Apache Http Client 4 Form Post Multi-part data

Question

I need to post some form parameters to a server through an HTTP request (one of which is a file). So I use Apache HTTP Client like so...

HttpPost httpPost = new HttpPost(urlStr);

params = []
params.add(new BasicNameValuePair("username", "bond"));
params.add(new BasicNameValuePair("password", "vesper"));
params.add(new BasicNameValuePair("file", payload));

httpPost.setEntity(new UrlEncodedFormEntity(params));
httpPost.setHeader("Content-type", "multipart/form-data");

CloseableHttpResponse response = httpclient.execute(httpPost);

The server returns an error, stack trace is..

the request was rejected because no multipart boundary was found
at org.apache.commons.fileupload.FileUploadBase$FileItemIteratorImpl.<init>(FileUploadBase.java:954)
at org.apache.commons.fileupload.FileUploadBase.getItemIterator(FileUploadBase.java:331)
at org.apache.commons.fileupload.FileUploadBase.parseRequest(FileUploadBase.java:351)
at org.apache.commons.fileupload.servlet.ServletFileUpload.parseRequest(ServletFileUpload.java:126)
at org.springframework.web.multipart.commons.CommonsMultipartResolver.parseRequest(CommonsMultipartResolver.java:156)

I understand from other posts that I need to somehow come up with a boundary, which is a string not found in the content. But how do I create this boundary in the code I have above? Should it be another parameter? Just a code sample is what I need.

Answer 1

I accepted gustf's answer because it got rid of the exception I was having and so I thought I was on the right track, but it was not complete. The below is what I did to finally get it to work...

File payload = new File("/Users/CasinoRoyaleBank")
MultipartEntity entity = new MultipartEntity( HttpMultipartMode.BROWSER_COMPATIBLE );
entity.addPart( "file", new FileBody(payload))
entity.addPart( "username", new StringBody("bond"))
entity.addPart( "password", new StringBody("vesper"))
httpPost.setEntity( entity );
CloseableHttpResponse response = httpclient.execute(httpPost);