Aru*_*kur 1 arrays android json
我正在研究json解析.我正在以JSON数组的形式获得响应,并在另一个JSON数组中获得另一个JSON数组.我没有得到热解析.在json数组中获取json数组对象我在json中是新的,所以请需要帮助.
这是我的回答: -
[
[
{
"Postcode": "BR6 0NH",
"Notes": null,
"Name": "Tesco Pharmacy (Orpington), Bromley, London",
"Phone": "01689 307447",
"Address": "10 AUGUSTUS LANE, ORPINGTON",
"ServiceId": 0,
"Capacity": null,
"URL": null,
"Error": null
},
{
"Postcode": "DA14 5BN",
"Notes": null,
"Name": "Tesco Pharmacy, (Edgington Way), Bromley, London",
"Phone": "0208 258 9447",
"Address": "EDGINGTON WAY, SIDCUP",
"ServiceId": 0,
"Capacity": null,
"URL": null,
"Error": null
}
]
]
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试试这种方式
try {
JSONArray mainArry = new JSONArray(strJsonResponse);
JSONArray contentArray = mainArry.getJSONArray(0);
for (int i =0 ;i<contentArray.length(); i++){
JSONObject item = contentArray.getJSONObject(0);
System.out.println("Postcode : " + item.getString("Postcode"));
System.out.println("Notes : " + item.getString("Notes"));
System.out.println("Phone : " + item.getString("Phone"));
System.out.println("Address : " + item.getString("Address"));
System.out.println("ServiceId : " + item.getString("ServiceId"));
System.out.println("Capacity : " + item.getString("Capacity"));
System.out.println("URL : " + item.getString("URL"));
System.out.println("Error : " + item.getString("Error"));
}
} catch (JSONException e) {
e.printStackTrace();
}
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