REGEX_Replace function

[cmohyi]

Hi, 

I've searched various posts but couldn't find an answer that worked. I have a string of data, "1/FundName-Class", and I want to extract everything left of the / using regex_replace. The numbers go from 1-47000 so a LEFT function won't work. Any help is appreciated. 

[Kenda · Accepted solution]

Hey @cmohyi!

When you want to use a symbol in a regex statement, put a \ before it. Below I created two new fields, one that keeps everything before your slash and one that keeps everything after the slash. Remember, this assumes you only have one slash in your Field1.

[MarqueeCrew · Accepted solution]

@cmohyi,

Here's 2 alternative solutions (I do see that @Kenda has answered while I was playing with your post):

REGEX_Replace([Field1], "(.*)\/.*", '$1')

left([Field1],FindString([Field1], "/"))

The first function will take anything left of the "last" / and provide you with that value.

The second function will take the leftmost characters that it finds before the first /.

Cheers,

Mark

[cmohyi]

@barnesk @marqueecrew Thank you both!

[Gamaliel]

Hello @Kenda .What if there are more than 2 slashes found? What expression should be used to delimit them to more than 2 columns. Thank you

[apathetichell]

Hi - this is a fairly old thread. Post some examples of your regex problem in a new thread and I'm sure someone can help you...