Contains function
I am having trouble with a formula that contains multiple "If(contains)" function. The following returns unparsable.
=IF(CONTAINS("52", [Account #]@row), "52", IF(CONTAINS("53", [Account #]@row), "53", IF(CONTAINS(“54”, [Account #]@row)”54”, "")))
Best Answers

Could you share a screenshot of the sheet?
Amber Eakin, MSLS, M.Ed.
Adult Education Specialist  Process Improvement Enthusiast

This is so strange. How about this? It looks the same to me, but my sheet is accepting it.
=IF(CONTAINS("52", [Account #]@row), "52", IF(CONTAINS("53", [Account #]@row), "53", IF(CONTAINS("54", [Account #]@row), "54")))
Amber Eakin, MSLS, M.Ed.
Adult Education Specialist  Process Improvement Enthusiast
Answers

Hello, @M Scrog  It looks like a syntax error. Does this work?
=IF(CONTAINS("52", [Account #]@row), "52", IF(CONTAINS("53", [Account #]@row), "53", IF(CONTAINS(“54”, [Account #]@row), "54")))
Amber Eakin, MSLS, M.Ed.
Adult Education Specialist  Process Improvement Enthusiast

Thank you @Amber Eakin  I am still receiving an "unparseable" error.

Could you share a screenshot of the sheet?
Amber Eakin, MSLS, M.Ed.
Adult Education Specialist  Process Improvement Enthusiast

This is so strange. How about this? It looks the same to me, but my sheet is accepting it.
=IF(CONTAINS("52", [Account #]@row), "52", IF(CONTAINS("53", [Account #]@row), "53", IF(CONTAINS("54", [Account #]@row), "54")))
Amber Eakin, MSLS, M.Ed.
Adult Education Specialist  Process Improvement Enthusiast

@Amber Eakin Thank you! That did it. I think there were some spacing issues based on my comparison.

Hi @M Scrog, if you just want the first two digits, you could do this instead:
=LEFT([Account #]@row, 2)
If you want the Series to be a number (right now your computer thinks it's a word) wrap my function in a value function like this:
=VALUE(LEFT[Account #]@row, 2))
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