CODEWITHSUNDEEP
rubyregex
Gavrilov Pavel
Gavrilov Pavel

Reputation: 71

Get last digits using regex

I have a string: 

10 101/12/201209/12/2012 8 75.00 600.00 2 1RPT/136185RAMADA HOTEL & SUITES

I want to get from this value from the string:

136185

I tried to use /([^\/]*)$/, but it returned:

136185RAMADA HOTEL

What should I do?

Upvotes: 0

Views: 491

Answers (5)

benstoltz
benstoltz

Reputation: 111

The inverse of using scan is to use split:

    s = '10 101/12/201209/12/2012 8 75.00 600.00 2 1RPT/136185RAMADA HOTEL & SUITES'
    s.split(/\D+/).last # Everything that isn't a number is a separator, take the last one
    => "136185"

If the rule is first number after the last '/', then this works even if the business has digits in its name:

    s = '10 101/12/201209/12/2012 8 75.00 600.00 2 1RPT/136185MOTEL 6'
    s.split(?/).last.to_i # => 136185

Note that "7 Seas Hotel" is going to still cause problems

Upvotes: 1

Darren Coxall
Darren Coxall

Reputation: 1208

I don't know if it's the shortest solution but (\d+)\D*\Z should work.

Upvotes: 1

the Tin Man
the Tin Man

Reputation: 160549

If the number is ALWAYS six digits, use:

'10 101/12/201209/12/2012 8 75.00 600.00 2 1RPT/136185RAMADA HOTEL & SUITES'.split[6].split('/').last[0, 6]
=> "136185"

The potential problem with any regex solution is the number + hotel name could cause solutions using \d+ to return bad values if the hotel's name is something like 7 Seas.

Upvotes: 0

sawa
sawa

Reputation: 168101

str = "10 101/12/201209/12/2012 8 75.00 600.00 2 1RPT/136185RAMADA HOTEL & SUITES"

str[/\d+(?=\D*\z)/]
# => "136185"

Upvotes: 1

akuhn
akuhn

Reputation: 27793

Try

string[/.*\b(\d+)/,1]

this matches as many characters as possible before matching a group of digits, starting with a word boundary to make sure the greedy .* doesn't eat up the begin of the digit group.

Or use

string.scan(/\d+/).last

which is more readable, but not a pure regex.

Upvotes: 0

Related Questions