我想建立一个原始的HTTP POST请求。我不想真正连接到服务器并发送邮件,但是。
我一直在打探将Apache HTTP库,希望我能创造一个HttpPost对象,设置实体,然后抓住它会创建消息。到目前为止,我可以转储实体,而不是整个请求,因为它会出现在服务器侧
任何想法?除了刚刚重新创建轮,当然。
解决方案我重构ShyJ的反应成一对静态类的,但是原始响应工作得很好。这里有两个类:
公共静态最后一类LoopbackPostMethod扩展的PostMethod { 私有静态最后弦乐STATUS_LINE =HTTP / 1.1 200 OK; @覆盖 保护无效readResponse(州的HttpState,康涅狄格州的HttpConnection)抛出IOException异常,HttpException { 状态行=新的状态行(STATUS_LINE); }}公共静态最后一类LoopbackHttpConnection扩展的HttpConnection { 私有静态最后弦乐HOST =127.0.0.1; 私有静态最终诠释PORT = 80; 私人最终的OutputStream fOutputStream; 公共LoopbackHttpConnection(OutputStream中的OutputStream){ 超(主机,端口); fOutputStream =的OutputStream; } @覆盖 公共无效flushRequestOutputStream()抛出IOException异常{/ *什么也不做* /} @覆盖 公众的OutputStream getRequestOutputStream()抛出IOException异常,IllegalStateException异常{ 返回fOutputStream; } @覆盖 公共无效写入(字节[]数据)抛出IOException异常,IllegalStateException异常{ fOutputStream.write(数据); }}下面是工厂方法,我使用了我自己的实现,作为一个例子:
私人的ByteBuffer createHtt prequest(ByteBuffer的数据)抛出HttpException,IOException异常{ LoopbackPostMethod的PostMethod =新LoopbackPostMethod(); 最后ByteArrayOutputStream的OutputStream =新ByteArrayOutputStream(); postMethod.setRequestEntity(新ByteArrayRequestEntity(data.array())); postMethod.execute(新的HttpState(),新LoopbackHttpConnection(OutputStream中)); 字节[]字节= outputStream.toByteArray(); ByteBuffer的缓冲= ByteBuffer.allocate(bytes.length); buffer.put(字节); 返回缓冲区;}解决方案
这可以和 achived HTTP客户端,并伪造一些方法。我用的 3.1 版本 HTTP客户端。
的示例的
这code:
进口java.io.ByteArrayOutputStream中;进口java.io.IOException异常;进口java.io.OutputStream中;进口org.apachemons.httpclient.HttpConnection;进口org.apachemons.httpclient.HttpException;进口org.apachemons.httpclient.HttpState;进口org.apachemons.httpclient.StatusLine;进口org.apachemons.httpclient.methods.PostMethod;公共类主要{ 公共静态无效的主要(字串[] args)抛出异常{ 最后ByteArrayOutputStream BAOS =新ByteArrayOutputStream(); 的PostMethod方法=新的PostMethod(){ @覆盖 保护无效readResponse(州的HttpState,康涅狄格州的HttpConnection) 抛出IOException异常,HttpException { 状态行=新状态行(HTTP / 1.1 200 OK); } }; method.addParameter(AA,B); method.execute(新的HttpState(),新的HttpConnection(www.google.abc/hi,80){ @覆盖 公共无效flushRequestOutputStream()抛出IOException } @覆盖 公众的OutputStream getRequestOutputStream()抛出IOException异常, IllegalStateException异常{ 返回BAOS; } @覆盖 公共无效写入(字节[]数据)抛出IOException异常, IllegalStateException异常{ baos.write(数据); } }); 最后弦乐postBody =新的String(baos.toByteArray()); 的System.out.println(postBody); }}将返回
POST / HTTP / 1.1用户代理:Jakarta Commons的-的HttpClient / 3.1主持人:www.google.abc/hi内容长度:4内容类型:应用程序/ x-WWW的形式urlen codeDAA = BI'm trying to build a raw HTTP POST request. I don't want to actually connect to the server and send the message, however.
I've been poking around the Apache HTTP libraries, hoping that I could just create an HttpPost object, set the entity, and then grab the message that it would have created. So far, I can dump the entity, but not the entire request as it'd appear on the server-side.
Any ideas? Aside from just re-creating the wheel, of course.
SolutionI refactored ShyJ's response into a pair of static classes, however the original response works just fine. Here are the two classes:
public static final class LoopbackPostMethod extends PostMethod { private static final String STATUS_LINE = "HTTP/1.1 200 OK"; @Override protected void readResponse(HttpState state, HttpConnection conn) throws IOException, HttpException { statusLine = new StatusLine (STATUS_LINE); } } public static final class LoopbackHttpConnection extends HttpConnection { private static final String HOST = "127.0.0.1"; private static final int PORT = 80; private final OutputStream fOutputStream; public LoopbackHttpConnection(OutputStream outputStream) { super(HOST, PORT); fOutputStream = outputStream; } @Override public void flushRequestOutputStream() throws IOException { /* do nothing */ } @Override public OutputStream getRequestOutputStream() throws IOException, IllegalStateException { return fOutputStream; } @Override public void write(byte[] data) throws IOException, IllegalStateException { fOutputStream.write(data); } }Here's the factory method that I'm using for my own implementation, as an example:
private ByteBuffer createHttpRequest(ByteBuffer data) throws HttpException, IOException { LoopbackPostMethod postMethod = new LoopbackPostMethod(); final ByteArrayOutputStream outputStream = new ByteArrayOutputStream(); postMethod.setRequestEntity(new ByteArrayRequestEntity(data.array())); postMethod.execute(new HttpState(), new LoopbackHttpConnection(outputStream)); byte[] bytes = outputStream.toByteArray(); ByteBuffer buffer = ByteBuffer.allocate(bytes.length); buffer.put(bytes); return buffer; }解决方案
This can be achived with http-client and faking some methods. I used the 3.1 version of http-client.
Example
This code:
import java.io.ByteArrayOutputStream; import java.io.IOException; import java.io.OutputStream; import org.apachemons.httpclient.HttpConnection; import org.apachemons.httpclient.HttpException; import org.apachemons.httpclient.HttpState; import org.apachemons.httpclient.StatusLine; import org.apachemons.httpclient.methods.PostMethod; public class Main { public static void main(String[] args) throws Exception { final ByteArrayOutputStream baos = new ByteArrayOutputStream(); PostMethod method = new PostMethod () { @Override protected void readResponse(HttpState state, HttpConnection conn) throws IOException, HttpException { statusLine = new StatusLine ("HTTP/1.1 200 OK"); } }; method.addParameter("aa", "b"); method.execute(new HttpState (), new HttpConnection("www.google.abc/hi", 80) { @Override public void flushRequestOutputStream() throws IOException { } @Override public OutputStream getRequestOutputStream() throws IOException, IllegalStateException { return baos; } @Override public void write(byte[] data) throws IOException, IllegalStateException { baos.write(data); } }); final String postBody = new String (baos.toByteArray()); System.out.println(postBody); } }will return
POST / HTTP/1.1 User-Agent: Jakarta Commons-HttpClient/3.1 Host: www.google.abc/hi Content-Length: 4 Content-Type: application/x-www-form-urlencoded aa=b
更多推荐
获取充分,原始的HTTP请求会从HttpPost发送消息
发布评论