Java 我的 Servlet 如何从 multipart/form-data 表单接收参数？
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How can my Servlet receive parameters from a multipart/form-data form?
提问by Bolaum
I have a page that has this piece of code:
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>
When I use request.getParameter("text1");in my Servlet it shows null. How can I make my Servlet receive the parameters?
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？
回答by Francisco Paulo
Use getParts()
回答by gorjusborg
All the request parameters are embedded into the multipart data. You'll have to extract them using something like Commons File Upload: http://commons.apache.org/fileupload/
所有请求参数都嵌入到多部分数据中。您必须使用类似 Commons File Upload 的方法提取它们：http: //commons.apache.org/fileupload/
回答by qiangbro
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 @MultipartConfigon the servlet class, then you can receive the text data by calling request.getParameter(), very easy.
Pleepleus 是对的，commons-fileupload 是一个不错的选择。
如果您在 中工作servlet 3.0+ environment，您还可以使用其多部分支持轻松完成多部分数据解析工作。只需@MultipartConfig在 servlet 类上添加一个，就可以通过调用 request 来接收文本数据。getParameter()， 好简单。
回答by DejanR
You need to send the parameter like this:
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();
And on servlet side, process the Form elements:
在 servlet 端，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();
}}
