从HttpURLConnection对象解析JSON

Question

我正在用HttpURLConnectionJava中的对象做基本的http auth .

        URL urlUse = new URL(url);
        HttpURLConnection conn = null;
        conn = (HttpURLConnection) urlUse.openConnection();
        conn.setRequestMethod("GET");
        conn.setRequestProperty("Content-length", "0");
        conn.setUseCaches(false);
        conn.setAllowUserInteraction(false);
        conn.setConnectTimeout(timeout);
        conn.setReadTimeout(timeout);
        conn.connect();

        if(conn.getResponseCode()==201 || conn.getResponseCode()==200)
        {
            success = true;
        }

我期待一个JSON对象,或者是有效JSON对象格式的字符串数据,或者是带有简单明文的HTML,它是有效的JSON.如何从HttpURLConnection返回响应后访问它？

Answer 1

您可以使用以下方法获取原始数据.顺便说一句,这种模式适用于Java 6.如果您使用的是Java 7或更高版本,请考虑使用try-with-resources模式.

public String getJSON(String url, int timeout) {
    HttpURLConnection c = null;
    try {
        URL u = new URL(url);
        c = (HttpURLConnection) u.openConnection();
        c.setRequestMethod("GET");
        c.setRequestProperty("Content-length", "0");
        c.setUseCaches(false);
        c.setAllowUserInteraction(false);
        c.setConnectTimeout(timeout);
        c.setReadTimeout(timeout);
        c.connect();
        int status = c.getResponseCode();

        switch (status) {
            case 200:
            case 201:
                BufferedReader br = new BufferedReader(new InputStreamReader(c.getInputStream()));
                StringBuilder sb = new StringBuilder();
                String line;
                while ((line = br.readLine()) != null) {
                    sb.append(line+"\n");
                }
                br.close();
                return sb.toString();
        }

    } catch (MalformedURLException ex) {
        Logger.getLogger(getClass().getName()).log(Level.SEVERE, null, ex);
    } catch (IOException ex) {
        Logger.getLogger(getClass().getName()).log(Level.SEVERE, null, ex);
    } finally {
       if (c != null) {
          try {
              c.disconnect();
          } catch (Exception ex) {
             Logger.getLogger(getClass().getName()).log(Level.SEVERE, null, ex);
          }
       }
    }
    return null;
}

然后你可以使用Google Gson返回的字符串将JSON映射到指定类的对象,如下所示:

String data = getJSON("http://localhost/authmanager.php");
AuthMsg msg = new Gson().fromJson(data, AuthMsg.class);
System.out.println(msg);

有一个AuthMsg类的示例:

public class AuthMsg {
    private int code;
    private String message;

    public int getCode() {
        return code;
    }
    public void setCode(int code) {
        this.code = code;
    }

    public String getMessage() {
        return message;
    }
    public void setMessage(String message) {
        this.message = message;
    }
}

http://localhost/authmanager.php返回的JSON 必须如下所示:

{"code":1,"message":"Logged in"}

问候

Answer 2

定义以下功能(不是我的,不知道我很久以前在哪里找到它):

private static String convertStreamToString(InputStream is) {

BufferedReader reader = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();

String line = null;
try {
    while ((line = reader.readLine()) != null) {
        sb.append(line + "\n");
    }
} catch (IOException e) {
    e.printStackTrace();
} finally {
    try {
        is.close();
    } catch (IOException e) {
        e.printStackTrace();
    }
}
return sb.toString();

}

然后:

String jsonReply;
if(conn.getResponseCode()==201 || conn.getResponseCode()==200)
    {
        success = true;
        InputStream response = conn.getInputStream();
        jsonReply = convertStreamToString(response);

        // Do JSON handling here....
    }