java post 消息体,如何使用Java进行POST并包含参数和原始请求体？

最新推荐文章于 2025-02-23 21:24:20 发布

转载最新推荐文章于 2025-02-23 21:24:20 发布 · 249 阅读

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本文档探讨了如何在Java中使用Apache HttpClient库向Web服务发送POST请求，同时包含参数和请求体。示例代码展示了一种方法，其中请求体能够正确传递，但添加参数后，请求体不再被接收。建议使用`setQueryString`方法将参数添加到URL中，或者考虑调整请求结构，如使用PUT请求保存资源并指定URI。

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I am communicating with a web service that expects a POST parameter and also expect Request body. I have confirmed that such a POST request can be done using a REST Console I have, but I am unable to make such a request in Java using Apache libraries.

In the code below, I am able to POST to the web service, and it correctly receives the contents of the variable raw_body. If I uncomment the first of the two commented lines, the web service receives the "fname" parameter, but it no longer receives the body of the POST.

import org.apache.commons.httpclient.HttpClient;

import org.apache.commons.httpclient.methods.PostMethod;

import org.apache.commons.httpclient.methods.RequestEntity;

...

HttpClient httpClient = new HttpClient();

String urlStr = "http://localhost:8080/MyRestWebService/save";

PostMethod method = new PostMethod(urlStr);

String raw_body = "This is a very long string, much too long to be just another parameter";

RequestEntity re = new StringRequestEntity(raw_body, "text/xml", "UTF-16");

//method.addParameter("fname", "test.txt");

//httpClient.getParams().setParameter("fname", "test.txt");

method.setRequestEntity(re);

How can I transmit both the parameter and the body?

解决方案

You could use the setQueryString method to add the parameters to the URL that is being POSTed to. From a RESTful perspective I'd argue you should normally not be doing that, however, since a POST should represent a call to a resource and anything that would qualify for a query parameter should be included in the representation that is being transferred in the request body...or it should represent qualification of the resource itself in which case it should be part of the path that is posted to which could then be extracted by the controller using @PathVariable/@PathParam or something similar. So in your case you could also be looking for something like POST /MyRestWebService/files/test.txt or more fittingly a PUT if you're saving the resource and know the URI. The code on the server could pull the filename out from a URL pattern.