Question

我有一个像Curl这样的Post请求程序。我该如何在Java中准确地完成这项工作？

curl -X POST \
-H "api_key: xxxxxxxxxxxxx" \
-H "speed: 0" \
-H "voice: male" \
-H "prosody: 1" \
-H "Cache-Control: no-cache" \
-d 'This is the text to transfer' \
"http://somewhere.com"

Answer 1

我使用Apache创建了POST请求的可运行示例

import org.apache.commons.io.IOUtils;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.ContentType;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.HttpClients;

import java.io.IOException;
import java.io.InputStream;
import java.nio.charset.StandardCharsets;
...

public static void main(String[] args) throws IOException {
    HttpClient httpclient = HttpClients.createDefault();
    HttpPost httppost = new HttpPost("https://reqres.in/api/register");

    // Simple data
    httppost.setEntity(
        new StringEntity("{\"email\":\"eve.holt@reqres.in\",\"password\":\"pistol\"}",ContentType.create("application/json", StandardCharsets.UTF_8)));

    // Simple Header
    httppost.setHeader("cache-control", "no-cache");

    // Execute and get the response.
    HttpResponse response = httpclient.execute(httppost);
    HttpEntity entity = response.getEntity();
    System.out.println("Status: " + response.getStatusLine().toString());
    if (entity != null) {
      try (InputStream instream = entity.getContent()) {
        System.out.println("response: " + IOUtils.toString(instream, StandardCharsets.UTF_8));
      }
    }
  }

来自System.out.println()

的回复

状态：HTTP / 1.1 200 OK

响应：{“ id”：4，“令牌”：“ QpwL5tke4Pnpja7X4”}

Answer 2

用Java转换代码

    String url = ""http://somewhere.com";

    HttpClient client = new DefaultHttpClient();
    HttpPost post = new HttpPost(url);

    // add header
    post.setHeader("api_key", "xxxxxxxxxxxxx");
    post.setHeader("speed", "0");
    post.setHeader("voice", "male");
    post.setHeader("prosody", "1");
    post.setHeader("Cache-Control", "no-cache");

    post.setEntity(new StringEntity("This is the text to transfer",ContentType.create("text/plain", "UTF-8")));

    HttpResponse response = client.execute(post);
    System.out.println("\nSending 'POST' request to URL : " + url);
    System.out.println("Post parameters : " + post.getEntity());
    System.out.println("Response Code : " + 
                                response.getStatusLine().getStatusCode());

    BufferedReader rd = new BufferedReader(
                    new InputStreamReader(response.getEntity().getContent()));

    StringBuffer result = new StringBuffer();
    String line = "";
    while ((line = rd.readLine()) != null) {
        result.append(line);
    }

    System.out.println(result.toString());

希望有帮助。

我怎样才能在Java中发出这样的Post请求？

2 个答案: