Hello, I am testing a query in Java code, but got “Exception in thread “main” java.io.IOException: Server returned HTTP response code: 400 for URL: https://api.twitter.com/1.1/search/tweets.json?q=iPhone&count=100”
What’s the right way to use in Java Code? Thanks.
String urlStr = "https://api.twitter.com/1.1/search/tweets.json?q=iPhone&count=100";
StringBuffer buff = new StringBuffer();
URL url = new URL(urlStr);
BufferedReader br = new BufferedReader(new InputStreamReader(url.openConnection().getInputStream()));
