Entity framework 4 选择两列的最佳实践

Entity framework 4 选择两列的最佳实践,entity-framework-4,linq-to-entities,Entity Framework 4,Linq To Entities,我有如下疑问 var email= (from c in dataContext.tblC where (c.AA == aa) select c.email).ToList(); string emails = email.Aggregate((a, b) => a + "," + b); 现在我需要另一个专栏SecEmail,我不能只是…选择c.email,c.SecEmail.ToList。 如果您正在使用动态对象,我可以得到任何建议

我有如下疑问

var email= (from c in dataContext.tblC
            where (c.AA == aa)
            select c.email).ToList();
string emails = email.Aggregate((a, b) => a + "," + b);
现在我需要另一个专栏SecEmail,我不能只是…选择c.email,c.SecEmail.ToList。
如果您正在使用动态对象,我可以得到任何建议,如email1、SecEmail1、email2、email3、SecEmail3、email4…

var email = (from c in dataContext.tblC
             where c.AA == aa
             select new {
               email = x.email,
               secemail = c.secEmail,
               // ...
             }).ToList(); // IList<dynamic> & IDE will know what properties
                          // you supplied in the `select new { }`
这将使:

+------------------+
| email            |
|------------------|
| foo@contoso.com  |
| bar@contoso.com  |
| baz@contoso.com  |
+------------------+
求助于:

foo@contoso.com, bar@contoso.com, baz@contoso.com
多列混凝土:

var email = (from c in dataContext.tblC
             where c.AA == AA
             select new { email = c.email, secEmail = c.secEmail }).AsEnumerable();
var to = String.Join(", ",
    email.Select(x => x.email)
        .Concat(email.Select(y => y.secEmail))
        // .Concat(eail.Select(y => x.thirdColumn))
        // ...
);

谢谢但如何将电子邮件转换为数组列表?所以我可以得到像email1,SecEmail1,email2,email3,SecEmail3,email4这样的列表。。。。实际上我需要一个字符串来发送电子邮件哦,我理解错了。要将列连接起来吗?只需简单地从dataContext.tblC中的c中选择String.Join,其中c.AA==AA选择c.email.AsEnumerable;谢谢,但您只需选择字符串。加入电子邮件,第二封电子邮件在哪里?有些人有电子邮件,有些人没有。谢谢。但是如何将secEmail添加到列表中?@MikeLong:很抱歉耽搁了;看看我答案的最后一部分
foo@contoso.com, bar@contoso.com, baz@contoso.com
var email = (from c in dataContext.tblC
             where c.AA == AA
             select new { email = c.email, secEmail = c.secEmail }).AsEnumerable();
var to = String.Join(", ",
    email.Select(x => x.email)
        .Concat(email.Select(y => y.secEmail))
        // .Concat(eail.Select(y => x.thirdColumn))
        // ...
);