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这是我在 ASP 中所做的：

在表格上：

<form id="form1" runat="server">
   ...
   <asp:Button ID="cmdGetFileToUpload" runat="server" Text="Send File" />
   ...
   <asp:Panel id=divUploadFile Visible="False" ...>
      <input id="inpUploadFile" runat="server" name="inpUploadFile" type="file" />
      <asp:Button ID="cmdUploadFile" runat="server" Text="Upload File" />
   </asp:Panel
   ...
</form>

在后面的代码中：

Protected Sub cmdGetFileToUpload_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles cmdGetFileToUpload.Click
   'This shows the upload file panel when the user clicks the Send Files button
   Me.divUploadFile.Visible = True

   ... hide other panels
End Sub

Protected Sub cmdUploadFile_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles cmdUploadFile.Click
   Dim oFile As HttpPostedFile
   Dim sFullPathName as String
   Dim sFileName as String

   oFile = Request.Files(0)
   If oFile.FileName <> "" Then
      sFullPathName = oFile.FileName
      sFileName = sFullPathName.Substring(sFullPathName.LastIndexOf("\") + 1)
      oFile.SaveAs(Server.MapPath(".\myFolder\") & sFileName)
   End If

   ... do stuff with the file
End Sub

我已经下载了 commons-fileupload.jar 及其依赖项并将它们放在我的项目中。但是我不清楚代码应该放在哪里。我有一个类方法，我在回发后设置我的页面时调用，所以把它放在那里，但 servletfileupload.ismultipartcontent(request) 总是返回 false。文档说这通常是因为请求已经被处理。因此，我将代码移至表单顶部，如下所示：

<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
    pageEncoding="ISO-8859-1"
    import="myClasses.myClass,
        org.apache.commons.fileupload.*,
        org.apache.commons.fileupload.servlet.*,
        org.apache.commons.fileupload.disk.*,
        org.apache.commons.fileupload.servlet.*,
        java.io.*, java.util.*"
%>
<!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=ISO-8859-1">
    <title>MyPageTile</title>
    <style>
        ... various styles
    </style>
    <script type="text/javascript">
        ... various scripts
    </script>
</head>
<body style="background-color:tan;" onload="window.history.forward();" >
    <%@ include file="subforms/my_banner.jsp" %>

<%  if (session.getAttribute("user")==null) { response.sendRedirect("index.jsp"); } 
    else {
        boolean isMultipart = ServletFileUpload.isMultipartContent(request);
        if (isMultipart) {
            DiskFileItemFactory factory = new DiskFileItemFactory();
            ServletContext servletContext = this.getServletConfig().getServletContext();
            File repository = (File) servletContext.getAttribute("javax.servlet.context.tempdir");
            factory.setRepository(repository);
            ServletFileUpload upload = new ServletFileUpload(factory);
            List<FileItem> items = upload.parseRequest(request);
            Iterator<FileItem> iter = items.iterator();
            while (iter.hasNext()) {
                FileItem item = iter.next();
                if ( ! item.isFormField() ) {
                    File uploadedFile = new File( "/uploads/" + session.getAttribute("user") + ".xlsm");
                    item.write(uploadedFile);
                }
            }
        }
... other stuff then my form
<form id="AdminForm" enctype="multipart/form-data">

但是，servletfileupload.ismultipartcontent(request) 再次返回 false。

我究竟做错了什么？

4

1 回答 1

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我发现了这个问题 - 我的表单标签没有将方法属性设置为post。我还有调试工作要做，但我至少已经过了这个障碍。希望这对其他人有用。

于 2013-06-16T12:30:51.227 回答