Library and examples of parsing multipart/form-dat

问题:

The response to one kind of HTTP request I send is a multipart/form-data looks something like:

--------boundary123
Content-Disposition: form-data; name="json"
Content-Type: application/json

{"some":"json"}
--------boundary123
Content-Disposition: form-data; name="bin"
Content-Type: application/octet-stream

<file data>

--------boundary123

I've been using apache to send and receive the HTTP requests, but I can't seem to find an easy way to use it to parse the above for easy access of the form fields.

I would prefer not to reinvent the wheel, so I'm looking for a library that allows me to do something similar to:

MultipartEntity multipart = new MultipartEntity(inputStream);
InputStream bin = multipart.get("bin");

Any suggestions?

回答1:

See Apache Commons File Upload

If you are using Spring MVC, see this http://static.springsource.org/spring/docs/3.0.0.M3/reference/html/ch16s08.html

回答2:

Example code using deprecated constructor:

import java.io.ByteArrayInputStream;

import org.apache.commons.fileupload.MultipartStream;

public class MultipartTest {

    // Lines should end with CRLF
    public static final String MULTIPART_BODY =
            "Content-Type: multipart/form-data; boundary=--AaB03x\r\n"
            + "\r\n"
            + "----AaB03x\r\n"
            + "Content-Disposition: form-data; name=\"submit-name\"\r\n"
            + "\r\n"
            + "Larry\r\n"
            + "----AaB03x\r\n"
            + "Content-Disposition: form-data; name=\"files\"; filename=\"file1.txt\"\r\n"
            + "Content-Type: text/plain\r\n"
            + "\r\n"
            + "HELLO WORLD!\r\n"
            + "----AaB03x--\r\n";

    public static void main(String[] args) throws Exception {

        byte[] boundary = "--AaB03x".getBytes();

        ByteArrayInputStream content = new ByteArrayInputStream(MULTIPART_BODY.getBytes());

        @SuppressWarnings("deprecation")
        MultipartStream multipartStream =
                new MultipartStream(content, boundary);

        boolean nextPart = multipartStream.skipPreamble();
        while (nextPart) {
            String header = multipartStream.readHeaders();
            System.out.println("");
            System.out.println("Headers:");
            System.out.println(header);
            System.out.println("Body:");
            multipartStream.readBodyData(System.out);
            System.out.println("");
            nextPart = multipartStream.readBoundary();
        }
    }
}

回答3:

Example code without using deprecated methods. import com.google.common.net.MediaType; import org.apache.commons.fileupload.RequestContext;

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.nio.charset.Charset;

public class SimpleRequestContext implements RequestContext {
    private final Charset charset;
    private final MediaType contentType;
    private final byte[] content;

    public SimpleRequestContext(Charset charset, MediaType contentType, byte[] content) {
        this.charset = charset;
        this.contentType = contentType;
        this.content = content;
    }

    public String getCharacterEncoding() {
        return charset.displayName();
    }

    public String getContentType() {
        return contentType.toString();
    }

    @Deprecated
    public int getContentLength() {
        return content.length;
    }

    public InputStream getInputStream() throws IOException {
        return new ByteArrayInputStream(content);
    }
}

{
  ...
  Charset encoding = UTF_8;
  RequestContext requestContext = new SimpleRequestContext(encoding, contentType, body.getBytes());
  FileUploadBase fileUploadBase = new PortletFileUpload();
  FileItemFactory fileItemFactory = new DiskFileItemFactory();
  fileUploadBase.setFileItemFactory(fileItemFactory);
  fileUploadBase.setHeaderEncoding(encoding.displayName());
  List<FileItem> fileItems = fileUploadBase.parseRequest(requestContext);
  ...
}