Based on the code above,list the number,name,and complete address of every customer located on a street that contains the letters "Oxford".
A) SELECT CustomerNum, CustomerName, Street, City, State, PostalCode FROM Customer WHERE Street LIKE "?Oxford" ;
B) SELECT CustomerNum, CustomerName, Street, City, State, PostalCode FROM Customer WHERE Street LIKE "%Oxford%" ;
C) SELECT CustomerNum, CustomerName, Street, City, State, PostalCode FROM Customer WHERE Street LIKE "@Oxford@" ;
D) SELECT CustomerNum, CustomerName, Street, City, State, PostalCode FROM Customer WHERE Street LIKE "Oxford" ;

Question

Based on the code above,list the number,name,and complete address of every customer located on a street that contains the letters "Oxford".​
A) ​SELECT CustomerNum, CustomerName, Street, City, State, PostalCode FROM Customer WHERE Street LIKE "?Oxford" ;
B) ​SELECT CustomerNum, CustomerName, Street, City, State, PostalCode FROM Customer WHERE Street LIKE "%Oxford%" ;
C) ​SELECT CustomerNum, CustomerName, Street, City, State, PostalCode FROM Customer WHERE Street LIKE "@Oxford@" ;
D) ​SELECT CustomerNum, CustomerName, Street, City, State, PostalCode FROM Customer WHERE Street LIKE "Oxford" ;

Quizplus · Accepted Answer

The correct syntax to search for a specific string within a column is to use the LIKE operator with the wildcard symbol "%", which represents any number of characters. Option B uses the correct syntax, searching for any street containing the letters "Oxford" anywhere in the string. Option A is incorrect because it uses a single wildcard symbol "?", which can only represent a single character. Option C is incorrect because it uses "@" symbols, which are not recognized as wildcard symbols in SQL. Option D is incorrect because it only searches for streets with the exact string "Oxford", rather than any streets containing those letters.

Based on the Code Above,list the Number,name,and Complete Address of Every