步骤二:发送异步POST请求 接下来,我们编写一个示例代码,演示如何发送异步POST请求。 importokhttp3.*;importjava.io.IOException;publicclassAsyncPostRequest{privatefinalOkHttpClientclient=newOkHttpClient();publicvoidsendPostRequest(Stringurl,StringjsonData){RequestBodybody=RequestBody.create(jsonData,MediaType.ge...
obj.sendPost(); } privatevoidsendGet()throwsException { HttpRequestrequest=HttpRequest.newBuilder() .GET() .uri(URI.create("你请求数据的url地址")) .setHeader("User-Agent","Java 11 HttpClient Bot") .build(); HttpResponse<String> response = httpClient.send(request, HttpResponse.BodyHandlers....
在Java中,我们可以使用HttpURLConnection类来发送HTTP请求。下面是一个简单的示例代码,演示了如何发送POST请求并传递HttpServletRequest对象。 importjava.io.BufferedReader;importjava.io.DataOutputStream;importjava.io.IOException;importjava.io.InputStreamReader;importjava.net.HttpURLConnection;importjava.net.URL;pub...
方法一: HttpClient publicvoidpostTest(HttpServletRequest request,Integer type,String phone,String passwd,String schoolld,String agent){ String url= "xxxxxxxxx";//发送请求的URL地址JSONObject jsonObject =newJSONObject();//封装参数 jsonObject.put("phone", "xxxx");jsonObject.put("agent", "xxxx")...
public static void main(String[] args) { //发送 GET 请求 String s=HttpRequest.sendGet("http://localhost:6144/Home/RequestString", "key=123&v=456"); System.out.println(s); //发送 POST 请求 String sr=HttpRequest.sendPost("http://localhost:6144/Home/RequestPostString", "key=123&v=...
Java-通过POST方法轻松发送HTTP参数 我正在成功地使用此代码发送HTTP具有某些参数的请求GET方法 void sendRequest(String request){ // i.e.: request = "http://example.com/index.php?param1=a¶m2=b¶m3=c"; URL url = new URL(request); ...
点击"Send" 按钮。 你应该会在 Postman 中看到响应:POST request received,并且在 IDE 的控制台中看到打印的请求体内容。 使用cURL 你也可以使用 cURL 命令行工具来测试 POST 请求: curl -X POST http://localhost:8080/api/post -d '{"name": "John", "age": 30}' ...
method=RequestMethod.POST)@ResponseBodypublicUsergetUserStream(@RequestBodyUseruser){System.out.println("请求参数:"+user);StringuserRet=HttpUtil.sendPostByURLConnection(urlStream,user);Userret=JSON.toJavaObject(JSONObject.parseObject(userRet),User.class);System.out.println("URLConnection方式发送POST...
@RequestMapping(value = "/getUserStream", method = RequestMethod.POST) @ResponseBody public User getUserStream(@RequestBody User user) { System.out.println("请求参数:"+user); String userRet = HttpUtil.sendPostByURLConnection(urlStream, user); ...
printStackTrace(); } } return result; } public static void main(String[] args) { //demo:代理访问 String url = "http://api.adf.ly/api.php"; String para = "key=youkeyid&youuid=uid&advert_type=int&domain=adf.ly&url=http://somewebsite.com"; String sr=HttpRequestUtil.sendPost(url...