解决当FORM的ENCTYPE="multipart/form-data" 时request.getParameter()获取不到值的方法

I cannot read the submitter using request.getParameter("submitter") (it returns null). ]

Situation:

javax.servlet.HttpServletRequest.getParameter(String) returns null when the ContentType is multipart/form-data

Solutions:

Solution A:

1. download http://www.servlets.com/cos/index.html
2. invoke getParameters() on com.oreilly.servlet.MultipartRequest

Solution B:

1. download http://jakarta.apache.org/commons/sandbox/fileupload/
2. invoke readHeaders() in
org.apache.commons.fileupload.MultipartStream

Solution C:

1. download http://users.boone.net/wbrameld/multipartformdata/
2. invoke getParameter on
com.bigfoot.bugar.servlet.http.MultipartFormData

Solution D:

Use Struts. Struts 1.1 handles this automatically.

说是不详细，接着往下看，另一种解决方法
> Solution B:> 1. download > http://jakarta.apache.org/commons/sandbox/fileupload/> 2. invoke readHeaders() in > org.apache.commons.fileupload.MultipartStreamThe Solution B as given by my dear friend is a bit hectic and a bit complex We can try the following solution which I found much simpler (at least in usage).1. Download one of the versions of UploadFile from http://jakarta.apache.org/commons/fileupload/2. Invoke parseRequest(request) on org.apache.commons.fileupload.FileUploadBase which returns list of org.apache.commons.fileupload.FileItem objects. 3. Invoke isFormField() on each of the FileItem objects. This determines whether the file item is a form paramater or stream of uploaded file. 4. Invoke getFieldName() to get parameter name and getString() to get parameter value on FileItem if it's a form parameter. Invoke write(java.io.File) on FileItem to save the uploaded file stream to a file if the FileItem is not a form parameter.
按照上面的步骤来，果然一切都ok，ＧＯＯＧＬＥ真不错，主要是getFieldName和getString