java如何接受curl -H "Content-Type: application/json" -X POST -d '{"user_id": "123", "coin":100, "success":1, "msg":"OK!" }'中-d中的POST内容

在Java中，可以使用Apache HttpClient库来发送HTTP请求，并接收cURL命令中-d参数的POST内容。以下是示例代码：

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.CloseableHttpClient;
import org.apache.http.impl.client.HttpClients;
import org.apache.http.util.EntityUtils;

import java.io.IOException;

public class CurlToJava {
    public static void main(String[] args) {
        CloseableHttpClient httpClient = HttpClients.createDefault();
        try {
            HttpPost httpPost = new HttpPost("http://example.com/api/endpoint");

            // 设置请求头
            httpPost.setHeader("Content-Type", "application/json");

            // 设置请求体
            String jsonBody = "{\"user_id\": \"123\", \"coin\":100, \"success\":1, \"msg\":\"OK!\" }";
            StringEntity entity = new StringEntity(jsonBody);
            httpPost.setEntity(entity);

            // 发送请求
            HttpResponse response = httpClient.execute(httpPost);

            // 获取响应内容
            HttpEntity responseEntity = response.getEntity();
            String responseBody = EntityUtils.toString(responseEntity);
            
            System.out.println(responseBody);
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                httpClient.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }
}

在上述代码中，我们创建了一个HttpPost对象，并设置了请求URL、请求头和请求体。通过执行`httpClient.execute(httpPost)`方法发送请求并获取响应。最后，我们将响应内容转换为字符串并进行处理。