将 CURL 请求转换为 HTTP 请求 Java

本文介绍了将 CURL 请求转换为 HTTP 请求 Java的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有以下 CURL 请求任何人都可以确认我什么是 subesquest HTTP 请求

I have the following CURL request Can anyone please confirm me what would be the subesquest HTTP Request

      curl -u "Login-dummy:password-dummy" -H "X-Requested-With: Curl" "https://qualysapi.qualys.eu/api/2.0/fo/report/?action=list" -k

会是这样吗?

    String url = "https://qualysapi.qualys.eu/api/2.0/fo/report/";
    URL obj = new URL(url);
    HttpURLConnection con = (HttpURLConnection) obj.openConnection();

    // optional default is GET
    con.setRequestMethod("GET"); ..... //incomplete

任何人都可以帮助我将上述 curl 请求完全转换为 httpreq.

Can anyone be kind enough to help me convert the above curl request completely to httpreq.

提前致谢.

苏维

推荐答案

有很多方法可以实现这一点.我认为下面是最简单的，同意它不是很灵活但有效.

There are numerous ways to achieve this. Below one is simplest in my opinion, Agree it isn't very flexible but works.

import java.io.IOException;
import java.io.InputStreamReader;
import java.net.URL;
import java.net.URLConnection;

import org.apache.commons.codec.binary.Base64;

public class HttpClient {

    public static void main(String args[]) throws IOException {
        String stringUrl = "https://qualysapi.qualys.eu/api/2.0/fo/report/?action=list";
        URL url = new URL(stringUrl);
        URLConnection uc = url.openConnection();

        uc.setRequestProperty("X-Requested-With", "Curl");

        String userpass = "username" + ":" + "password";
        String basicAuth = "Basic " + new String(new Base64().encode(userpass.getBytes()));
        uc.setRequestProperty("Authorization", basicAuth);

        InputStreamReader inputStreamReader = new InputStreamReader(uc.getInputStream());
        // read this input

    }
}

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