如何从此示例 HTTP 请求中获取 body？

Question

我试图在 Java 中找到解析 RFC-822 文档的最简单方法。假设我有一个存储 HTTP 消息的 message-queue。请求和响应。因此，它们不会以 "normal" 方式检索，方法是将 socket-connection 发送到 - 比如说 - 端口 80 和 sending/retrieving 来自那里的消息。

在下面的代码中，我故意将 "mail" headers 与 HTTP 消息混合在一起。这是为了证明两者并没有太大的不同。但这不是重点。这是代码：

package httpexample;

import java.io.ByteArrayInputStream;
import java.io.IOException;
import org.apache.http.Header;
import org.apache.http.HttpException;
import org.apache.http.HttpRequest;
import org.apache.http.impl.io.DefaultHttpRequestParser;
import org.apache.http.impl.io.HttpTransportMetricsImpl;
import org.apache.http.impl.io.SessionInputBufferImpl;
import org.apache.http.io.HttpMessageParser;
import org.apache.http.message.BasicHttpEntityEnclosingRequest;

public class HttpExample {

    // RFC 822

    public static void main(String[] args) throws IOException, HttpException {
        String str = "POST http://localhost:8080/foobar/1234567 HTTP/1.1\n" +
            "Message-ID: <19815303.1075861029555.JavaMail.ss@kk>\n" +
            "Date: Wed, 6 Mar 2010 12:32:20 -0800 (PST)\n" +
            "From: someone@someotherplace.com\n" +
            "To: someone@someplace.com\n" +
            "Subject: some subject\n" +
            "Mime-Version: 1.0\n" +
            "Content-Type: text/plain; charset=us-ascii\n" +
            "Content-Transfer-Encoding: 7bit\n" +
            "X-From: one, some <some.one@someotherplace.com>\n" +
            "X-To: one\n" +
            "X-cc: \n" +
            "X-bcc: \n" +
            "X-Origin: Bob-R\n" +
            "X-FileName: rbob (Non-Privileged).pst\n" +
            "\n" +
            "some message\n";
        ByteArrayInputStream fakeStream = new ByteArrayInputStream(
                str.getBytes());
        HttpTransportMetricsImpl metrics = new HttpTransportMetricsImpl();
        SessionInputBufferImpl inbuffer = new SessionInputBufferImpl(metrics, 1024);

        inbuffer.bind(fakeStream);
        HttpMessageParser<HttpRequest> requestParser =
                new DefaultHttpRequestParser(inbuffer);
        BasicHttpEntityEnclosingRequest request = (BasicHttpEntityEnclosingRequest)requestParser.parse();

        for (Header hdr : request.getAllHeaders()) {
            System.out.println(String.format("%-30s = %s", hdr.getName(), hdr.getValue()));
        }
        System.out.println(String.format("Request Line: %s", request.getRequestLine()));
        System.out.println(String.format("Body\n------------------\n%s",
                request.getEntity()));
    }

}

输出如下所示：

Message-ID                     = <19815303.1075861029555.JavaMail.ss@kk>
Date                           = Wed, 6 Mar 2010 12:32:20 -0800 (PST)
From                           = someone@someotherplace.com
To                             = someone@someplace.com
Subject                        = some subject
Mime-Version                   = 1.0
Content-Type                   = text/plain; charset=us-ascii
Content-Transfer-Encoding      = 7bit
X-From                         = one, some <some.one@someotherplace.com>
X-To                           = one
X-cc                           = 
X-bcc                          = 
X-Origin                       = Bob-R
X-FileName                     = rbob (Non-Privileged).pst
Request Line: POST http://localhost:8080/foobar/1234567 HTTP/1.1
Body
------------------
null

我想不通的是如何访问消息的 body。

我希望它具有 some message\n

的内容

我在 BasicHttpEntityEnclosingRequest 中找不到任何方法可以给我这个值。在我使用的早期版本中

HttpRequest request = requestParser.parse();

而不是

BasicHttpEntityEnclosingRequest request = 
    (BasicHttpEntityEnclosingRequest) requestParser.parse();

我将其更改为 BasicHttpEntityEnclosingRequest，因为它具有 getEntity 方法。但是那个 returns null.

所以我有点迷茫。

我在哪里可以找到 body？

Answer 1

我认为问题可能是从您的消息中不清楚 headers body 的长度是多少，因此接收方只是忽略它。 HTTP specification 定义了几个关于如何传达此信息的选项，其中 none 似乎适用于此处：

Content-Transfer-Encoding 必须是 Transfer-Encoding
7bit 不在 the standard options.
当您使用 str.getBytes() 时，它会为您提供未在 Content-Type.

us-ascii

所以，我会稍微更改一下您的要求：

使用headerContent-Type: text/plain; charset=UTF-16
删除headerContent-Transfer-Encoding
添加 Content-Lenght: 28（28 为 "some message\n".getBytes().length()）。

Answer 2

查看 DefaultHttpRequestParser 的源代码，它似乎只解析请求行和 headers，它不会尝试解析 body。

This thread is discussing the same topic. There are few solution proposals as well.

Answer 3

我添加了Content-Lengthheader，否则解析器会忽略POSTbody。我修改了你的代码，现在它解析 body 就好了：

package org.apache.http.examples;

import java.io.ByteArrayInputStream;
import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.net.Socket;

import org.apache.http.Header;
import org.apache.http.HttpException;
import org.apache.http.message.BasicHttpEntityEnclosingRequest;
import org.apache.http.util.EntityUtils;

public class HttpExample {

    // RFC 822

    public static void main(String[] args) throws IOException, HttpException {
        String str = "POST http://localhost:8080/foobar/1234567 HTTP/1.1\n" +
            "Message-ID: <19815303.1075861029555.JavaMail.ss@kk>\n" +
            "Date: Wed, 6 Mar 2010 12:32:20 -0800 (PST)\n" +
            "From: someone@someotherplace.com\n" +
            "To: someone@someplace.com\n" +
            "Subject: some subject\n" +
            "Mime-Version: 1.0\n" +
            "Content-Type: text/plain; charset=us-ascii\n" +
            "Content-Transfer-Encoding: 7bit\n" +
            "X-From: one, some <some.one@someotherplace.com>\n" +
            "X-To: one\n" +
            "X-cc: \n" +
            "X-bcc: \n" +
            "X-Origin: Bob-R\n" +
            "X-FileName: rbob (Non-Privileged).pst\n" +
            "Content-Length: 13\n" +
            "\n" +
            "some message\n";
        ByteArrayInputStream fakeStream = new ByteArrayInputStream(
                str.getBytes());

        BHttpConnectionBaseImpl b = new BHttpConnectionBaseImpl(fakeStream);

        BasicHttpEntityEnclosingRequest request1 = (BasicHttpEntityEnclosingRequest) b.receiveRequestHeader();
        b.receiveRequestEntity(request1);


        for (Header hdr : request1.getAllHeaders()) {
            System.out.println(String.format("%-30s = %s", hdr.getName(), hdr.getValue()));
        }
        System.out.println(String.format("Request Line: %s", request1.getRequestLine()));
        System.out.println(String.format("Body\n------------------\n%s",
                EntityUtils.toString( request1.getEntity() ) ));
    }

}

class BHttpConnectionBaseImpl extends  org.apache.http.impl.DefaultBHttpServerConnection{

    private InputStream inputStream;

    public BHttpConnectionBaseImpl(final InputStream inputStream) {
        super(4048);
        this.inputStream = inputStream;
        try {
            super.bind(new Socket());
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    @Override
    protected InputStream getSocketInputStream(final Socket socket) throws IOException {
        return inputStream;
    }

    @Override
    protected OutputStream getSocketOutputStream(final Socket socket) throws IOException {
        return new ByteArrayOutputStream();
    }
}

POST body 的解析发生在 org.apache.http.impl.BHttpConnectionBase.prepareInput(HttpMessage) 中，无论其唯一的构造函数是 protected 并且需要很多参数。 child org.apache.http.impl.DefaultBHttpServerConnection 有一个方便的 public 构造函数，并在 receiveRequestHeader() 中进行 header 解析。我重载的方法需要绕过一些错误检查，例如如果 Socket == null 并且能够从 fakeStream

读取请求

另一种可能有效的方法是覆盖 Socket，尤其是它的 getInputStream() 和 getOutputStream()，尽管我还没有测试过。然后创建一个 DefaultBHttpServerConnection 的实例并调用它的 bind 方法。其余的应该是一样的。

Answer 4

通过覆盖行解析器自定义解析头：

inbuffer = new SessionInputBufferImpl(new HttpTransportMetricsImpl(), reqDataLength);
inbuffer.bind(input);
HttpMessageParser<org.apache.http.HttpRequest> requestParser = new DefaultHttpRequestParser(
                inbuffer,
                new LineParser(),
                new DefaultHttpRequestFactory(),
                MessageConstraints.DEFAULT
        );

获取实体主体如下：

        HttpEntityEnclosingRequest ereq = (HttpEntityEnclosingRequest) req;
        ContentLengthStrategy contentLengthStrategy =
                    StrictContentLengthStrategy.INSTANCE;
        long len = contentLengthStrategy.determineLength(req);
        InputStream contentStream = null;
        if (len == ContentLengthStrategy.CHUNKED) {
            contentStream = new ChunkedInputStream(buf);
        } else if (len == ContentLengthStrategy.IDENTITY) {
            contentStream = new IdentityInputStream(buf);
        } else {
            contentStream = new ContentLengthInputStream(buf, len);
        }
        BasicHttpEntity ent = new BasicHttpEntity();
        ent.setContent(contentStream);
        ereq.setEntity(ent);
        return ereq;

如何从此示例 HTTP 请求中获取 body？

How do I get the body from this example HTTP request?

java

http

apache-httpcomponents