java - 无法解析有效的 JSON 字符串

Question

我正在尝试解析一个 JSON 字符串并将其映射到 HashMap ，我从服务器获得了一个有效的 JSONString，但是当我遍历它时，我只能得到第一个结果。

 JSONArray peoples = null;
 ArrayList<HashMap<String, String>> personList = new ArrayList<HashMap<String,String>>();

JSONObject jsonObj = new JSONObject(value);
                Log.d("Jello",jsonObj.toString());
                peoples = jsonObj.getJSONArray("product");//check here
                Log.d("Jello",peoples.toString());


                for(int i=0;i<peoples.length();i++){
                    JSONObject c = peoples.getJSONObject(i);
                    String service_group = c.getString("sgroup");
                    String service = c.getString("service");
                    String value = c.getString("value");
                    String updated_at = c.getString("updated_at");

                    HashMap<String,String> persons = new HashMap<String,String>();

                    persons.put("service_group",service_group);
                    persons.put("service",service);
                    persons.put("value",value);
                    persons.put("updated_at",updated_at);


                    personList.add(persons);
                }

我的 JSON 字符串是:

{"product":[{"sgroup":"Dummy_BIG_ONE","service":"Dummy_UNDER_BIG_ONE","code":"128","value":"0","updated_at":"2015-12-04 21:21:00"}]}{"product":[{"sgroup":"Hello Monkey","service":"Do u work","code":"123","value":"0","updated_at":"2015-12-04 21:27:51"}]}{"product":[{"sgroup":"Checking from Android Device","service":"Monkey","code":"12345","value":"0","updated_at":"2015-12-04 22:55:39"}]}{"product":[{"sgroup":"Checking from Android Device","service":"Monkey","code":"12345","value":"0","updated_at":"2015-12-04 22:55:40"}]}{"product":[{"sgroup":"Checking from Android Device","service":"Monkey","code":"12345","value":"0","updated_at":"2015-12-04 22:55:42"}]}{"product":[{"sgroup":"Hello World","service":"Donkey","code":"24411","value":"0","updated_at":"2015-12-04 22:57:05"}]}{"product":[{"sgroup":"lkfnhjdiofho","service":"dfjdifj","code":"1101","value":"0","updated_at":"2015-12-05 01:15:49"}]}{"product":[{"sgroup":"Baal","service":"Saal","code":"1234","value":"21","updated_at":"2015-12-05 01:34:59"}]}{"product":[{"sgroup":"Inis","service":"Mona","code":"1234","value":"1001","updated_at":"2015-12-05 01:39:51"}]}{"product":[{"sgroup":"Medical Treatment Loan","service":"Number of referral slip","code":"128","value":"0","updated_at":"2015-12-05 01:50:42"}]}{"product":[{"sgroup":"Medical Treatment Loan","service":"Number of referral slip","code":"128","value":"0","updated_at":"2015-12-05 01:55:12"}]}{"product":[{"sgroup":"Medical Treatment Loan","service":"Number of referral slip","code":"128","value":"1000","updated_at":"2015-12-05 01:56:10"}]}

或HERE

这是我发送 JSON 的方式

while($row = mysql_fetch_assoc($output))
        {
            $product = array();
            $product["sgroup"] = $row["service_group"];
            $product["service"] = $row["service"];
            $product["code"] = $row["code"];
            $product["value"] = $row["amount"];
            $product["updated_at"] = $row["updated_at"];

            // user node
            $response["product"] = array();

            array_push($response["product"], $product);

            // echoing JSON response
            echo json_encode($response);
        }

我想将所有来自 JSON 的数据添加到我的 ArrayList 中，以便我以后可以使用它。感谢您的帮助。

Answer 1

首先，有online tools可用于确定您的 JSON 是否有效。

以下是格式化 JSON 数据的更好方法的一般示例:

<?php 
    $bigArray = array();

    for ($x = 0; $x <= 10; $x++) {
            $product = array();
            $product["sgroup"] = $x;
            $product["service"] = $x;
            $product["code"] = $x;
            $product["value"] = $x;
            $product["updated_at"] = $x;

            array_push($bigArray, $product);
        }
      // echoing JSON response
      echo json_encode($bigArray);      
?>

它给出了简单易解析的有效 JSON 响应:

[
   {
      "sgroup":0,
      "service":0,
      "code":0,
      "value":0,
      "updated_at":0
   },
   {
      "sgroup":1,
      "service":1,
      "code":1,
      "value":1,
      "updated_at":1
   },
   {
      "sgroup":2,
      "service":2,
      "code":2,
      "value":2,
      "updated_at":2
   },
   {
      "sgroup":3,
      "service":3,
      "code":3,
      "value":3,
      "updated_at":3
   },
   {
      "sgroup":4,
      "service":4,
      "code":4,
      "value":4,
      "updated_at":4
   },
   {
      "sgroup":5,
      "service":5,
      "code":5,
      "value":5,
      "updated_at":5
   },
   {
      "sgroup":6,
      "service":6,
      "code":6,
      "value":6,
      "updated_at":6
   },
   {
      "sgroup":7,
      "service":7,
      "code":7,
      "value":7,
      "updated_at":7
   },
   {
      "sgroup":8,
      "service":8,
      "code":8,
      "value":8,
      "updated_at":8
   },
   {
      "sgroup":9,
      "service":9,
      "code":9,
      "value":9,
      "updated_at":9
   },
   {
      "sgroup":10,
      "service":10,
      "code":10,
      "value":10,
      "updated_at":10
   }
]

至于解析，使用HashMap 并不是最好的方法。创建 Person POJO 对象列表。

首先定义Person类:

class Person {
    public String group;
    public String service;
    public String value;
    public String updated;
    public Person(String g, String s, String v, String u) {
        group = g;
        service = s;
        value = v;
        updated = u;
    }
}

然后，解析就相当简单了:

    List<Person> personList = new ArrayList<>();
    JSONArray jsonArr;
    try {
        jsonArr = new JSONArray(response);
        for(int i=0;i<jsonArr.length();i++){
            JSONObject c = jsonArr.getJSONObject(i);
            String service_group = c.getString("sgroup");
            String service = c.getString("service");
            String value = c.getString("value");
            String updated_at = c.getString("updated_at");

            Person p = new Person(service_group, service, value, updated_at);

            personList.add(p);
        }
    } catch (JSONException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }