Clear Queue" onclick="$('#file_upload').uploadifyClearQueue();"/>
    <input type="button" value="Submit Queue" onclick="$('#file_upload').uploadifyUpload();"/>
</body>




而我的servlet代码是

protected void doPost(HttpServletRequest request, HttpServletResponse response)               throws ServletException, IOException {

    if (ServletFileUpload.isMultipartContent(request)){
          // Parse the HTTP request...
        try {
        ServletFileUpload servletFileUpload = new ServletFileUpload(new DiskFileItemFactory());
        List fileItemsList = servletFileUpload.parseRequest(request);

        Iterator itr = fileItemsList.iterator();

        while(itr.hasNext()) {
            FileItem item = (FileItem) itr.next();

        // check if the current item is a form field or an uploaded file
            if(item.isFormField()) {

        // get the name of the field
            String fieldName = item.getFieldName();

    // if it is name, we can set it in request to thank the user
        if(fieldName.equals("name"))
        request.setAttribute("msg", "Thank You: " + item.getString());

            } else {


                File fullFile  = new File(item.getName());


                String filename = item.getName();



                 File targetdir = new File("E:/");
            File savedFile = new File(targetdir,fullFile.getName());
                 item.write(savedFile);

            }
        }
        }
           catch(Exception e)
           {
               e.printStackTrace();
           }
        } //end servletFileUpload


谢谢,

最佳答案

我只是使用swfupload ...使用小提琴手,我看到我的基本示例只是使用uploadify提交了1个文件,所以我放弃了。

关于java - 使用Uploadify时出错,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/1726564/

10-12 12:29
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