I'm opening a csv file and I'm trying to check for quotation marks within the string so that I can remove them.
For example in the following "Trial "A" 123" I want to get it so that it's "Trial A 123"
I thought I could analyze each individual character (and if a character was a quotation mark then I was going to remove it.) But that's no good because VB identifies the 2nd quotation mark as the end of the string. In other words, all I get when I input the string is "Trial "
Ah yes of course.... I did the line input, got rid of those quotation characters and now I'm looking at my new csv file with the correct characters/words. And the incorrect csv file is still intact. Nice. Thanks DataMiser!