JSON解析。 位置2出现意外字符（t）。JAVA

Question

I'm trying to parse JSON data from a google maps search. I've tryed both JACKSON and and now I'm Trying JSON SIMPLE. Both of them gives the same error.

First of all I'm doing an search on Google maps.

String urlString = "http://maps.google.com/maps?f=q&source=s_q&output=json&start=0&q="+ "Stockholm" + "+Gym";

Gives me JSON while(1);{title:"stockholm Gym - Google Maps",url:"/maps?f=q\\x26source=s_q\\x26start=0\\x26q=stockholm+Gym\\x26ie=UTF8\\x26hq=Gym.............. and so on. I'm replacing the while(1); with ""; before i return the string.

To the problem when I'm trying to parse it

JSONParser parser = new JSONParser();

    String jsonString = "";

// UriHandler.mapSearchJson is the method that returns the jsonString.

    String jsonData = UriHandler.mapSearchJSON(jsonString);

    Object obj = "";
    try {

        obj = parser.parse(jsonData);

    } catch (ParseException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

    JSONObject jsonObj = (JSONObject) obj;

    String title = (String) jsonObj.get("title");
    System.out.println(title);

This gives me the exception. Unexpected character (t) at position 2.

When I'm debbuging it. comes all the way to when it's trying to parse the string. then the obj is = null.

What in thw world am I doing wrong.

Thanks!

Answer 1

As the others already mentioned, a nonquoted field name is not standard JSON. However, Jackson (and maybe others) has a set of option settings that allow it to work with nonstandard, but common JSON derivatives:

JsonParser.Feature.ALLOW_UNQUOTED_FIELD_NAMES

will enable processing of unquoted field names.

Answer 2

响应不是有效的JSON，因为键名未用双引号引起来。

Answer 3

{title:"stockholm Gym"

is invalid JSON, it should be this:

{"title":"stockholm Gym" 

Notice how title is surrounded by " double quotes

Answer 4

You are pulling back Javascript code that is meant for the maps.google.com site to use.

There could be any Javascript code in that response, not just the JSON that happens to be returned as part of the search.

You need to request from their maps API instead:

http://maps.googleapis.com/maps/api/geocode/json?address=Stockholm+Gym&sensor=false

This will return you only the JSON data.

Have a look the Google Maps API for more options.

Answer 5

I faced this error when trying to parse the json returned from kafka (kafka twitter producer).

The message returned was including some extra text other than json (KeyedMessage(twitter-test_english,null,null). Because of that I was facing this error.

KeyedMessage(twitter-test_english,null,null,{"created_at":"Sat Apr 23 18:31:10 +0000 2016","id":723942306777337856,"id_str":"723942306777337856"}

Pass only the message part from returned json and convert it into string.

{"created_at":"Sat Apr 23 18:31:10 +0000 2016","id":723942306777337856,"id_str":"723942306777337856"}

message = new KeyedMessage("twitter-test_english", (String)queue.take());
//System.out.println("This is message"+message.message());
String message_string = message.message().toString();
JsonParse.toParseJson(message_string);