Ruby String processing

Question

I am trying to learn ruby. What is an elegant way to convert '"abcd" "efg"' to a ['abcd', 'efg'] in Ruby?

Thanks.

tokland · Accepted Answer · 2012-06-10 08:44:27Z

3

Try this:

require 'shellwords'
'"abcd" "efg"'.shellsplit
#=> ["abcd", "efg"]

answered Jun 10, 2012 at 8:38

tokland

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1 Comment

Thanks! What can I use in Ruby 1.8.7?

kiddorails · Accepted Answer · 2012-06-10 16:54:15Z

1

You can also do it by removing " character and then splitting by space

'"abcd" "efg"'.tr('"','').split
'"abcd" "efg"'.delete('"').split

answered Jun 10, 2012 at 16:54

kiddorails

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Nigel Thorne · Accepted Answer · 2012-06-11 23:15:50Z

0

You can use scan which finds all occurrences of a pattern within a string.

'"aaa" "bbb"'.scan(/"([^"]*)"/)
=> [["aaa"], ["bbb"]]

Explained:

/ something/ is a regular expression (which can match strings)
/" something "/ is a regular expression that matches strings starting and ending in "
[^"] matches against any character that isn't a "
[^"]* matches as many of these characters as it can.
The ( and ) tell it want the bits I want in the results.

answered Jun 11, 2012 at 23:15

Nigel Thorne

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Note: This doesn't deal with escaped " characters.