Question

大家好，这是我的json回复：

 [
  {
  "Woa": [
    "Seo",
    "Rikjeo",
    "JDa"
   ]
  },
 "Aha",
 "Aad",
 "Char"
 ]

我想添加Woa的列表字符串：

{
  "Woa": [
    "Seo",
    "Rikjeo",
    "JDa"
   ]
  }

这是我到目前为止所做的：

JSONObject object = new JSONObject(result);
            JSONArray a = object.getJSONArray("Woa");
            for (int i = 0; i < a.length(); ++i) {
                listWoa.add(a.getString(i));
            }

但是我收到了这个错误：

type org.json.JSONArray cannot be converted to JSONObject

任何想法，为什么我没有得到任何字符串，也无法转换为JSONObject。

Answer 1

解析json响应之上。请尝试以下代码：

        JSONArray jsonArray = new JSONArray(result);
        JSONObject jsonObjWoa = jsonArray.getJSONObject(0);
        JSONArray jsonArrayWoa = jsonObjWoa.getJSONArray("Woa");
        for (int i = 0; i < jsonArrayWoa.length(); ++i) {
            listWoa.add(jsonArrayWoa.getString(i));
        }

Answer 2

您的JSON是一个数组（“事物”列表），其中第一个项目是对象

 [                             << This is an array (let's call it A)
  {                                 << This is A[0]
  "Woa": [
    "Seo",
    "Rikjeo",
    "JDa"
   ]
  },
 "Aha",                             << This is A[1] == "AhA"
 "Aad",                             << This is A[2] == "aad"
 "Char"                             << This is A[3] == "Char"
 ]

因此，A [0]是对象：

  {                           << This is an object ( A[0] )
  "Woa": [                       << This is A[0].Woa (it's an array)
    "Seo",                          << This is A[0].Woa[0] == "Seo"
    "Rikjeo",                       << This is A[0].Woa[1] == "Rikjeo"
    "JDa"                           << This is A[0].Woa[2] == "JDa"
   ]
  }

不在JSON中混合数组和对象的简单方法是：

[...]    is an array
{...}    is an object

Answer 3

做这样的事情：

JSONArray jsonArray = new JSONArray(result);
JSONObject object = new JSONObject(jsonArray.get(0).toString());// if you have only one element then get 0th index
JSONArray a = object.getJSONArray("Woa");
 for (int i = 0; i < a.length(); ++i) {
      listWoa.add(a.getString(i));
  }

Answer 4

从我的角度来看：我会使用GSON自动解析JSON和这个站点来生成POJO。

http://www.jsonschema2pojo.org/ 确保单击：GSON作为注释样式，并使用JSON而不是JSON Scheme

Android Jsonarray无法转换为json对象

4 个答案: