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IF(OR Formula is being Other(Wise... Please Help

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AlexiosKit
4 - Data Explorer
4 - Data Explorer

I am trying to incorporate a formula column which looks for entries which may include swear words, but I cant seem to get this formula to work. Here is a snippet (I've taken out a lot of the middle which is just 3230 different swear words) but the beginning and end are there. Any ideas?

 

IF((OR(SEARCH(2 girls 1 cup”, ARRAYJOIN({Lower})) > 0, SEARCH(2g1c”, ARRAYJOIN({Lower})) > 0, SEARCH(4r5e”, ARRAYJOIN({Lower})) > 0, SEARCH(5h1t”, ARRAYJOIN({Lower})) > 0, SEARCH(5hit”, ARRAYJOIN({Lower})) > 0, SEARCH(“a fanabla”, ARRAYJOIN({Lower})) > 0, SEARCH(“a pecorina”, ARRAYJOIN({Lower})) > 0, SEARCH(“a tomar por culo”, ARRAYJOIN({Lower})) > 0, SEARCH(“a tomar por saco”, ARRAYJOIN({Lower})) > 0, SEARCH(“a$$”, ARRAYJOIN({Lower})) > 0, SEARCH(“a$$hole”, ARRAYJOIN({Lower})) > 0, SEARCH(“a_s_s”, ARRAYJOIN({Lower})) > 0, SEARCH(“a2m”, ARRAYJOIN({Lower})) > 0, SEARCH(“a54”, ARRAYJOIN({Lower})) > 0, SEARCH(“a55”, ARRAYJOIN({Lower})) > 0, SEARCH(“zut”, ARRAYJOIN({Lower})) > 0, SEARCH(“zut alors”, ARRAYJOIN({Lower})) > 0,"!","")
 
In total it is 146,000 characters.
1 Solution

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Sho
11 - Venus
11 - Venus

I don't know if it could actually handle that number of words, but this regex formula seems better.

 

IF(REGEX_MATCH({Lower}&"", "2 girls 1 cup|2g1c|4r5e"),"!")

 

 

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1 Reply 1
Sho
11 - Venus
11 - Venus

I don't know if it could actually handle that number of words, but this regex formula seems better.

 

IF(REGEX_MATCH({Lower}&"", "2 girls 1 cup|2g1c|4r5e"),"!")