Hi I am using the following query:

"SELECT * FROM members WHERE name LIKE '$searchKey%';"

Where $searchKey is a variable that contains the string to be searched. The problem is that the code only searches until it has found a space and even if characters after space match the value, it returns null.

Database has two entries:
Andy Johnson
Paul Ford

1. $searchKey = "An" - The code works fine and returns 'Andy Johnson'
2. $searchKey = "For" - Returns Null; however I want it to return "Paul Ford"


put a % symbol BEFORE and AFTER the variable name:

"SELECT * FROM members WHERE name LIKE '%$searchKey%';"

It works!

Thanks a lot!

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