Problem:
Column A contains E-mail addresses, from which we would like to pull out the user names of Gmail users only (addresses containing ""@gmail"").
In case the address does belong to a gmail user, a string indicating that conclusion should be returned (instead of a #VALUE! Error).
Solution:
"Use the LEFT and FIND functions to find the requested substring in the address,
and the ISNUMBER and IF functions to determine the value to return in case the substring could not be found.
Following is the formula:
=IF(ISNUMBER(FIND(""@gmail"",A2)),LEFT(A2,FIND(""@gmail"",A2)-1),""Not a Gmail user"")
Example:
Emails_______________Gmail Users
User1@gmail.com______User1
User2@hotmail.com____Not a Gmail user
User3@yahoo.com______Not a Gmail user
User4@gmail.com______User4
Screenshot // Altering results returned in case of an error.
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