我正在制作我的Discord bot所具有的调平系统的排行榜。该列表将列出具有最多XP的人员,并从最高到最低金额订购。我已经实现了这个目标,但我只能在排行榜的XP数量旁边显示用户的ID。如何将此用户ID转换为用户名?
foreach (ulong n in DbContext.Experiences.OrderByDescending(x =>
x.XP).Select(x => x.ID))
{
Context.Guild.GetUser(n).ToString()
}
var leaderboard = string.Concat(DbContext.Experiences.OrderByDescending(x =>
x.XP).Select(x => $"Level {x.LevelNumber} with {x.XP} xp
{//username must be here}\n"));
await ReplyAsync(leaderboard.ToString());
查看代码中的注释:
//What exactly is this for loop meant to do?
//You appear to be getting a user based on ID and doing nothing with that value.
//This should be removed
foreach (ulong n in DbContext.Experiences.OrderByDescending(x =>
x.XP).Select(x => x.ID))
{
//While this is how to get a user by ID, you aren't actually doing anything with this once it's retrieved.
Context.Guild.GetUser(n).ToString()
}
//You can simply fetch the username here, given that you have access to the ID
var leaderboard = string.Concat(DbContext.Experiences.OrderByDescending(x =>
x.XP).Select(x => $"Level {x.LevelNumber} with {x.XP} xp
{//username must be here}\n"));
await ReplyAsync(leaderboard.ToString());
您修改后的代码应如下所示:
var leaderboard = string.Join("\n", DbContext.Experiences
.OrderByDescending(x => x.XP)
.Select(x => $"Level {x.LevelNumber} with {x.XP} xp {Context.Guild.GetUser(x.ID).ToString()}"));
await ReplyAsync(leaderboard);
注意:我用string.Concat
替换了string.Join
,因为它是一种更有效的String构建方法。
假设您已拥有所需的用户ID
Context.Guild.GetUser(THE_USER_ID_HERE).Username
如果存在,则返回用户的用户名。
如果您的排行榜是全局的(用户可能不在执行命令的服务器中)
你可以使用client.GetUser(THE_USER_ID_HERE).Username
,其中client
是你机器人当前的套接字客户端。
(或者,如果您希望显示绑定到服务器的用户名,请访问Nickname
属性)
我期望看到的内容(从未看过Discord,如果这是数据库结构的来源)将是一个表:
我希望dbContext.Users有类似的东西:
UserId | UserName
然后一个公会表有类似的东西:
GuildId | GuildName | UserId
和经验看起来像:
ExperienceId | UserId
我将继续在这里做一些假设:
Context.Guild.GetUser(n).ToString();
对我来说,这看起来像一个转换为SQL的EF Core查询:
select UserId from Guild
通过这种方式,我看到了一些潜在的问题:
首先,仅返回字符串或长的Guild方法很奇怪。如果那是您的实现,则返回一个对象。
更重要的是,您可以在1个查询中执行此操作:
在Sql中:
Select g.GuildId, e.Experiece, u.UserId, u.UserName from Guild g
left join Users u on g.UserId = u.UserId
left join Experiences e on u.UserId = e.UserId
where g.GuildId = @myGuildId
order by e.Experience Desc
这会给你回行如下:1 | 1500 | 10 | BitesSpoons 1 | 1450 | 51 | LostElbows 1 | 1121 | 98 | EarthSkyFire 1 | 990 | 15 | GoldenGoose
我会做什么:创建一个视图类,并使用查询中的AutoMapper或者实现实例时映射它:
上课:
public class LeaderView
{
public string UserName {get; set;}
public long UserId {get; set;}
public long GuildId {get; set; }
public int Experience { get; set; }
}
然后一个linq到Sql查询如:
var leaders = from g in context.Guilds
join u in context.Users on g.UserId = u.UserId
join e in context.Experience on u.UserId = e.UserId
select new LeaderView
{
UserName = u.UserName,
UserId = u.UserId,
GuildId = g.UserId,
Experience = e.Experience
};
leaders = leaders.OrderByDescending(o => o.Experience);
return leaders.ToList();