我有一个文件中的以下文字:
{"players":[{"i":11,"p":0,"a":3186,"n":"IanHx","f":1,"ps":0,"pd":0,"bc":0},{"i":12,"p":0,"a":115,"n":"LoZtamnik","f":1,"ps":0,"pd":0,"bc":0},{"i":58,"p":0,"a":156,"n":"Mr701","f":2,"ps":0,"pd":0,"bc":0},{"i":59,"p":0,"a":156,"n":"B0NE4","f":2,"ps":0,"pd":0,"bc":0},{"i":64,"p":0,"a":324,"n":"5teveJ","f":1,"ps":0,"pd":0,"bc":0}],[.......
我要做的是解析文本,最后得到一个数组,用于{...}
之间的每一位数据所以最终结果如下:
i=11
p=0
a=3186
n=IanHx
f=1
ps=0
pd=0
bc=0
然后我可以将它们存储在数据库中
到目前为止我有这样的事情:
string contents = System.IO.File.ReadAllText("local text file"); //load string contents with text file
Regex regex1 = new Regex("\"players\":\\[(?<players>.*)\\]"); //find the players section "players":[.......]
Match match1 = regex1.Match(contents); //load match1
Regex regex2 = new Regex("{(?<player>([^}]*))}"); // then break down each player {....}
MatchCollection match2 = regex2.Matches(match1.Groups["players"].Value); //load match2 with each player
然后我试图以某种方式分割匹配字符串[],看着它可能会使它过于复杂?
指向数据解析的更简单解决方案的任何指针
由于
答案 0 :(得分:4)
尝试使用NuGet中的Json.Net 3}进行反序列化会更好
定义一些类以匹配您的文件结构:
public class Root
{
public List<Something> players {get; set;}
}
public class Something
{
public string i {get; set;}
public string p {get; set;}
public string a {get; set;}
public string n {get; set;}
public string f {get; set;}
public string ps {get; set;}
public string pd {get; set;}
public string bc {get; set;}
}
使用Json.Net进行处理:
var json = @"{""players"":[{""i"":11,""p"":0,""a"":3186,""n"":""IanHx"",""f"":1,""ps"":0,""pd"":0,""bc"":0},{""i"":12,""p"":0,""a"":115,""n"":""LoZtamnik"",""f"":1,""ps"":0,""pd"":0,""bc"":0},{""i"":58,""p"":0,""a"":156,""n"":""Mr701"",""f"":2,""ps"":0,""pd"":0,""bc"":0},{""i"":59,""p"":0,""a"":156,""n"":""B0NE4"",""f"":2,""ps"":0,""pd"":0,""bc"":0},{""i"":64,""p"":0,""a"":324,""n"":""5teveJ"",""f"":1,""ps"":0,""pd"":0,""bc"":0}]}";
var data = JsonConvert.DeserializeObject<Root>(json);
答案 1 :(得分:3)
您文件中包含的数据为JSON
格式。如果格式正确,JSON很容易阅读。如果我重新格式化您的输入,结构就会变得更清晰:
{
"players": [
{
"i": 11,
"p": 0,
"a": 3186,
"n": "IanHx",
"f": 1,
"ps": 0,
"pd": 0,
"bc": 0
},
{
"i": 12,
"p": 0,
"a": 115,
"n": "LoZtamnik",
"f": 1,
"ps": 0,
"pd": 0,
"bc": 0
},
{
"i": 58,
"p": 0,
"a": 156,
"n": "Mr701",
"f": 2,
"ps": 0,
"pd": 0,
"bc": 0
},
{
"i": 59,
"p": 0,
"a": 156,
"n": "B0NE4",
"f": 2,
"ps": 0,
"pd": 0,
"bc": 0
},
{
"i": 64,
"p": 0,
"a": 324,
"n": "5teveJ",
"f": 1,
"ps": 0,
"pd": 0,
"bc": 0
}
]
}
在JSON中,[ ]
中包含的任何内容都表示一个集合,{ }
中包含的任何内容都表示一个对象。因此,您可以看到您有一个名为players
的集合,其中包含5个对象(因为{ }
中有5对players [ ]
),每个对应8个属性。如果您在C#术语中考虑这些,那么您将拥有一个名为Player
的类,其中包含这8个属性,以及List<Player>
来保存每个Player
实例。然后,您可以获取JSON数据并将它们反序列化为C#对应项,以便您可以按照自己的需要操作它们,正如Dave Bish在其答案中指出的那样。
有一种非常简单的方法可以自动从JSON数据创建C#类:
Edit -> Paste Special -> Paste JSON As Classes
瞧。 Visual Studio得到了回报。你现在应该看到以下几点:
public class Rootobject
{
public Player[] players { get; set; }
}
public class Player
{
public int i { get; set; }
public int p { get; set; }
public int a { get; set; }
public string n { get; set; }
public int f { get; set; }
public int ps { get; set; }
public int pd { get; set; }
public int bc { get; set; }
}
然后,您可以做出最适合您情景的事情,例如:添加System.Collections.Generic
命名空间,以便Player[]
改为List<Player>
,而不是等等。
现在,要操纵JSON数据并将它们反序列化为我们刚刚创建的C#类,您可以使用优秀的Json.NET
库。要添加它,请右键单击解决方案资源管理器中的应用程序,然后单击"Manage NuGet Packages..."
。在搜索框中输入"Json.NET"
并进行安装。
准备好后,添加Newtonsoft.Json
命名空间,您就可以了。您现在可以使用Json.NET的DeserializeObject<T>()
方法将JSON数据反序列化为我们创建的C#类:
//i've hardcoded the JSON data here, obviously you would extract them from your file
var jsonData = @"{""players"":[{""i"":11,""p"":0,""a"":3186,""n"":""IanHx"",""f"":1,""ps"":0,""pd"":0,""bc"":0},{""i"":12,""p"":0,""a"":115,""n"":""LoZtamnik"",""f"":1,""ps"":0,""pd"":0,""bc"":0},{""i"":58,""p"":0,""a"":156,""n"":""Mr701"",""f"":2,""ps"":0,""pd"":0,""bc"":0},{""i"":59,""p"":0,""a"":156,""n"":""B0NE4"",""f"":2,""ps"":0,""pd"":0,""bc"":0},{""i"":64,""p"":0,""a"":324,""n"":""5teveJ"",""f"":1,""ps"":0,""pd"":0,""bc"":0}]}";
//deserialize the JSON into the C# class we created and store it in myPlayerData
var myPlayerData = JsonConvert.DeserializeObject<Rootobject>(jsonData);
//you can now do stuff such as..
foreach(Player player in myPlayerData.players)
{
MessageBox.Show(string.Format("Player {0} has an i of {1} and an a of {2}", player.n, player.i, player.a));
}