Question

我的问题是如何加入具有相同ID但联系人不同的所选行

这是我正在做的输出

|     Name     |    Address   |     Cellphone    |     Email      | 
|     John     |     NY       |       n/a        |   johndoe@y.c  |
|     John     |     NY       |    123456781     |       n/a      |

我希望我的输出是一个组合

|     Name     |    Address   |     Cellphone    |     Email      | 
|     John     |     NY       |     123456781    |   johndoe@y.c  |

这是我的Linq

var an = (from a in db.Info
          join b in db.Contact on a.ID equals b.InfoID
          where b.ContactTypeID == 56
             || b.ContactTypeID == 59
          select new
          {
              a.ID,
              a.LastName,
              a.FirstName,
              a.MiddleName,
              b.ContactTypeID,
              b.Values
          }).ToList();

List<InfoList> wlist = new List<InfoList>();

foreach (var row in an)
{
    InfoList ci = new InfoList
        {
            ID = row.ID,
            Name = row.FirstName + " " + row.MiddleName + " " + row.LastName,
            ContactType = GetLookupDisplayValById(row.ContactTypeID),
            ContactValue = row.Values
        };
    wlist.Add(ci);
}

return Json(wlist.ToList(), JsonRequestBehavior.AllowGet);

我希望有人能帮助我。

Answer 1

对于该数据和仅该数据，自联接可行。这是你可以转换为linq的sql。

select name, address
, case when t1.cellphone <> 'n/a' then t1.cellphone else t2.cellphone end cellphone
, case when t1.email <> 'n/a' then t1.email else t2.email end email
from yourtable t1 join your table t2 on t1.name = t2.name
where t1.name = 'John'

但是，如果数据看起来像这样，

|     Name     |    Address   |     Cellphone    |     Email      | 
|     John     |     NY       |    365418794     |   johndoe@y.c  |
|     John     |     NY       |    123456781     |       n/a      |

您想要什么样的输出？

Answer 2

select max(Name),MAX(Address),MAX(Cellphone),MAX(Email) from grp  group by Name

使用相同的Id连接2列

2 个答案: