0

I try to parse this valid JSON

I use new JSONObject(resultString); but i have JSONException

How can i parse his Json

Help me please!

Thanks

[{
    "id": "8637F7F78C8C1",
    "from_account_id": "1025630",
    "to_account_id": "1025635",
    "transaction_type_id": "15",
    "transaction_mode_id": "2",
    "transaction_status_id": "1",
    "amount": "1000.00",
    "real_amount": "1000.00",
    "promote_amount": "0.00",
    "from_fix_fee_amount": "0.00",
    "from_percent_fee_amount": "50.00",
    "to_fix_fee_amount": "0.00",
    "to_percent_fee_amount": "0.00",
    "description": "demo",
    "created_on": "2012-07-16 10:04:29",
    "activated_on": "0000-00-00 00:00:00",
    "executed_on": "0000-00-00 00:00:00",
    "verify_transaction_by": "o",
    "from_account_name": "xxxxxxxxxxxx",
    "from_account_email": "xxxxxxx@gmail.com",
    "from_account_phone_no": "xxcxxxxxx",
    "to_account_name": "yyyyyyy",
    "to_account_email": "yyyyy@gmail.com",
    "to_account_phone_no": "yyyyyyyy"
},
{
    "id": "A26BF7F75534B",
    "from_account_id": "1014635",
    "to_account_id": "1054630",
    "transaction_type_id": "5",
    "transaction_mode_id": "2",
    "transaction_status_id": "4",
    "amount": "1000.00",
    "real_amount": "1000.00",
    "promote_amount": "0.00",
    "from_fix_fee_amount": "0.00",
    "from_percent_fee_amount": "0.00",
    "to_fix_fee_amount": "0.00",
    "to_percent_fee_amount": "0.00",
    "description": "",
    "created_on": "2012-07-15 00:52:40",
    "activated_on": "2012-07-15 00:53:19",
    "executed_on": "2012-07-15 00:54:57",
    "verify_transaction_by": "o",
    "from_account_name": "yyyyyy",
    "from_account_email": "yyyyyy@gmail.com",
    "from_account_phone_no": "yyyyyyy",
    "to_account_name": "wwwwwww",
    "to_account_email": "yywywyyy@gmail.com",
    "to_account_phone_no": "yyyyyyyyy"
}]
4

4 回答 4

3

它JSONArray不是一个JSONObject..
Android/Java 中的 JSON 数组迭代 

JSONArray values = new JSONArray(resultString);
for(int i = 0 ; i < values.length(); i++){
    JSONObject object = (JSONObject) values.get(i); 
    String id = object.getString("id");
    //the same for the rest
}
于 2012-07-16T13:00:08.227 回答
1

这不是 JSON 对象——它是一个包含两个对象的 JSON 数组。

先使用new JSONArray(resultString);然后从中获取对象。

于 2012-07-16T13:00:39.853 回答
1

Hi [] 表示 JSON 中的数组，因此在解析后尝试将其放入 JSONArray...new JSONArray(resultString).getJSONObject(int index);

通过这种方式，您可以获得 JSON 对象..

于 2012-07-16T13:05:17.387 回答
1

为了解析 JSON，我使用了 GSON。

JsonElement jelement = new JsonParser().parse("json text");

JsonArray 数组 = jobject.getAsJsonArray("myarray");
for (JsonElement item : array) {...}

于 2012-07-16T13:06:37.893 回答