You want to display details of all customers who reside in cities starting with the letter D followed by at least two characters. Which query can be used?
there is no valid answer here
correct would be (followed by at least two characters):
SELECT * FROM EMPLOYEES WHERE FIRST_NAME LIKE 'D__%';
but A is the most correct
I assume there's been some issue with copy-pasting the text, and the first ' as been replaced with "~
I think it's supposed to be A anyway, but you're certainly correct that they're all wrong written as they are.
Many comments here saying that 2 underscores must be there but I think single underscore is enough because 'D_%' means that after underscore % means that there must be a character or more to display the result. Hence, A option is accurate and correct.
A is the most close to be valid answer.
LIKE must be used when comparing text strings (B, D are incorrect), metacharacter % must be used to accept strings of variable length (C is incorrect), metacharacter _ stands for any character.
Underscore _ should be doubled to fulfill the request of two additional characters that follows initial letter D, so the query should end with LIKE ‘D__%’
the table name CUSTOMERS table, from where you came with EMPLOYEES ?
ex: WHERE CustomerName LIKE 'a_%' --> Finds any values that start with "a" and are at least 2 characters in length
https://www.w3schools.com/sql/sql_like.asp
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