Forum Discussion
Calculated Column using SEARCH
I have an Excel table with a column $P of comma-separated labels [@Labels]. The table is linked to an external data source I want to create a column which can be used for filtering, where I am l...
Mike Williams It's the ,0 part at the end of the SEARCH that causes the #VALUE! error (also in the formula with the direct cell reference). The start number should be at least 1. But if you don't want to skip any characters, just leave it out. Change that and it will work, however, it will still give a #VALUE! error when the word "security" is not found (see picture).
To avoid the error you can write the formula as follows:
=IFERROR(IF(SEARCH("security",[@Labels]),"Y"),"")
- Hi. I don't get an error with ,0 in formula with direct cell reference (which is why I wrote it like that in my post), but I did work out that it does cause the error with column labels.
Mike Williams That's odd. I get an error with both methods. But happy you got it to work within a structured table.
