Get Information in JSON format

Question

I want to get/parse Information from these websites(Freebase/Wikipedia...) in JSON format in Android.My aim is to post variable like "Obama" and get information about him in JSON.I searched for it but just found some things for XML (APIs). I allready programmed my JSON Parser but I need the source ^^

Are there any ways to get it easily ? It doesn't have to be Wikipedia or Freebase IF there are other alternatives. I only know these websites to get Info.

Thanks for everyone !

You are Chuck Norris, you have just to look at the code and it will be compiled by itself — Giorgio Antonioli, Mar 25 '15 at 16:39
I suggest you to write XML parser and parse the information using XML nodes. Otherwise, you can use information from XML parser and use in JSON. — eracube, Mar 25 '15 at 16:39
Check this - http://stackoverflow.com/questions/10467492/how-to-extract-data-from-a-wikipedia-article — Surender Kumar, Mar 25 '15 at 16:42

score 0 · Answer 1 · edited May 23 '17 at 11:43

One way i can think of is first get the HTML content

    URL yahoo = new URL(url);
    URLConnection yc = yahoo.openConnection();
    BufferedReader in = new BufferedReader(new InputStreamReader(
            yc.getInputStream(), "UTF-8"));
    String inputLine;
    StringBuilder a = new StringBuilder();
    while ((inputLine = in.readLine()) != null)
        a.append(inputLine);
    in.close();

Then convert the HTML to json using Get data from HTML and convert it to JSON

score 0 · Answer 2 · answered Mar 25 '15 at 16:48

First you can convert HTML to XML Then use framework like Jackson json parser like XML to JSON.

ObjectMapper xmlMapper = new XmlMapper();
String xml = xmlMapper.writeValueAsString(new Simple());
xmlMapper.writeValue(new File("/tmp/stuff.json"), new Simple());

Samples :https://github.com/FasterXML/jackson-dataformat-xml

Get Information in JSON format

2 Answers2