如何使用JSONObject解析此響應

Question

我收到從服務器返回的此響應。 我想解析它並從其中獲取出醫院名稱。 我將如何處理？

[
    {
        "Hospital": {
            "id": "63083",
            "hospital_name": "Colorado Mental Health Inst",
            "hospital_add_1": "1600 W 24th St",
            "hospital_add_2": null,
            "hospital_city": "Pueblo",
            "hospital_state": "CO",
            "hospital_zip": "81003",
            "hospital_phone": "719-546-4000\r",
            "hospital_fax": null,
            "hospital_description": null,
            "callcenter_agent_approval": "0",
            "hospital_site": "",
            "mdpocket_approval": "0",
            "facebook": ""
        },
        "Floor": [],
        "Department": [],
        "Image": [],
        "Notes": []
    },
    {
        "Hospital": {
            "id": "63084",
            "hospital_name": "Parkview Medical Center",
            "hospital_add_1": "400 W 16th St",
            "hospital_add_2": null,
            "hospital_city": "Pueblo",
            "hospital_state": "CO",
            "hospital_zip": "81003",
            "hospital_phone": "719-584-4000\r",
            "hospital_fax": null,
            "hospital_description": null,
            "callcenter_agent_approval": "0",
            "hospital_site": "",
            "mdpocket_approval": "0",
            "facebook": ""
        },
        "Floor": [],
        "Department": [],
        "Image": [],
        "Notes": []
    },
    {
        "Hospital": {
            "id": "63085",
            "hospital_name": "St Mary-Corwin Medical Center",
            "hospital_add_1": "1008 Minnequa Ave",
            "hospital_add_2": null,
            "hospital_city": "Pueblo",
            "hospital_state": "CO",
            "hospital_zip": "81004",
            "hospital_phone": "719-560-4000\r",
            "hospital_fax": null,
            "hospital_description": null,
            "callcenter_agent_approval": "0",
            "hospital_site": "",
            "mdpocket_approval": "0",
            "facebook": ""
        },
        "Floor": [],
        "Department": [],
        "Image": [],
        "Notes": []
    }
]

編輯JSON * 更新JSON *

Answer 1

[ // json array node 
{  // json object node 
"Hospital": { // json object Hospital 

解析

JSONArray jr = new JSONArray("jsonstring");
for(int i=0;i<jr.length();i++)
{
   JSONObject jb = (JSONObject)jr.getJSONObject(i);
   JSONObject jb1 =(JSONObject) jb.getJSONObject("Hospital");
   String name =  jb1.getString("hospital_name");
   Log.i("name....",name);
}

日志記錄

02-18 03:09:43.950: I/name....(951): Colorado Mental Health Inst
02-18 03:09:43.950: I/name....(951): Parkview Medical Center
02-18 03:09:43.950: I/name....(951): St Mary-Corwin Medical Center

Answer 2

您不會-無效的JSON。 您缺少字段名稱中的大部分“”。

Answer 3

嘗試這個..

JSONArray tot_array = new JSONArray(response);

for(int i = 0; i< tot_array.length(); i++){
    JSONObject obj = tot_array.getJSONObject(i);
    JSONObject hospital_obj = obj.getJSONObject("Hospital");
    String hospital_name =  hospital_obj.getString("hospital_name");
}

Answer 4

我建議您使用fastjson（ https://github.com/alibaba/fastjson ）。

Answer 5

簽出這個很棒的庫，用於在Android中解析JSOn，它名為GSON https://code.google.com/p/google-gson/ 。 通過這個解析，這非常簡單。

Answer 6

您的字符串不是有效的JSON對象。 在嘗試將字符串解析為JSON之前，請先檢查jsonlint 。 之后，您可以閱讀有關在Android中解析JSON的信息 。 使用內置的org.json足夠容易，但是如果您使用那里的許多Java庫之一進一步簡化它，那么應該會容易得多。 您可以研究Jackson或google-gson ，這兩種功能最強大的實用程序可滿足您的需求。