is there a better way to write this frankenstein LINQ query that searches for values in a child tabl
Posted
by MRV
on Stack Overflow
See other posts from Stack Overflow
or by MRV
Published on 2010-03-11T19:45:00Z
Indexed on
2010/03/11
19:49 UTC
Read the original article
Hit count: 253
I have a table of Users and a one to many UserSkills table. I need to be able to search for users based on skills. This query takes a list of desired skills and searches for users who have those skills. I want to sort the users based on the number of desired skills they posses. So if a users only has 1 of 3 desired skills he will be further down the list than the user who has 3 of 3 desired skills.
I start with my comma separated list of skill IDs that are being searched for:
List<short> searchedSkillsRaw = skills.Value.Split(',').Select(i => short.Parse(i)).ToList();
I then filter out only the types of users that are searchable:
List<User> users = (from u in db.Users
where
u.Verified == true &&
u.Level > 0 &&
u.Type == 1 &&
(u.UserDetail.City == city.SelectedValue || u.UserDetail.City == null)
select u).ToList();
and then comes the crazy part:
var fUsers = from u in users
select new
{
u.Id,
u.FirstName,
u.LastName,
u.UserName,
UserPhone = u.UserDetail.Phone,
UserSkills = (from uskills in u.UserSkills
join skillsJoin in configSkills on uskills.SkillId equals skillsJoin.ValueIdInt into tempSkills
from skillsJoin in tempSkills.DefaultIfEmpty()
where uskills.UserId == u.Id
select new
{
SkillId = uskills.SkillId,
SkillName = skillsJoin.Name,
SkillNameFound = searchedSkillsRaw.Contains(uskills.SkillId)
}),
UserSkillsFound = (from uskills in u.UserSkills
where uskills.UserId == u.Id && searchedSkillsRaw.Contains(uskills.SkillId)
select uskills.UserId).Count()
} into userResults
where userResults.UserSkillsFound > 0
orderby userResults.UserSkillsFound descending
select userResults;
and this works! But it seems super bloated and inefficient to me. Especially the secondary part that counts the number of skills found.
Thanks for any advice you can give.
--r
© Stack Overflow or respective owner