Ruby regex not validating in condition?

Question

I wrote my first regex and an if condition to check a passed def argument loop. It doesn't seem to be working properly, and I am guessing I have something wrong with the regex. The condition runs, but it says the value doesn't match the regex when it does match the pattern I want, and vice-versa.

The pattern I need is like this: 123456_12345_ABC_123:

6 digits 
underscore
5 digits (may include 0 padding)
underscore
3 upper case letters
underscore
3 digits (may include 0 padding)

Here is my code:

if ev_val.match(/[0-9]{6}(_)[0-9]{5}(_)[a-z]{3}(_)[0-9]{3}/)

    puts "#{ev_val} matches required batch naming convention"
else
    puts "#{ev_val} doesn't match required batch naming convention\nSTOPPING SCRIPT!!!"
    exit
end

If anyone has any thoughts, I would greatly appreciate it!

the Tin Man · Accepted Answer · 2013-10-24 22:05:36Z

2

Instead of:

/[0-9]{6}(_)[0-9]{5}(_)[a-z]{3}(_)[0-9]{3}/

You can use:

/\d{6}_\d{5}_[A-Z]{3}_\d{3}/

\d is a digit, AKA [0-9], so you should get used to using that short-hand.
(_) doesn't accomplish anything in your code, so don't capture the underscore, instead simply use _.
[a-z] captures all lower-case letters. You want upper-case so use [A-Z].

answered Oct 24, 2013 at 22:05

the Tin Man

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4 Comments

pguardiario Over a year ago

Good but you should add 0*'s where he says may include 0 padding (unless I'm reading that wrong.)

Courtney Fay Over a year ago

Thank you for both of your responses! I had the \d first, but I had read it was exactly 1 digit, so that I thought that was why it wasn't working. I had also read parentheses looks for a literal character, which is why I put the underscore in them. Looks like I have much to learn for regex! I've been avoiding it, but I know that learning it will give me a great advantage in coding. Yes, any of the digits are 0 padded (all the way up to the first digit). I will try to look that up. but if someone knows the shorthand, I am much obliged.

Courtney Fay Over a year ago

Never mind. I figured it out! /0*\d{6}_0*\d{5}_[A-Z]{3}_0*\d{3}/ Thanks!

Courtney Fay Over a year ago

Just as a side note. This does find the string, even if there are characters on the end. It passed every test except 123456_12345_ABC_123zz That shouldn't pass, because of the zz, but it does. This is in a csv with double quotes around it. By the time the script is evaluating it, the double quotes are stripped, but I don't know why it allows the extra characters on the end.

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