Soundex

2005
2008
2012
2014
2016
2017
2019
2022

2005, 2008, 2012, 2014, 2016, 2017, 2019, 2022

Soundex String Function

Use the Soundex function to evaluate the similarity of two strings.

Syntax

db.fx.Soundex({expression})

Arguments

expression: – The value to use for computation of the four character result.

Returns

string or string?

(depending on the nullability of `expression`)

Examples

Select Statement

Select the soundex value of a person's last name.

IEnumerable<string> result = db.SelectMany(
		db.fx.Soundex(dbo.Person.LastName)
	)
	.From(dbo.Person)
	.Execute();

Where Clause

Select persons where their last name is similar to their first name.

IEnumerable<Person> result = db.SelectMany<Person>()
    .From(dbo.Person)
	.Where(db.fx.Soundex(dbo.Person.LastName) != dbo.Person.LastName)
	.Execute();

Order By Clause

Select a list of persons, ordered by the soundex value of their last name.

IEnumerable<Person> result = db.SelectMany<Person>()
	.From(dbo.Person)
	.OrderBy(db.fx.Soundex(dbo.Person.LastName))
	.Execute();

Group By Clause

Select a list of address values grouped by address type and the similarity of the names of the city they live in (distinct list of similar sonding cities).

IEnumerable<dynamic> values = db.SelectMany(
		dbo.Address.AddressType,
		db.fx.Soundex(dbo.Address.City).As("city")
	)
	.From(dbo.Address)
	.GroupBy(
		dbo.Address.AddressType,
		db.fx.Soundex(dbo.Address.City)
	)
	.Execute();

Having Clause

Select a list of address data, grouped by address type and city where the value of city with leading and trailing spaces removed is "Chicago".

IEnumerable<dynamic> addresses = db.SelectMany(
		db.fx.Count().As("count"),
		dbo.Address.AddressType
	)
	.From(dbo.Address)
	.GroupBy(
		dbo.Address.AddressType,
		dbo.Address.City
	)
	.Having(
		db.fx.Soundex(dbo.Address.City) == "G452"
	)
	.Execute();