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.
