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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The contest is open till next blog post shows up at which is next day GTM+2.5.

Reference:  Pinal Dave (https://blog.sqlauthority.com)

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334 Comments. Leave new

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

    Reply
  • hi
    i have created sqllite database(question) and i save this my desktop. now i want to access(question) using html5. pls help me i am new in html5 its urgent.

    thanks in advance

    Reply
  • The Correct answer is :

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

    Reply
  • The Correct answer is : 5

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

    Country is Canada

    Reply
  • The correct choice is 5. I am very late to read this article.

    Reply
  • SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’ , united states of america

    Reply
  • Vinit Prajapati
    August 16, 2011 2:47 pm

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

    Option is : 5

    I am from India

    Reply
  • Nepolean Arumugam
    August 16, 2011 8:38 pm

    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?

    Ans:

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

    Reply
  • Nepolean Arumugam
    August 16, 2011 8:47 pm

    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?

    The correct answer is

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

    Reply
  • Correct Answer : 5

    Reply
  • Bryant Griffith
    August 18, 2011 7:57 am

    My Answer: (option 5)
    5.SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’

    Explanation
    *Without the “where” clause you will get all the records in the Customer table
    *The “where” is used as a filter
    *In option one, the syntax is incorrect; the operator “<=" can be used for <= 'm'
    but not accompanied by "%"; you need "Like" for a wildcard.
    * Option 2 has an incorrect syntax as well; you need "like" when using a wildcard and you need the brackets for a range
    *Option 3 will give you an empty set, because there is no names that starts with 'a-m'
    *Option 4 is an incorrect syntax; though the bracket is used, you can't use the "=" operator with a wildcard.
    *Option 5 is the only one that answers the question

    Country of Resident: USA

    If I am selected for a free book; I don't have volumes 4 & 5. Thanks.

    Reply
  • Answer is #5

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

    Thanks,
    Wayne

    Reply
  • answer is #3
    select * from customer
    where firstname like ‘a-m%’

    xplanation:- if we want to search specific pattern matching in the database then we use the like operator ……..here we have to find the first name that start with a-m so.. we have to place a-m before the % sign it will return a first name column that start with’ A-M’

    i’m atul singh
    Country of Resident: india(mumbai)

    Reply
  • Correct Answer is 5 –
    SELECT * FROM Customer WHERE Firstname like ‘[a-m]%’

    Explanation: This command will select all the Record from Customer table whose firstname start between a to m latters.

    Reply
  • Rajesh Varma Kalidindi
    August 20, 2011 4:36 pm

    The correct answer is 5 – SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’

    Explanation :
    Because strings in SQL SERVER are compared using LIKE operator and [A-M] denotes all the alphabets between A and M .

    Regards,
    Rajesh Varma ,
    India

    Reply
  • Good post. The correct answer is #5.

    I’m from India

    Reply
  • Anjana Chandrabose
    August 22, 2011 10:23 am

    Answer: Option 5
    SELECT * FROM Customer
    WHERE Firstname like ‘[a-m]%’

    Anjana ChandraBose(India)

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

    Reply
  • Answer : Option 5

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

    Reply

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