如何反在安卓code JSON [英] how to decode json in android

问题描述

这是我的JSON数据。

this is my json data

 [{"FeedbackId":"1","Phoneid":"9774d56d682e549c","feedbackdate":"2011\/9 \/24","GuestName":"sdf","Address":"sdf","Phone":"456","Email":"sdf","suggestion":"sdf","grievances":"sdf","visitAgain":"0","purposeofvisit":"sdf","flag":"1"},{"FeedbackId":"2","Phoneid":"9774d56d682e549c","feedbackdate":"","GuestName":"sdf","Address":"sdf","Phone":"456","Email":"sdf","suggestion":"sdf","grievances":"sdf","visitAgain":"0","purposeofvisit":"sdf","flag":"1"},{"FeedbackId":"3","Phoneid":"9774d56d682e549c","feedbackdate":"","GuestName":"sdf","Address":"sdf","Phone":"456","Email":"sdf","suggestion":"sdf","grievances":"sdf","visitAgain":"0","purposeofvisit":"sdf","flag":"1"},{"FeedbackId":"4","Phoneid":"9774d56d682e549c","feedbackdate":"2011\/9 \/24","GuestName":"sdf","Address":"dsf","Phone":"456","Email":"sdf","suggestion":"sf","grievances":"sdf","visitAgain":"0","purposeofvisit":"sdf","flag":"1"},{"FeedbackId":"5","Phoneid":"9774d56d682e549c","feedbackdate":"2011\/9 \/24","GuestName":"sdf","Address":"dsf","Phone":"456","Email":"sdf","suggestion":"sf","grievances":"sdf","visitAgain":"0","purposeofvisit":"sdf","flag":"1"},{"FeedbackId":"6","Phoneid":"9774d56d682e549c","feedbackdate":"","GuestName":"xcv","Address":"xcv","Phone":"89","Email":"xcv","suggestion":"xcv","grievances":"xcv","visitAgain":"1","purposeofvisit":"","flag":"1"},{"FeedbackId":"7","Phoneid":"9774d56d682e549c","feedbackdate":"","GuestName":"gfhj","Address":"ghj6678","Phone":"678","Email":"ghjgh","suggestion":"678fgh","grievances":"fgh","visitAgain":"0","purposeofvisit":"sdf","flag":"1"}]

如何去$ C C这个$？

how to decode this?

推荐答案

您应该能够使用的 org.json 包。从这里：

You should be able to decode using the org.json package. From here:

String json = "{"
         + "  \"query\": \"Pizza\", "
         + "  \"locations\": [ 94043, 90210 ] "
         + "}";

 JSONObject object = (JSONObject) new JSONTokener(json).nextValue();
 String query = object.getString("query");
 JSONArray locations = object.getJSONArray("locations");

只是用，而不是他们自己的JSON。

Just use your own JSON instead of theirs.

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