java - 将 JAVA String 转换为 JSON 对象时出现 NullPointerException

Question

我正在编写一个 Java Web 应用程序来使用 Twitter API 检索推文。我有一个搜索方法并试图在 jsp 页面调用该方法。

public class SearchTwitter {

private static final String CONSUMER_KEY = "*************";
private static final String CONSUMER_SECRET = "****************";
private static final String OAUTH_TOKEN = "*********************";
private static final String OAUTH_TOKEN_SECRET = "*****************";

public String Search(String query) throws Exception {
    OAuthConsumer consumer = new DefaultOAuthConsumer(CONSUMER_KEY,
            CONSUMER_SECRET);
    consumer.setTokenWithSecret(OAUTH_TOKEN, OAUTH_TOKEN_SECRET);
    query = query.replaceAll(" ", "%20");
    String twitterURL = "https://api.twitter.com/1.1/search/tweets.json?q=" + query + "&src=typd";
    URL url = new URL(twitterURL);
    HttpURLConnection request = (HttpURLConnection) url.openConnection();
    request.setDoOutput(true);
    request.setRequestProperty("Content-Type", "application/json");
    request.setRequestProperty("Accept", "application/json");
    request.setRequestMethod("GET");
    consumer.sign(request);
    request.connect();
    StringBuilder text = new StringBuilder();
    InputStreamReader in = new InputStreamReader((InputStream) request.getContent());
    BufferedReader buff = new BufferedReader(in);
    System.out.println("Getting data ...");
    String line;
    do {
        line = buff.readLine();
        text.append(line);
    } while (line != null);
    String strResponse = text.toString();
    return strResponse;
}

Search 方法返回一个字符串

            <%
            SearchTwitter stw = new SearchTwitter();
            String tweet = request.getParameter("query");

            JSONObject result = new JSONObject(stw.Search(tweet));
            JSONArray statuses = result.getJSONArray("statuses");
            for (int i = 0; i < statuses.length(); i++) {
                String time = statuses.getJSONObject(i).getString("created_at");
                String user = statuses.getJSONObject(i).getJSONObject("user").getString("screen_name");
                String text = statuses.getJSONObject(i).getString("text");
                System.out.println(time.toString());
                System.out.println(user.toString());
                System.out.println(text.toString());
            }
        %>

我正在尝试将字符串转换为 JSON 对象，但它显示 HTTP 500 NullPointerException 错误我不知道我哪里出错了，因为我是 JAVA 的新手。有人可以帮忙吗？非常感谢!

Answer 1

你的错误很可能发生是因为

String tweet = request.getParameter("query");

返回null，即找不到参数，返回null。

之后您尝试在以下位置对 null 执行操作:

query = query.replaceAll(" ", "%20");

但是因为 null 不能执行 .replaceAll 它抛出一个 NullPointerException

有不同的方法来处理这个问题:

例如，您可以在检索 tweet 变量后立即检查 null，如下所示:

    <%
        SearchTwitter stw = new SearchTwitter();
        String tweet = request.getParameter("query");

        if(tweet!=null){
            JSONObject result = new JSONObject(stw.Search(tweet));
            JSONArray statuses = result.getJSONArray("statuses");
            for (int i = 0; i < statuses.length(); i++) {
                String time = statuses.getJSONObject(i).getString("created_at");
                String user = statuses.getJSONObject(i).getJSONObject("user").getString("screen_name");
                String text = statuses.getJSONObject(i).getString("text");
                System.out.println(time.toString());
                System.out.println(user.toString());
                System.out.println(text.toString());
            }
        } else {
            System.out.println("Unable to retrieve query!");
        }
    %>

因此，如果 tweet 变量出现 null，那么您将打印“Unable to retrieve query!”，而不是时间用户和文本。等等