com.google.gson.JsonParseException: Failed parsing JSON source: java.io.BufferedReader to Json - perplexed with this problem

Question

I am trying to parse a json object using gson in Android application ,the test passed quite gracefully in emulator ,while i used the actual device ,the problems started I am getting the exception as follows

.....: com.google.gson.JsonParseException: Failed parsing JSON source: java.io.BufferedReader@44848f38 to Json
......: print stack Failed parsing JSON source: java.io.BufferedReader@44848f38 to Json
.......:com.google.gson.JsonParseException: Failed parsing JSON source: java.io.BufferedReader@44848f38 to Json
.......:    at com.google.gson.JsonParser.parse(JsonParser.java:59)
........:    at com...xx..allmenus.DiningDetail$3.run(DiningDetail.java:361)
........: Caused by: com.google.gson.ParseException: Encountered " <IDENTIFIER_SANS_EXPONENT> "HTTP "" at line 1, column 1.
07-29 11:33:55.186: WARN/System.err(2110): Was expecting one of:
07-29 11:33:55.186: WARN/System.err(2110):     <EOF> 
07-29 11:33:55.186: WARN/System.err(2110):     <DIGITS> ...
07-29 11:33:55.186: WARN/System.err(2110):     "null" ...
07-29 11:33:55.186: WARN/System.err(2110):     "NaN" ...
07-29 11:33:55.194: WARN/System.err(2110):     "Infinity" ...
07-29 11:33:55.194: WARN/System.err(2110):     <BOOLEAN> ...
07-29 11:33:55.194: WARN/System.err(2110):     <SINGLE_QUOTE_LITERAL> ...
07-29 11:33:55.194: WARN/System.err(2110):     <DOUBLE_QUOTE_LITERAL> ...
07-29 11:33:55.202: WARN/System.err(2110):     ")]}\'\n" ...
07-29 11:33:55.210: WARN/System.err(2110):     "{" ...
07-29 11:33:55.210: WARN/System.err(2110):     "[" ...
07-29 11:33:55.217: WARN/System.err(2110):     "-" ...
07-29 11:33:55.217: WARN/System.err(2110):     
07-29 11:33:55.217: WARN/System.err(2110):     at   com.google.gson.JsonParserJavacc.generateParseException(JsonParserJavacc.java:705)
......:   at com.google.gson.JsonParserJavacc.jj_consume_token(JsonParserJavacc.java:587)
......:     at com.google.gson.JsonParserJavacc.parse(JsonParserJavacc.java:62)
......:    at com.google.gson.JsonParser.parse(JsonParser.java:54)
.....:     ... 1 more

Please suggest a solution ,I am using gson 1.4 version.The code i use is

            URLsource = new URL("http://.........);
      in = new BufferedReader(new UnicodeReader(URLsource.openStream(), "UTF-8"));
       jse = new JsonParser().parse(in);
    in.close(); 
            JsonArray info = jse.getAsJsonObject().getAsJsonArray("info");

the o/p of the url as in the browser is

    {"info":[{"restaurant_id":"","name":"Vie de France","campus_id":"","address":"600 Maryland Ave SW","city":"WASHINGTON","state":"DC","neighborhood":"","neighborhood_urlname":"","zipcode":"20024","average_entree":"6.29","chain_id":"994","phone":"(202) 554-7870","is_voicestar":"0","menu_type":"1","url":"","logo":"","details":"","latitude":"38.886877","longitude":"-77.019913","properties":"rmwa","reviews":"y","yelp_phone":"2025547870","reservation_id":"","reservation_url":"","price":"","cross_street":"","public_transit":"","restaurant_description":"","payment_methods":"","hours":"","next_open":"","restaurant_status":"","is_preorder":"","citysearch_hours":" ","minimum_order":"","minimum_order_2":"","delivery_rate":"","delivery_time":"","delivery_info":"","misc_info":"","delivery_charge":"","delivery_charge_2":"","is_pickup":"","is_delivery":"","minimum_tip_amount":"","minimum_tip_percent":"","restaurant_account_id":"","urlname":"\/dc\/washington\/239256-vie-de-france\/","facebook_fan_page":"","twitter_page":"","is_dz":"0","loyalty":"","delivery_hours":""}],"cuisines":[{"mastercuisineid":"143","cuisine":"American","urlname":"american"},{"mastercuisineid":"90","cuisine":"Bakery & Pastries","urlname":"bakery-pastries"},{"mastercuisineid":"57","cuisine":"French","urlname":"french"}]}

thanks

Sephy · Accepted Answer · 2010-07-29 06:50:12Z

1

If your JSON is on a server and you access it through an HTTP request, you need to parse only the content, not the whole file with the headers.
Try doing an HTTP client, use getResponse.getEntity or getResponse.getContent, and then your parser on one of these.

answered Jul 29, 2010 at 6:50

Sephy

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CommonsWare · Accepted Answer · 2010-07-29 06:40:23Z

1

You appear to be attempting to parse HTTP as JSON. HTTP has a status line and headers before any JSON entity.

answered Jul 29, 2010 at 6:40

CommonsWare

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2 Comments

ganesh Over a year ago

thanks Commonsware ,is there any way to skip status line and headers?

CommonsWare Over a year ago

@ganesh: As Sephy indicated, use HttpClient and BasicResponseHandler instead of URL.

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