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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T13:23:42+00:00 2026-06-11T13:23:42+00:00

I have a problem with a project. I’m trying to get a list of

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I have a problem with a project. I’m trying to get a list of companies that are located in a specific country or city.

Table structure:

Company

CompanyID
CompanyName
etc…

CompanyAddressDetails (relation table)

Company_CompanyID
CorrespondingAddress_AddressID

CorrespondingAddress:

AddressID
StreetName
RegionID
etc…

Region

RegionID
RegionName
RegionRegionTypeID

RegionDetails (relation table)

RegionParent
RegionChild

So to find an address in example Stockholm (which has ID 1198 in the Region table), I would do:

var addresses = from c in db.CorrespondingAddress select c;
addresses = addresses.Where(s => s.RegionID.Equals(1198));

And to find a company in Stockholm I would do:

companyModel = from c in db.Company select c;
companyModel =  companyModel.Where(s => s.CorrespondingAddress.Any(x => x.RegionID.Equals(1198)));

But now I want to take into account the RegionDetails table (which has a parent, and child, for example: 1 (Sweden) is parent, and 1198 (Stockholm) is child etc)

How can I do to find a company which is located in Sweden, but has the ID 1198 (Stockholm) in its address row?

In plain SQL I would maybe do something like:

SELECT CompanyName FROM Company
LEFT JOIN CompanyAddressDetails ON (Company.CompanyID = CompanyAddressDetails.Company_CompanyID)
LEFT JOIN CorrespondingAddress ON (CompanyAddressDetails.CorrespondingAddress_AddressID = CorrespondingAddress.AddressID)
LEFT JOIN Region ON (CorrespondingAddress.RegionID = Region.RegionID)
WHERE CorrespondingAddress IN (SELECT RegionChild FROM RegionDetails WHERE RegionParent = 1)
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T13:23:44+00:00Added an answer on June 11, 2026 at 1:23 pm

    since there is no model for the details tables (many-to-many relationship tables), I solved it by using this method:

    companyModel = companyModel.Where(s => s.CorrespondingAddress.Any(x => x.Region.RegionParent.Any(d => d.RegionID == region)));
    
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