CODEWITHSUNDEEP
regexruby
John Smith
John Smith

Reputation: 1876

How to extract the year via regex from a string in Ruby

I'm trying to extract the year from a string with this format:

dataset_name = 'ALTVALLEDAOSTA000020191001.json'

I tried:

dataset_name[/<\b(19|20)\d{2}\b>/, 1]
/\b(19|20)\d{2}\b/.match(dataset_name)

I'm still reading the docs but so far I'm not able to achieve the result I want. I'm really bad at regex.

Upvotes: 1

Views: 242

Answers (2)

the Tin Man
the Tin Man

Reputation: 160551

There are many ways to get to Rome.

Starting with:

foo = 'ALTVALLEDAOSTA000020191001.json'

Stripping the extended filename + extension to its basename then using a regex:

ymd = /(\d{4})(\d{2})(\d{2})$/
ext = File.extname(foo)
File.basename(foo, ext) # => "ALTVALLEDAOSTA000020191001"

File.basename(foo, ext)[ymd, 1] # => "2019"
File.basename(foo, ext)[ymd, 2] # => "10"
File.basename(foo, ext)[ymd, 3] # => "01"

Using a regex against the entire filename to grab just the year:

ymd = /^.*(\d{4})/
foo[ymd, 1] # => "1001"

or extracting the year, month and day:

ymd = /^.*(\d{4})(\d{2})(\d{2})/
foo[ymd, 1] # => "2019"
foo[ymd, 2] # => "10"
foo[ymd, 3] # => "01"

Using String's unpack:

ymd = '@18A4'
foo.unpack(ymd)  # => ["2019"]

or:

ymd = '@18A4A2A2'
foo.unpack(ymd)  # => ["2019", "10", "01"]

If the strings are consistent length and format, then I'd work with unpack, because, if I remember right, it is the fastest, followed by String slicing, with anchored, then unanchored regular expressions trailing.

Upvotes: 1

ggorlen
ggorlen

Reputation: 56865

Since your dataset name always ends in yyyymmdd.json, you can take a slice of the last 13-9 characters counting from the rear:

irb(main):001:0> dataset_name = 'ALTVALLEDAOSTA000020191001.json'
irb(main):002:0> dataset_name[-13...-9]
=> "2019"

You can also use a regex if you want a bit more precision:

irb(main):003:0> dataset_name =~ /(\d{4})\d{4}\.json$/
=> 18
irb(main):004:0> $1
=> "2019"

Upvotes: 1

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