string test_json = "{\"name\":\"tom\",\"nickname\":\"tony\",\"sex\":\"male\",\"age\":20,\"email\":\"123@123.com\"}"; var o = JObject.Parse(yourJsonString); foreach (JToken child in o.Children()) { var property1 = child as JProperty; MessageBox.Show(property1.Name + ":
C#直接解析Json键值对
最新推荐文章于 2023-09-26 16:43:23 发布