Java JSONObject给我错误

我有一个json从服务器返回给我。我试图迭代它，但这一行的gving me错误

 JSONObject obj = jsonHospitals.getJSONObject(i);

错误以红色下划线显示。getJSONObject（i）并提供了两个选项 1-更改为optJSONObject（…） 2-将“i”的类型更改为“String”

如何使用JSONObject并迭代json。我是android新手，所以请大家耐心点！ 下面是更多的代码

String responseData = new String(charArray);

                    JSONObject jsonHospitals = new JSONObject(responseData);

                    for(int i = 0; i< jsonHospitals.length(); i++){
                        JSONObject obj = jsonHospitals.getJSONObject(i);
                        JSONObject hospital_obj = obj.getJSONObject("Hospital");
                        String hospital_name =  hospital_obj.getString("hospital_name");
                        Log.v(TAG, hospital_name);
                    }

//CATLOG已完成

//这不是CATLOG的一部分 JSON

[{"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":[]}]

---

没有将int作为参数的

getJSONObject

因此出现了错误。您需要发布json以获得进一步帮助

您的json似乎与前面问题中的json相似，我已经回答了这个问题@

编辑：

试试看
{
HttpClient HttpClient=新的DefaultHttpClient（）；
httpclient.getParams（）.setParameter（CoreProtocolPNames.PROTOCOL_VERSION，HttpVersion.HTTP_1_1）；
HttpGet请求=新建HttpGet（“您的url”）；
HttpResponse response=httpclient.execute（请求）；
HttpEntity当前性=response.getEntity（）；
字符串_response=EntityUtils.toString（最近性）；
JSONArray jr=新JSONArray（_响应）；
对于（int i=0；i变化：

JSONObject obj = jsonHospitals.getJSONObject(i);

致：

原因是函数getJSONObject需要一个字符串作为其参数-
您正在尝试迭代JSONArray
更改此行：

JSONObject jsonHospitals=新JSONObject（responseData）

对此

JSONAray jsonHospitals=新的JSONAray（responseData）

如果仍然不起作用，您能否提供您的JSON解决方案。
如果您的输入JSON类型为JSON array，则应将其视为JSON array（请参阅）：

然后您可以使用以下命令对其进行迭代：

for(int i = 0; i< array.length(); i++) {
  array.getJSONObject(i);
}

for（int i=0；i
你好，Autolycus，
我认为您必须将responseData解析为如下所示的JSONArray
**JSONArray jsonHospitals=新的JSONArray（responseData）**
for（int i=0；i
@BrianRoach可能吧。不在这里。请查看我的更新。我已经添加了关于CatLog所说内容的详细信息etc@Autolycus你能发布url以便我们能看到完整的json吗？我会尝试复制粘贴问题中的完整json。这是一个内部网站，因此外人无法访问它。给我5minutes@Autolycus你的厕所ks与您的previos问题中的一个完全相同。因此，该答案应该适用于您。您有错误，因为您没有阅读完整的json。请发布一个您试图解析的json示例。不清楚您是否真的有json数组或对象。@BrianRoach我猜它与他之前的问题相关@，答案告诉请看更新！谢谢！我看到catlog在抱怨解析
public JSONObject getJSONObject (String name)

Added in API level 1
Returns the value mapped by name if it exists and is a JSONObject.

Throws
JSONException   if the mapping doesn't exist or is not a JSONObject.

    try
    {
     HttpClient httpclient = new DefaultHttpClient();
     httpclient.getParams().setParameter(CoreProtocolPNames.PROTOCOL_VERSION, HttpVersion.HTTP_1_1);
     HttpGet request = new HttpGet("your url ");  
     HttpResponse response = httpclient.execute(request);
     HttpEntity resEntity = response.getEntity();
     String _response=EntityUtils.toString(resEntity); 
         JSONArray jr = new JSONArray(_response);
         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);
         }
    }catch(Exception e)
    {
        e.printStackTrace();

JSONObject obj = jsonHospitals.getJSONObject(i);

JSONObject obj = jsonHospitals.getJSONObject(String.valueof(i));

JSONArray array = new JSONArray(responseData);

for(int i = 0; i< array.length(); i++) {
  array.getJSONObject(i);
}

 Hi Autolycus,
             I think you have to parse  responseData as JSONArray like below

            **JSONArray jsonHospitals = new JSONArray(responseData);**

          for(int i = 0; i< jsonHospitals.length(); i++){
                    JSONObject obj = jsonHospitals.getJSONObject(i);
                    JSONObject hospital_obj = obj.getJSONObject("Hospital");
                    String hospital_name =  hospital_obj.getString("hospital_name");
                    Log.v(TAG, hospital_name);
                }

       Hope it will work for you :)