开发者

LINQ Where Exists GROUP BY

开发者 https://www.devze.com 2023-02-21 23:13 出处:网络
How can i convert this in LINQ? SELECT B.SENDER, B.SENDNUMBER, B.SMSTIME, B.SMSTEXT FROM MESSAGES B WHERE EXISTS ( SELECT A.SENDER

How can i convert this in LINQ?

SELECT B.SENDER, B.SENDNUMBER, B.SMSTIME, B.SMSTEXT
FROM MESSAGES B
WHERE EXISTS ( SELECT A.SENDER
        FROM MESSAGES A
    WHERE A.SENDER = B.SENDER
    GROUP BY A.SENDER
    开发者_如何学编程HAVING B.SMSTIME = MAX( A.SMSTIME))
GROUP BY B.SENDER, B.SENDNUMBER, B.SMSTIME, B.SMSTEXT ;

Thanks a lot :)

EDIT!!

Resolved with:

  var Condition = "order by SMSTime desc";      

  IEnumerable<ClassMessaggio> messaggi = Database.Select<ClassMessaggio>(Condition);; // Load all but sorted

  ElencoConversazioni =  messaggi.GroupBy(m => new { m.Number })
                .Select(g => g.OrderByDescending(m => m.SMSTime).First()).ToObservableCollection();


Try

db.Messages.Where(b => b.SmsTime == Messages.Where(a => a.Sender == b.Sender)
                                            .Max(a => a.SmsTime))

Or

db.Messsages.GroupBy(m => new { m.Sender, m.SendNumber })
            .Select(g => g.OrderByDescending(m => m.SmsTime).First())

Where db is your DataContext.


To show the list of conversations along with "the last message of every contact", you could try something like:

// a query to find the last Sms time per sender
var lastSmsQuery = from m in db.messages
                 group m by m.Sender into grouping
                 select new
                 {
                     Sender = grouping.Key,
                     LastSmsTime = grouping.Max(x => x.SmsTime)
                 };

// a query to find the messages linked to the last Sms per sender
var lastMessageQuery = from m in db.messages
                       join l in lastSmsQuery on new { m.Sender, m.SmsTime } equals new { l.Sender, l.LastSmsTime }
                       select m;

The 2-query method used is similar to that of this question from earlier today - Convert SQL Sub Query to In to Linq Lambda

0

精彩评论

暂无评论...
验证码 换一张
取 消