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Java HTTP请求_Java_Http Post_Http Request - Fatal编程技术网

Java HTTP请求

Java HTTP请求,java,http-post,http-request,Java,Http Post,Http Request,我正在尝试使用java设置httprequest 比如说我想要这样的东西: POST /TEST/PATH HTTP/1.1 [...] POST / HTTP/1.1 [...] SOAPServerAddress = "http://example.com/TEST/PATH"; // setup HTTP POST HttpClient httpClient = HttpClients.custom().build(); HttpPost httpPost = new HttpPo

我正在尝试使用java设置httprequest

比如说我想要这样的东西:

POST /TEST/PATH HTTP/1.1
[...]
POST / HTTP/1.1
[...]
SOAPServerAddress = "http://example.com/TEST/PATH";

// setup HTTP POST
HttpClient httpClient = HttpClients.custom().build();

HttpPost httpPost = new HttpPost(SOAPServerAddress);

// execute http request 
HttpResponse response = httpClient.execute(httpPost);
我想获得“POST”后面的/TEST/PATH。 我没有找到任何二传者来定义它。 当我执行请求时,请求发送如下所示:

POST /TEST/PATH HTTP/1.1
[...]
POST / HTTP/1.1
[...]
SOAPServerAddress = "http://example.com/TEST/PATH";

// setup HTTP POST
HttpClient httpClient = HttpClients.custom().build();

HttpPost httpPost = new HttpPost(SOAPServerAddress);

// execute http request 
HttpResponse response = httpClient.execute(httpPost);
我的代码如下所示:

POST /TEST/PATH HTTP/1.1
[...]
POST / HTTP/1.1
[...]
SOAPServerAddress = "http://example.com/TEST/PATH";

// setup HTTP POST
HttpClient httpClient = HttpClients.custom().build();

HttpPost httpPost = new HttpPost(SOAPServerAddress);

// execute http request 
HttpResponse response = httpClient.execute(httpPost);

看来你对每件事都做了正确的定义。您使用的是
HttpPost
,因此HTTP方法实际上是POST,您的URL与您想要获得的URL匹配。你试过运行你的代码吗?如果没有,试试看。如果是,有什么问题?我希望标题如下所示:POST/TEST/PATH。但我得到的一切都是POST/。你知道一个设定者来定义这个吗?有什么问题吗?您想说您的请求发送到了错误的URL,即发送到
/
而不是
TEST/PATH
?您如何知道这是您得到的?你测试得怎么样?@Alex:没有。请求被发送到正确的URL,但标题必须看起来像上面的那个。