Question

原始Json数据

 [  
   {  
      "worker.1":{  
         "last_share":1443639029,
         "score":"3722204.62578",
         "alive":true,
         "shares":1124332,
         "hashrate":1047253
      },
      "worker.2":{  
         "last_share":1443639029,
         "score":"3794755.69049",
         "alive":true,
         "shares":1069332,
         "hashrate":1012070
      },
      "worker.3":{  
         "last_share":1440778690,
         "score":"0.0",
         "alive":false,
         "shares":0,
         "hashrate":0
      },
      "worker.4":{  
         "last_share":1443638222,
         "score":"940190.67723",
         "alive":true,
         "shares":449932,
         "hashrate":404772
      }
      "worker.nth":{  
         "last_share":1443638222,
         "score":"940190.67723",
         "alive":true,
         "shares":449932,
         "hashrate":404772
      }
   }
]

从主要问题我能够缩小结果范围并设法添加Square括号。现在我上面有一个Json数组，所以有人可以告诉我如何处理这个Json数组。另请注意，最终可能会有1000名工人。

请帮助我在这个问题上突破了2天另请注意，实际的Json数据很复杂且有1000个条目，包含的json数据只是一个概述。

编辑：我用来处理上述Json的代码

try{
                        JSONArray jArray = new JSONArray(jsonObj.getJSONObject("workers"));
                        if(jArray.length() == 0){
                            Log.d("Json Err", "JSON string has no entries");
                            return null;
                        }
                        JSONObject jObject = jArray.getJSONObject(0);

                        // To get one specific element:
                        JSONObject worker1 = jObject.getJSONObject("worker.1");
                        JSONObject worker2 = jObject.getJSONObject("worker.2");

                        // Iterate over all elements
                        Iterator<?> keys = jObject.keys();
                        while( keys.hasNext() ) {
                            String key = (String)keys.next(); // "worker.1", etc
                            if ( jObject.get(key) instanceof JSONObject ) {
                                JSONObject entry = jObject.getJSONObject(key);
                                // Do something with your entry here like:
                                String score = entry.optString("score", "");
                                Boolean alive = entry.optBoolean("alive", false);
                                Integer shares = entry.optInt("shares", 0);
                                Log.d("Json Success", score );
                            }
                        }
                    } catch (JSONException e){
                        Log.e("Json Err", "Failure converting JSON String", e);
                    }

我收到错误：

E2570（下午5:10）： org.json.JSONException：Value {“worker.1”：{“last_share”：1443694390，“score”：“15.6018529132”，“alive”：true，“shares”：59880，“hashrate”：0}，“worker 0.2 “：{” last_share “：1443694180，” 分数 “：” 2.97304689833" ， “活着”：真正的 “股”：2048， “hashrate”：0}， “worker.3”：{ “last_share”：1440778690 ， “分数”： “0.0”， “活的”：假， “股份”：0， “hashrate”：0}， “ivonme.ant4”：{ “last_share”：1443701343， “得分”： “8688.78118933”，”活着 “：真正的” 股 “：203088，” hashrate“：78633}} 类型org.json.JSONObject无法转换为JSONArray 在org.json.JSON.typeMismatch（JSON.java）在org.json.JSONArray。（JSONArray.java）在org.json.JSONArray。（JSONArray.java）

Answer 1

所以你有......

   {  //JSONObject
      "worker.1":{  
         "last_share":1443639029,
         "score":"3722204.62578",
         "alive":true,
         "shares":1124332,
         "hashrate":1047253
      },
      "worker.2":{  
         "last_share":1443639029,
         "score":"3794755.69049",
         "alive":true,
         "shares":1069332,
         "hashrate":1012070
      } //,...
   }

为了得到你所有的工人......

try{
   JSONObject jObject = new JSONObject(yourStringFromYourQuestion);

   // To get one specific element:
   JSONObject worker1 = jObject.getJSONObject("worker.1");
   JSONObject worker2 = jObject.getJSONObject("worker.2");

   // Iterate over all elements
   Iterator<?> keys = jObject.keys();
   while( keys.hasNext() ) {
      String key = (String)keys.next(); // "worker.1", etc
      if ( jObject.get(key) instanceof JSONObject ) {
         JSONObject entry = jObject.getJSONObject(key);
         // Do something with your entry here like:
         String score = entry.optString("score", "");
         Boolean alive = entry.optBoolean("alive", false);
         Integer shares = entry.optInteger("shares", 0);
      }
   }
} catch (JSONException e){
   Log.e(LOG_TAG, "Failure converting JSON String", e);
}

Answer 2

试试这个：

而不是

JSONArray jArray=jsonObj.getJSONArray("workers");

错误是因为你将json数组作为对象。我想是的。

如何在没有方括号的android中处理Json数组

2 个答案: