Question

I probably won’t be able to explain this well enough, but here it goes…

I have a table similar to this:

Id | Foreign_Key_Id | A_Number | Some_Other_Info
1      1                100          Red
2      1                200          Blue
3      1                300          Orange
4      2                100          Green
5      2                200          Yellow
6      3                100          Brown

I want to get the max “A_Number” for a particular “Foreign_Key_Id” in this table, but also return the rest of the record so I have info from the “Some_Other_Info” column.

I’ve found I can accomplish this via a query like this:

from x in This_Table
group x by x.Foreign_Key_Id into g
orderby g.Max(x => x.A_Number) descending
select g.Where (x => x.Foreign_Key_Id == g.Key).OrderByDescending (x => x.A_Number).First()

Edit: Or possibly this more concise query:

from x in This_Table
group x by x.Foreign_Key_Id into g
let maxANumber = g.Max(x => x.A_Number)
orderby maxANumber descending
select g.Where(x => x.A_Number == maxANumber).First()

Although I’m not quite sure which record it would take if one Foreign_Key_Id had two records with the same “A_Number” value (which is possible in the table I am referring to, I would just need to take the most recent one).

Which in my example I believe would return:

Id | Foreign_Key_Id | A_Number | Some_Other_Info
3      1                300          Orange
5      2                200          Yellow
6      3                100          Brown

It seems like there has to be a way to do this in an easier fashion (that makes for a faster query than the one I came up with), I just can’t figure out how.

Thanks in advance.

Answer 1

The following also works. You can order inside the group:

from x in This_Table
group x by x.Foreign_Key_Id into g
select g.OrderByDescending(y => y.A_Number).First();

Taking the most recent record when A_Numbers are equal is as simple as adding another OrderBy to the query.