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1,875 PointsSecond attempt.
Now I feel like I'm way off. Here's what I got:
SELECT REPLACE(email, '%@%', "<a>") FROM customers WHERE REPLACE(email, '%@%', "<a>") = "<a>";
1 Answer
KRIS NIKOLAISEN
54,971 PointsSome hints with the following syntax for replace REPLACE(string,pattern,replacement)
:
- pattern doesn't need wildcards
- replacement is <at> not <a>
- there is no WHERE condition needed
- you are missing the
obfuscated_email
alias