java - 如何将简单的 Curl 调用转换为 Java HttpURLConnection

Question

有效的简单curl调用是:curl https://api.github.com/search/repositories\?q\=reactive

我现在在 Java 中尝试的内容

                URL url = new URL("https://api.github.com/search/repositories\\?q\\=reactive");
    HttpURLConnection con = (HttpURLConnection) url.openConnection();
    con.setRequestMethod("GET");
    con.setRequestProperty("Content-Type", "application/json");
    con.setConnectTimeout(5000);
    con.setReadTimeout(5000);

    int isError = con.getResponseCode();
    System.out.println("Error code is " + isError);

    InputStream error = con.getErrorStream();

    System.out.println("Error is " + error.toString() + " \n");

    BufferedReader in = new BufferedReader(
        new InputStreamReader(con.getErrorStream()));
    String inputLine;
    StringBuffer content = new StringBuffer();
    while ((inputLine = in.readLine()) != null) {
        content.append(inputLine);
    }
    System.out.println("Error: " + content.toString());

错误代码:404

错误消息:错误:{"message":"Not Found","documentation_url":"https://developer.github.com/v3"}

为什么我这样搜索却找不到呢？

谢谢!

Answer 1

您在java代码中使用的URL与用于curl的转义字符错误。删除转义字符并使用 url https://api.github.com/search/repositories?q=reactive:

public static void main(String[] args) throws IOException {
        URL url = new URL("https://api.github.com/search/repositories?q=reactive");
        HttpURLConnection con = (HttpURLConnection) url.openConnection();
        con.setRequestMethod("GET");
        con.setRequestProperty("Content-Type", "application/json");
        con.setConnectTimeout(5000);
        con.setReadTimeout(5000);

        System.out.println(con.getResponseMessage());
}

输出:

OK