java - 我的 Servlet 如何从 multipart/form-data 表单接收参数？

Question

我有一个包含这段代码的页面：

<form action="Servlet" enctype="multipart/form-data">
<input type="file" name="file">
<input type="text" name="text1">
<input type="text" name="text2">
</form>

当我request.getParameter("text1");在我的 Servlet 中使用时，它显示为空。如何让我的 Servlet 接收参数？

Answer 1

所有请求参数都嵌入到多部分数据中。您必须使用 Commons File Upload 之类的方法来提取它们：http: //commons.apache.org/fileupload/

Answer 2

Pleepleus is right, commons-fileupload is a good choice.
If you are working in servlet 3.0+ environment, you can also use its multipart support to easily finish the multipart-data parsing job. Simply add an @MultipartConfig on the servlet class, then you can receive the text data by calling request.getParameter(), very easy.

Tutorial - Uploading Files with Java Servlet Technology

Answer 3

使用getParts()

Answer 4

您需要像这样发送参数：

writer.append("--" + boundary).append(CRLF);
writer.append("Content-Disposition: form-data; name=\"" + urlParameterName + "\"" )
                .append(CRLF);
writer.append("Content-Type: text/plain; charset=" + charset).append(CRLF);
writer.append(CRLF);
writer.append(urlParameterValue).append(CRLF);
writer.flush();

在 servlet 方面，处理 Form 元素：

items = upload.parseRequest(request);
Iterator iter = items.iterator();
while (iter.hasNext()) {
       item = (FileItem) iter.next();
       if (item.isFormField()) {
          name = item.getFieldName(); 
          value = item.getString();

   }}