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如何使用c中的dictionary属性

properties linq c#

Canacourse · 技术社区 · 15 年前

虽然我的用户现在可以理解如何使用字符串等简单类型获取和设置词典准备好了吗?

我有一个小的DOS程序正在尝试这样做。下面的代码片段。

# User.cs
namespace LDIFMod
{
    public class User
    {

        public string UserHash { get; set; }
        public string UserID { get; set; }
        public Dictionary<string, string> UserDict { get; set; }  <- how to do this???
}

}

在我的程序中。

var query = from line in File.ReadAllLines(args[0])
let UserRecord = line.Split(',')
select new User()

{

 UserHash = UserRecord[2].Trim() +UserRecord[3].Trim(),
 UserID = UserRecord[4].Trim(),
 UserDict.????   // userDict should contain key = UserRecord[5] & value = UserRecord[9]

}

2 回复 | 直到 15 年前

Scott Chamberlain 15 年前

private set

# User.cs
namespace LDIFMod
{
    public class User
    {
        User()
        {
            UserDict = new Dictionary<string, string>()
        }
        public string UserHash { get; set; }
        public string UserID { get; set; }
        public Dictionary<string, string> UserDict { get; private set; }
    }
}

var query = from line in File.ReadAllLines(args[0])
let UserRecord = line.Split(',')
select new User()
{
     UserHash = UserRecord[2].Trim() +UserRecord[3].Trim(),
     UserID = UserRecord[4].Trim(),
     UserDict.Add(UserRecord[5],UserRecord[9]);
}

User

# User.cs
namespace LDIFMod
{
    public class User
    {
        public string UserHash { get; set; }
        public string UserID { get; set; }
        public readonly string[] SourceData {get; private set;}
    }
}

var query = from line in File.ReadAllLines(args[0])
let UserRecord = line.Split(',')
select new User()
{   
     UserHash = UserRecord[2].Trim() + UserRecord[3].Trim(),
     UserID = UserRecord[4].Trim(),
     SourceData = UserRecord;
}
var UserLookp = query.ToDictionary((user) => user.SourceData[5], (user) => user.SourceData[9]);

Etienne de Martel 15 年前

{
    UserHash = UserRecord[2].Trim() +UserRecord[3].Trim(),
    UserID = UserRecord[4].Trim(),
    UserDict = new Dictionary<string, string> { { UserRecord[5],  UserRecord[9] } }
}

http://msdn.microsoft.com/en-us/library/bb531208.aspx