SQL SERVER – Tips from the SQL Joes 2 Pros Development Series – Wildcard Basics Recap – Day 1 of 35

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Wildcard ranges

If you have ever been to a convention where they have a morning registration desk that must handle thousands of people in a short time you know they must put some pre-planning thought into how to handle this burst of volume. In fact often they will have many registration desks running in parallel to make things run faster. The first registration desk might handle all customer last names starting from A to K. Desk 2 will handle names from L to Q and the third desk will handle from R to Z.

With my last name being Morelan I would naturally head to registration desk #2 knowing that desk has the list with my name on it and the other desks don’t. Now let’s say you are in charge of creating these three separate lists and sending them out to the right registration workers. You know how to sort but how can you separate this lists using wildcards?

Wildcard Basics Recap

Lets start off with something most of us know already. Most SQL folks understand the usefulness and power of the basic uses of wildcards. Using wildcards allows you to do pattern matches in a column. In this case our criteria does not want to use the = sign to find a pattern match. The operator that allows you to do approximate predicates is LIKE. The LIKE operator allows you to do special relative searches to filter your result set.

--Find all LastNames that start with the letter A
SELECT *
FROM Employee
WHERE LastName LIKE 'A%'

To find everyone whose last name starts with the letter B, you need “B” to be the first letter. After the letter B you can have any number of characters. Using B% in single quotes after the LIKE operator gets all last names starting with the letter B.

--Find all LastNames that start with the letter B
SELECT *
FROM Employee
WHERE LastName LIKE 'B%'

Wildcard ranges or set specifiers

If you want to find all LastName values starting with the letters A or B you can use two predicates in your WHERE clause. You need to separate them with the OR operator.

--Find all LastNames that start with the letter B
SELECT *
FROM Employee
WHERE LastName LIKE 'A%'
OR LastName LIKE 'B%'

Finding names beginning with A or B is easy. How about the registration desk example where want the names ranging from A-K? This works well until you want a range of A-K as in the example below:

--Find all LastNames ranging from A-K
SELECT *
FROM Employee
WHERE LastName LIKE 'A%'
OR LastName LIKE 'B%'
OR LastName LIKE 'C%'
OR LastName LIKE 'D%'
OR LastName LIKE 'E%'
OR LastName LIKE 'F%'
OR LastName LIKE 'G%'
OR LastName LIKE 'H%'
OR LastName LIKE 'I%'
OR LastName LIKE 'J%'
OR LastName LIKE 'K%'

The previous query does find LastName values starting from A-K. However, if you need a range of letters, the LIKE operator has many better options. We only really care about the first letter of the last name and there a several first letters that fit with what were looking for. The first letter of the last name can be A,B,C,D,E,F,G,H,I,J or K. Simply list all the choices you want for the first letter inside a set of square brackets.

--LastNames ranging from A to K using a set of 11 letters
SELECT *
FROM Employee
WHERE LastName LIKE '[ABCDEFGHIJK]%'

Square brackets with wildcards enclose ranges or sets for 1 position. In this case the first position is a set of 11 different possible letters. This is not a series of letter but a multiple choice of letters. For example this works regardless of the order you put your letters in. This code sample below does the exact same thing.

--LastNames ranging from A to K using a set of 11 letters
SELECT *
FROM Employee
WHERE LastName LIKE '[KBCDEFGHIJA]%'

Again the set is how many letters you put in the square brackets. The code below is a logical mistake where you won’t get A to K but you just get A or K for the first letter.

--Find all LastNames starting with A or K (Mistake
SELECT *
FROM Employee
WHERE LastName LIKE '[AK]%'

Since we’re looking for the first letter to be within a range  from A to K, we specify that range in square brackets. This is even easier than using a set. The wildcard after the brackets allows any number of characters after the range.

--LastNames ranging from A to K using a range
SELECT *
FROM Employee
WHERE LastName LIKE '[A-K]%'

Note: this range will not work if your LIKE was changed to an equal (=) sign. The following code will not return any records to your result set:

--Bad query (it won’t error but returns no records)
SELECT *
FROM Employee
WHERE LastName = '[A-K]%'

Note: If you want to setup the sample JProCo database on your system you can watch this video.

Question 1

Q. 1) You want to find all first names that start with the letters A-M in your Customer table. Which SQL code would you use?

  1. SELECT * FROM Customer
    WHERE Firstname <= ‘m%’
  2. SELECT * FROM Customer
    WHERE Firstname  = ‘a-m%’
  3. SELECT * FROM Customer
    WHERE Firstname like ‘a-m%’
  4. SELECT * FROM Customer
    WHERE Firstname  = ‘[a-m]% ‘
  5. SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’

Please post your answer in comment section to win Joes 2 Pros books.

Rules:

Please leave your answer in comment section below with correct option, explanation and your country of resident.
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A valid answer must contain country of residence of answerer.
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Reference:  Pinal Dave (https://blog.sqlauthority.com)

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334 Comments. Leave new

  • Rama Krishna, Hyderabad , India

    Answer 5 is correct if we want to get the data that starts with ‘A’ to ‘M’

    If we want the data only letters with ‘A’ and ‘M’

    SELECT * FROM Customer WHERE Firstname like ‘[a&M]%’

    Reply
  • Jagdish Prajapati
    August 2, 2011 3:01 pm

    good post
    Ans is
    5. SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’

    Country: INDIA

    Reply
  • the correct answer is “answer 5″
    SELECT * FROM Customer WHERE Firstname like ‘[a-m]%’

    iT will return all record of range “a to m”.
    i am from india

    Reply
  • Answer) option 5 is the correct answer

    SELECT * FROM Customer WHERE Firstname like ‘[a-m]%’

    Country : India

    Reply
  • Option: 5
    SELECT * FROM Customer
    WHERE FirstName like ‘[a-m]%’

    Sudeepta, India

    Reply
  • Hi Pinal,

    Option 5 is correct answer,the reason is that as option 1,2 and 4th doesnt obey pattern matching and remaining is 3rd and 5th, but 5th optin has included within the braces so it works finee

    Reply
  • Logan Destefani Merazzi
    August 2, 2011 3:31 pm

    Number 5 is the right choice.
    I’m from Brazil. :-)

    Best regards.

    Reply
  • Option 5 is correct.
    SELECT * FROM Customer WHERE FirstName like ‘[a-m]%’

    Kapil Madan
    India

    Reply
  • Dayanand Singh
    August 2, 2011 3:43 pm

    Correct Answer is Option 5

    SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’

    Country – INDIA (Gujarat)

    Reply
  • SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’‘

    Option 5 is the correct answer

    Country of Resident – INDIA

    Reply
  • Correct Answer is Option 5 because it provide the set of all characters to LIKE keyword.

    India

    Reply
  • correct answer : 5

    SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’

    Reply
  • 1. DISCARDED
    2,3,4. = will find exact match instead of pattern match => DISCARDED
    5. In Square Bracket, range is provided from A to M and afterwards % sign is given. Hence using LIKE operator it will result all the records having first letter between a and m.

    Thus, 5 is the correct answer.

    India

    Reply
  • Nandish J Reddy
    August 2, 2011 4:34 pm

    SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’

    Country: India

    Reply
  • Option 4

    SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’

    It is Giving the range ([a-m]in Like operator and expecting that a and A,m And M are same

    Reply
  • kadamannil satisan
    August 2, 2011 4:47 pm

    SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’‘

    Option : 5 is correct.

    Reply
  • Uday Kumar B R
    August 2, 2011 4:48 pm

    Question 2:

    Q 2.) You want to find all grant names that have and Underscore as the second letter. Which SQL code would you use?
    Answer is
    1. SELECT * FROM [Grant]
    WHERE GrantName like ‘_[_]% ‘

    Because ‘_[_]% ‘ represents

    _ —> Represents first character of the word that may be any character

    [_] —> ‘_’ character in ‘[”]‘ square braces represents looking for this
    character the square brace will nullify(ignores) its(special character)
    meaning.

    % —> any number of character after [_]

    Country-India

    Reply
  • Answer is option 5 :

    SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’

    Country : USA

    Reply
  • Prashant Ingole
    August 2, 2011 5:07 pm

    Ans is 5) SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’
    Prashant Ingole
    India(Mumbai)

    Reply
  • correct answer is
    5.SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’

    Reply

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