Ruby中的正则表达式匹配字符串的第一和第二个空格

Question

I´m trying to get a regex to match first and second space and ":" from second column of strings like below in order to replace them with "|". I built the regex below but matches the opposite, since matches any word, not first and second space, nor ":". Maybe someone could give a hand with this.

(\S+\s*) (\S+\s*) (\S+\s*)  # My current regex

Strings are like these.

Usw 12:12 Desktop
Usw 1:2 Netbooks
Usw 1:345 Servers, mainframes and supercomputers

I´d like to convert these strings from above to this

Usw|12|12|Desktop
Usw|1|2|Netbooks
Usw|1|345|Servers, mainframes and supercomputers

Answer 1

We can use gsub here:

input = String.new("Usw 1:345 Servers, mainframes and supercomputers")
puts input.gsub(/(\S+)\s*(\d+):(\d+)\s*(.*)/, '\1|\2|\3|\4')

Usw|1|345|Servers, mainframes and supercomputers

Demo

The basic idea here is that we capture the non space content before the first space, after the first space, and the remainder of the string after the second space. Then we build the output string you want using those capture groups.

Answer 2

I believe the solution might be even simpler:

'Usw 12:12 Desktop'.split(/\s|:/, 4).join('|')
#⇒ "Usw|12|12|Desktop"

The above will fail if the first column has colons, but I am pretty sure this is not the case.

Answer 3

You can use String#sub to replace the characters step by step:

str.sub(/\\s/,'|').sub(/:/,'|').sub(/\\s/,'|')

Tests

test = [ "Usw 12:12 Desktop",
         "Usw 1:2 Netbooks",
         "Usw 1:345 Servers, mainframes and supercomputers" ]

test.map { |str| str.sub(/\s/,'|').sub(/:/,'|').sub(/\s/,'|') }                                                                      
 #=> [ "Usw|12|12|Desktop",
 #     "Usw|1|2|Netbooks",
 #     "Usw|1|345|Servers, mainframes and supercomputers" ]