Jak*_*ake 3 java android json json-simple
我有如下JSON:
[{"0":"1","id":"1","1":"abc","name":"abc"},{"0":"2","id":"2","1":"xyz","name":"xyz"}]
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它是一个对象数组.
我需要使用Java解析它.我在以下网址使用该库:http: //code.google.com/p/json-simple/downloads/list
此链接的示例1近似于我的要求:http: //code.google.com/p/json-simple/wiki/DecodingExamples
我有以下代码:
/** Decode JSON */
// Assuming the JSON string is stored in jsonResult (String)
Object obj = JSONValue.parse(jsonResult);
JSONArray array = (JSONArray)obj;
JSONObject jsonObj = null;
for (int i=0;i<array.length();i++){
try {
jsonObj = (JSONObject) array.get(i);
} catch (JSONException e) {
e.printStackTrace();
}
try {
Log.d(TAG,"Object no." + (i+1) + " field1: " + jsonObj.get("0") + " field2: " + jsonObj.get("1"));
} catch (JSONException e) {
e.printStackTrace();
}
}
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我收到以下异常:
java.lang.ClassCastException: org.json.simple.JSONArray
// at JSONArray array = (JSONArray)obj;
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有人可以帮忙吗?
谢谢.
Lal*_*ani 11
而不是将您的对象转换为JSONArray,您应该这样做:
JSONArray mJsonArray = new JSONArray(jsonString);
JSONObject mJsonObject = new JSONObject();
for (int i = 0; i < mJsonArray.length(); i++) {
mJsonObject = mJsonArray.getJSONObject(i);
mJsonObject.getString("0");
mJsonObject.getString("id");
mJsonObject.getString("1");
mJsonObject.getString("name");
}
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