SQL Server wildcard

In some cases, familiarity with the use of SQL Server wildcard can help us solve many problems.

 

-- Use the _ operator to search for the names of the three letters ending with an in the person table

Useadventureworks2012;

Go

Select firstname, lastname

From person. Person

Where firstname like '_'

Order by firstname;

 

--- Use the [^] operator to search for all persons whose names start with Al and whose third letter is not a in the contact table

Useadventureworks2012;

Go

Select firstname, lastname

From person. Person

Where firstname like 'al [^ A] %'

Order by firstname;

 

--- Use the [] operator to find the ID and name of all adventure works employees whose addresses contain four postal codes

Useadventureworks2012;

Go

Select E. businessentityid, P. firstname, P. lastname,
A. postalcode

Fromhumanresources. employeeas E

Inner join person. personas Pon E. businessentityid =
P. businessentityid

Inner join person. businessentityaddressas eaon E. businessentityid = EA. businessentityid

Inner join person. addressas Aon A. addressid =
EA. addressid

Where a. postalcodelike '[0-9] [0-9] [0-9] [0-9]';

 

Result set:

Employeeid firstname lastname postalcode
--------------------------------------
290 Lynn tsoflias
3000

 

-- Name a table in Chinese and English (refer to the Code in the Forum)

Create Table Tb (namenvarchar (20 ))

 

Insert into tbvalues ('kevin ')

Insert into tbvalues ('kevin Liu ')

Insert into tbvalues ('Liu ')

 

Select *, 'eng' from tbwherepatindex ('% [A-Z] %', name)> 0and (patindex ('% [A-Sit] %', name) = 0)

Union all

Select *, 'cn' from tbwherepatindex ('% [sitting] %', name)> 0 andpatindex ('% [A-Z] %', name) = 0

Union all

Select *, 'eng & cn' from tbwhere (patindex ('% [sitting] %', name)> 0) andpatindex ('% [A-Z] % ', name)> 0

 

Result set:

Name

--------------------------

Kevin Eng

Liu CN

Kevin Liu ENG & CN

 

(3 row (s) affected)