使用Apache HTTP Components的一种简单方法是

Request.Post("http://www.example.com/page.php")
            .bodyForm(Form.form().add("id", "10").build())
            .execute()
            .returnContent();

看一看Fluent API

更新后的答案

由于原始答案中的一些类在Apache HTTP组件的新版本中已弃用，所以我发布了此更新。

顺便说一下，您可以在这里访问更多示例的完整文档。

HttpClient httpclient = HttpClients.createDefault();
HttpPost httppost = new HttpPost("http://www.a-domain.example/foo/");

// Request parameters and other properties.
List<NameValuePair> params = new ArrayList<NameValuePair>(2);
params.add(new BasicNameValuePair("param-1", "12345"));
params.add(new BasicNameValuePair("param-2", "Hello!"));
httppost.setEntity(new UrlEncodedFormEntity(params, "UTF-8"));

//Execute and get the response.
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();

if (entity != null) {
    try (InputStream instream = entity.getContent()) {
        // do something useful
    }
}

原来的答案

我推荐使用Apache HttpClient。它更快更容易实现。

HttpPost post = new HttpPost("http://jakarata.apache.org/");
NameValuePair[] data = {
    new NameValuePair("user", "joe"),
    new NameValuePair("password", "bloggs")
};
post.setRequestBody(data);
// execute method and handle any error responses.
...
InputStream in = post.getResponseBodyAsStream();
// handle response.

欲了解更多信息，请查看此URL: http://hc.apache.org/

第一个答案很好，但我必须添加try/catch以避免Java编译器错误。 另外，我在弄清楚如何使用Java库读取HttpResponse时遇到了麻烦。

以下是更完整的代码:

/*
 * Create the POST request
 */
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://example.com/");
// Request parameters and other properties.
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("user", "Bob"));
try {
    httpPost.setEntity(new UrlEncodedFormEntity(params, "UTF-8"));
} catch (UnsupportedEncodingException e) {
    // writing error to Log
    e.printStackTrace();
}
/*
 * Execute the HTTP Request
 */
try {
    HttpResponse response = httpClient.execute(httpPost);
    HttpEntity respEntity = response.getEntity();

    if (respEntity != null) {
        // EntityUtils to get the response content
        String content =  EntityUtils.toString(respEntity);
    }
} catch (ClientProtocolException e) {
    // writing exception to log
    e.printStackTrace();
} catch (IOException e) {
    // writing exception to log
    e.printStackTrace();
}

我建议使用http-request建立在apache http api上。

HttpRequest<String> httpRequest = HttpRequestBuilder.createPost("http://www.example.com/page.php", String.class)
.responseDeserializer(ResponseDeserializer.ignorableDeserializer()).build();

public void send(){
   String response = httpRequest.execute("id", "10").get();
}

用post请求发送参数的最简单方法:

String postURL = "http://www.example.com/page.php";

HttpPost post = new HttpPost(postURL);

List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("id", "10"));

UrlEncodedFormEntity ent = new UrlEncodedFormEntity(params, "UTF-8");
post.setEntity(ent);

HttpClient client = new DefaultHttpClient();
HttpResponse responsePOST = client.execute(post);

你做到了。现在您可以使用responsePOST。 获取响应内容为字符串:

BufferedReader reader = new BufferedReader(new  InputStreamReader(responsePOST.getEntity().getContent()), 2048);

if (responsePOST != null) {
    StringBuilder sb = new StringBuilder();
    String line;
    while ((line = reader.readLine()) != null) {
        System.out.println(" line : " + line);
        sb.append(line);
    }
    String getResponseString = "";
    getResponseString = sb.toString();
//use server output getResponseString as string value.
}

我建议使用Postman来生成请求代码。简单地使用Postman进行请求，然后点击code选项卡:

然后你会看到下面的窗口，选择你想要的请求代码的语言: