get NullPointerException when converting JAVA String to JSON object

Question

I'm writing a java web application to retrieve tweets by using Twitter API. I got a Search method and trying to call the method at jsp page.

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;
}

}

The Search method returns a string

            <%
            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());
            }
        %>

And i'm trying to convert the String to JSON object, but it shows HTTP 500 NullPointerException error I don't know where i'm getting wrong because i'm new to JAVA. Could anybody help? Really appreciate!

Put up a stacktrace and show which line of code it points to — Dragondraikk
– Dragondraikk, Commented Mar 27, 2015 at 13:39
Just wondering: what makes you think that writing a java web application is a good starting point for a Java newbie? I mean: if you want to learn the language; why don't you start with tutorials/books that especially target newbies? If you are not doing this to learn java - why are you using java then (instead of a language in which you are more skilled?) — GhostCat
– GhostCat, Commented Mar 27, 2015 at 13:42
query = query.replaceAll(" ", "%20"); It points to this line — Evan Yeung
– Evan Yeung, Commented Mar 27, 2015 at 13:43
I know it is not a good starting point for Java... but this is one of my uni class i'm taking. Seems it is little bit hard for me :( — Evan Yeung
– Evan Yeung, Commented Mar 27, 2015 at 13:48

Ceiling Gecko · Accepted Answer · 2015-03-27 14:03:15Z

1

Your error most likely happens because

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

returns null, that is, it cannot find the parameter and returns null.

After that you attempt to perform actions on null at:

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

but since null cannot perform .replaceAll it throws a NullPointerException

There are different ways to deal with this:

For example, you could check for null right after retrieving the tweet variable like so:

    <%
        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!");
        }
    %>

So if the tweet variable comes up null then instead of time user and text you'll print "Unable to retrieve query!" etc.

edited Mar 27, 2015 at 14:03

answered Mar 27, 2015 at 13:51

Ceiling Gecko

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3 Comments

Evan Yeung Over a year ago

Yes but i got an input text form in jsp page, and query is the name. <input type="text" name="query" placeholder="Enter your keyword"> why would this error happened since i have no chance to give input. It throws NullPointerException directly.

Evan Yeung Over a year ago

I got those tweets!! Thank you so much. But i am still wondering why the results are retrieved after an if check. Why it returns null without check

Ceiling Gecko Over a year ago

That is strange... But glad you got it working in the end.

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