com.google.gson.JsonParseException：解析 JSON 源失败：java.io.BufferedReader 到 Json - 对这个问题感到困惑

Question

我正在尝试在 Android 应用程序中使用 gson 解析 json 对象，测试在模拟器中非常顺利地通过，而我使用实际设备时，问题开始了 我得到如下异常

.....: 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

请提出解决方案，我使用的是gson 1.4版本。我使用的代码是

            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");

浏览器中url的o/p是

    {"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"}]}

谢谢

Answer 1

如果您的 JSON 位于服务器上并且您通过 HTTP 请求访问它，则只需解析内容，而不是带有标头的整个文件。
尝试做一个 HTTP 客户端，使用 getResponse.getEntity 或 getResponse.getContent，然后在其中之一上使用您的解析器。

Answer 2

您似乎正在尝试将 HTTP 解析为 JSON。 HTTP 在任何 JSON 实体之前有一个 status line 和标头。