Android 转换json时出错
我需要帮助转换json 这是我的密码:Android 转换json时出错,android,json,Android,Json,我需要帮助转换json 这是我的密码: try { JSONObject mainJson22 = new JSONObject(reply); JSONArray jsonArray22 = mainJson22.getJSONArray("UserListByBusinessAreaContextResult"); Log.i("mainjson234","" + jsonArray22); for (int i2 = 0; i2 < jsonAr
try {
JSONObject mainJson22 = new JSONObject(reply);
JSONArray jsonArray22 = mainJson22.getJSONArray("UserListByBusinessAreaContextResult");
Log.i("mainjson234","" + jsonArray22);
for (int i2 = 0; i2 < jsonArray22.length(); i2++) {
JSONObject objJson22 = jsonArray22.getJSONObject(i2);
// JSONArray innerJsonArray = jsonArray.getJSONArray(i);
// JSONObject objJson = innerJsonArray.getJSONObject(0);
int id22 = objJson22.getInt("UserID");
String username22 =objJson22.getString("Username");
String name = objJson22.getString("Name");
reply = {"UserListByBusinessAreaContextResult":"{\"tableData\":[{\"UserID\":30,\"Username\":\"Teste\",\"Name\":\"Teste\"}]}"}
UserListByBusinessAreaContextResult
不是数组。这是一个JSONObject
tableData是您的JSONArray。因此,你必须:
try {
JSONObject mainJson22 = new JSONObject(reply);
JSONArray jsonResult = mainJson22.getJSONObject("UserListByBusinessAreaContextResult");
JSONArray jsonArray22 = jsonResult.getJSONArray("tableData");
Log.i("mainjson234","" + jsonArray22);
for (int i2 = 0; i2 < jsonArray22.length(); i2++) {
JSONObject objJson22 = jsonArray22.getJSONObject(i2);
...
试试{
JSONObject mainJson22=新JSONObject(回复);
JSONArray jsonResult=mainJson22.getJSONObject(“UserListByBusinessAreaContextResult”);
JSONArray jsonArray22=jsonResult.getJSONArray(“tableData”);
Log.i(“mainjson234”和“+jsonArray22”);
for(int i2=0;i2
您需要更新标题,以反映您无法将字符串转换为JSON数组。例外情况是正确的-这是一个对象,而不是数组。请进一步阅读JSON及其结构。
try {
JSONObject mainJson22 = new JSONObject(reply);
JSONArray jsonResult = mainJson22.getJSONObject("UserListByBusinessAreaContextResult");
JSONArray jsonArray22 = jsonResult.getJSONArray("tableData");
Log.i("mainjson234","" + jsonArray22);
for (int i2 = 0; i2 < jsonArray22.length(); i2++) {
JSONObject objJson22 = jsonArray22.getJSONObject(i2);
...