Question

起作用的简单curl调用是： curl https://api.github.com/search/repositories\?q\=reactive

我现在正在用Java进行的尝试

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

为什么以这种方式搜索它时找不到它？

谢谢！

Answer 1

您在Java代码中使用的URL与用于curl的转义符错误。删除转义符并使用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

如何将简单的Curl调用转换为Java HttpURLConnection

1 个答案: