CODEWITHSUNDEEP
regexstringgroovy
Anonymous Human
Anonymous Human

Reputation: 1948

Groovy extract substring before character

I have a bunch of strings like

asdf v1.0
jkl v3.04

all my strings are a group of characters followed by a space then 'v' then a number. I want to extract from each string, the part that comes before the 'v' and the space character preceeding the 'v' so that:

asdf v1.0 becomes 'asdf'
jkl v3.04 becomes 'jkl'

just need some help with this. I tried takeWhile { it != 'v' } but that ends up including the space before the 'v' in the result string which I don't want.

Upvotes: 23

Views: 60874

Answers (4)

Joxebus
Joxebus

Reputation: 205

Here is an example using regular expressions

String regex = /(v\d)(\d|\.)*/

def clean = { text ->
   text?.replaceAll(regex,'')?.trim()
}

assert 'v1.11' ==~ regex

assert clean(null) == null
assert clean('asdf v1.0') == 'asdf'
assert clean('jkl v3.04') == 'jkl'
assert clean('John Kramer v1.1') == 'John Kramer'
assert clean('Kramer v Kramer v9.51') == 'Kramer v Kramer'
assert clean('A very hungry caterpillar v2.6') == 'A very hungry caterpillar'

Upvotes: 2

Abdennour TOUMI
Abdennour TOUMI

Reputation: 93173

input[0..(input.lastIndexOf('v') - 1)].trim()

details

String str = 'jkl v3.04'

// index of last "v" char is 4
assert str[4..-1] == 'v3.04'
assert str[0..(4 - 1 )].trim() == 'jkl'

You do not have to be surprised: i am a very former groovy programmer

Upvotes: 2

M A
M A

Reputation: 72844

You can simply extract the substring:

input.substring(0, input.lastIndexOf(" v"))

To get the part after the v:

input.substring(input.lastIndexOf(" v") + 2)

Upvotes: 49

Roger Glover
Roger Glover

Reputation: 3256

I think, given the criteria as stated, that the following would give correct results in places where the chosen solution would not:

String stringParser(String inputString) {
    inputString ? inputString.split(/ v\d/)[0] : ''
}

Some sample tests:

assert stringParser('John Kramer v1.1') == 'John Kramer'
assert stringParser('Kramer v Kramer v9.51') == 'Kramer v Kramer'
assert stringParser('A very hungry caterpillar v2.6') == 'A very hungry caterpillar'

Upvotes: 5

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