我有一个像这样的文件
{ "_id": { "$oid" : "51776bca40bcc60038000001" },
"username": "domi55",
"Password": "test",
"Character": { "Job": "Warrior",
"Level": 1,
"Skill": { "SkillID": "1001",
"SkillName": "Blade Dance",
"LevelRequirment": 1
}
}
}
}
如何在C#中获取“作业”值和“SkillName”值? 我正在使用MongoDB和MongoDB C#驱动程序
答案 0 :(得分:1)
使用Json.Net
dynamic obj = JsonConvert.DeserializeObject(yourDoc);
Console.WriteLine("{0} {1}", obj.Character.Job, obj.Character.Skill.SkillName);
var obj = new JavaScriptSerializer().Deserialize<dynamic>(json);
Console.WriteLine("{0} {1}",obj["Character"]["Job"],obj["Character"]["Skill"]["SkillName"]);