Linq to SQL:具有匿名类型的DISTINCT [英] Linq to SQL: DISTINCT with Anonymous Types

问题描述

给出以下代码:

dgIPs.DataSource = 
    from act in Master.dc.Activities
    where act.Session.UID == Master.u.ID
    select new
    {
      Address = act.Session.IP.Address,
      Domain = act.Session.IP.Domain,
      FirstAccess = act.Session.IP.FirstAccess,
      LastAccess = act.Session.IP.LastAccess,
      IsSpider = act.Session.IP.isSpider,
      NumberProblems = act.Session.IP.NumProblems,
      NumberSessions = act.Session.IP.Sessions.Count()
    };

如何仅基于不同的地址提取Distinct()?也就是说，如果我仅添加Distinct()，它将整个行评估为不同的行，从而无法找到任何重复项.我想为每个act.Session.IP对象只返回一行.

How do I pull the Distinct() based on distinct Address only? That is, if I simply add Distinct(), it evaluates the whole row as being distinct and thusly fails to find any duplicates. I want to return exactly one row for each act.Session.IP object.

我已经找到了此答案，但这似乎是不同的情况.另外，如果我只选择act.Session.IP，则Distinct()可以很好地工作，但是我希望避免检索它的列，而我宁愿不必通过手动绑定我的datagrid列来做到这一点.

I've already found this answer, but it seems to be a different situation. Also, Distinct() works fine if I just select act.Session.IP, but it has a column I wish to avoid retrieving and I'd rather not have to do this by manually binding my datagrid columns.

推荐答案

dgIPs.DataSource = 
    from act in Master.dc.Activities
    where act.Session.UID == Master.u.ID
    group act by act.Session.IP.Address into g
    let ip = g.First().Session.IP
    select new
    {
      Address = ip.Address,
      Domain = ip.Domain,
      FirstAccess = ip.FirstAccess,
      LastAccess = ip.LastAccess,
      IsSpider = ip.isSpider,
      NumberProblems = ip.NumProblems,
      NumberSessions = ip.Sessions.Count()
    };

或者:

dgIPs.DataSource = 
    from act in Master.dc.Activities
    where act.Session.UID == Master.u.ID
    group act.Session.IP by act.Session.IP.Address into g
    let ip = g.First()
    select new
    {
      Address = ip.Address,
      Domain = ip.Domain,
      FirstAccess = ip.FirstAccess,
      LastAccess = ip.LastAccess,
      IsSpider = ip.isSpider,
      NumberProblems = ip.NumProblems,
      NumberSessions = ip.Sessions.Count()
    };

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