我正在尝试构建原始 HTTP POST 请求。但是,我不想实际连接到服务器并发送消息。
我一直在研究 Apache HTTP 库,希望我可以创建一个 HttpPost 对象,设置实体,然后获取它创建的消息。到目前为止,我可以转储实体,但不能转储出现在服务器端的整个请求。
有什么想法吗?当然,除了重新创建轮子。
解决方案
我将 ShyJ 的响应重构为一对静态类,但原始响应运行良好。这是两个类:
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);
}
}
下面是我用于自己实现的工厂方法,例如:
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;
}
最佳答案
这可以通过 http-client 实现并伪造一些方法。我使用了 3.1
版本的 http-client
。
示例
这段代码:
import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.io.OutputStream;
import org.apache.commons.httpclient.HttpConnection;
import org.apache.commons.httpclient.HttpException;
import org.apache.commons.httpclient.HttpState;
import org.apache.commons.httpclient.StatusLine;
import org.apache.commons.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("http://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);
}
}
会回来
POST / HTTP/1.1
User-Agent: Jakarta Commons-HttpClient/3.1
Host: http://www.google.abc/hi
Content-Length: 4
Content-Type: application/x-www-form-urlencoded
aa=b
关于java - 获取将从 HttpPost 发送的完整的原始 HTTP 请求消息,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13502198/