Question

我只是尝试获取存储在我的JSON文件中的值并将其保存到sqlite数据库中：

这是我的JSON文件：

{
  "list": {
    "meta": {
      "count": 132, 
      "start": 0, 
      "type": "resource-list"
    }, 
    "resources": [
      {
        "resource": {
          "classname": "Quote", 
          "fields": {
            "date": "2017-03-16", 
            "price": 3.6720000000000002, 
            "type": "currency", 
            "symbol": "AED=X"
          }
        }
      }, 
      {
        "resource": {
          "classname": "Quote", 
          "fields": {
            "date": "2017-03-16", 
            "price": 65.075000000000003, 
            "type": "currency", 
            "symbol": "AFN=X"
          }
        }
      }, 
      {
.............
}
............

我试过这样但得到例外：

JSONObject mainObj = null;
try {
    mainObj = new JSONObject(JSON);
    JSONObject getSth = mainObj.getJSONObject("list");
    if(mainObj != null){
        JSONArray list = getSth.getJSONArray("resources");
        if(list != null){
            for(int i = 0; i < list.length();i++){
                JSONObject elem = list.getJSONObject(i);
                if(elem != null){
                    JSONObject prods = elem.getJSONObject("fields");
                    Object level = prods.get("type");
                    Toast.makeText(getApplicationContext(),""+level.toString(),Toast.LENGTH_LONG).show();
                }
            }
        }
    }
}catch (Exception e){
    Toast.makeText(getApplicationContext(),""+e.toString(),Toast.LENGTH_LONG).show();
}

我得到了例外：字段中没有值......

请给出一些建议，将这些值存储在（行字段）名称，奖品，符号和类型的数据库表（matrotable）中，我可以尝试通过制作字符串数组并检索和存储sqlite的值，是否有任何其他简单选择...... 感谢

Answer 1

您的fields个对象位于resource对象内，所以

  for(int i = 0; i < list.length();i++){
                JSONObject elem = list.getJSONObject(i);
                if(elem != null){
                    JSONObject prods = elem.getJSONObject("resource")
                                           .getJSONObject("fields");

                    Object level = prods.get("type");
                    Toast.makeText(getApplicationContext(),""+level.toString(),Toast.LENGTH_LONG).show();
                }
            }

"resources": [                      // resources list
  {                                 // object i
    "resource": {                   // fields are inside "resource" object 
      "classname": "Quote", 
      "fields": {
        "date": "2017-03-16", 
        "price": 3.6720000000000002, 
        "type": "currency", 
        "symbol": "AED=X"
      }
    }
  }

Answer 2

您缺少resource JOSNObject解析...

 for(int i = 0; i < list.length();i++){
                JSONObject elem = list.getJSONObject(i);
                JSONObject resource = elem.getJSONObject("resource");
                if(resource != null){
                    JSONObject prods = resource.getJSONObject("fields");
                    Object level = prods.get("type");
                    Toast.makeText(getApplicationContext(),""+level.toString(),Toast.LENGTH_LONG).show();
                }
            }

Answer 3

我建议您使用最简单，最简单的方法来解析json响应以避免此类问题：

1-使用此工具生成模型类：http://www.jsonschema2pojo.org/下载并将生成的类添加到模型包中。

2-将此依赖项添加到您的gradle文件中：

compile 'com.google.code.gson:gson:2.4'

3-调用此方法来解析您的回复：

Gson gson = new Gson();
ResponseModel responseModel = gson.fromJson(json, ResponseModel.class);

解析嵌套的Json对象并存储在Android中的数据库中

3 个答案: