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Formatting strings, removing leading characters

I have a line like this: 00:11:40 or 00:02:40 how I form so that I can always get rid of the leading zero and colon (s), so it looks like 11:40 or 2:40

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ruby
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5 answers

We call these “leading” characters not final, as they are at the beginning, but the regex is very simple

x.sub(/^[0:]*/,"")

This works the way you formulated it: from the beginning of the line, delete all 0 and .s.

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You can use something like Peter, but correctly:

 s = "00:11:40" s = s[3..-1] # 11:40

Another approach is to use the split method:

 s = "00:11:40".split(":")[1,2].join(":")

Although I find it even more confusing and complicated.

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EDIT: OP wanted this from the start:

 seconds = 11*60+40 Time.at(seconds.to_i).gmtime.strftime('%M:%S') # gives '11:40'

or see man strftime for more formatting options.

EDIT: The inclusion of all discussions is recommended here. It also eliminates the need for a time challenge.

 seconds = seconds.to_i if seconds >= 60 "#{seconds/60}:#{seconds%60}" else "#{seconds}" end
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You might want to try the positive regex expression. Nice link

 it "should look-behind for zeros" do time = remove_behind_zeroes("ta:da:na") time.should be_nil time = remove_behind_zeroes("22:43:20") time.should == "22:43:20" time = remove_behind_zeroes("00:12:30") time.should == "12:30" time = remove_behind_zeroes("00:11:40") time.should == "11:40" time = remove_behind_zeroes("00:02:40") time.should == "2:40" time = remove_behind_zeroes("00:00:26") time.should == "26"

end

 def remove_behind_zeroes(value) exp = /(?<=00:00:)\d\d/ match = exp.match(value) if match then return match[0] end exp = /(?<=00:0)\d:\d\d/ match = exp.match(value) if match then return match[0] end exp = /(?<=00:)\d\d:\d\d/ match = exp.match(value) if match then return match[0] end exp = /\d\d:\d\d:\d\d/ match = exp.match(value) if match then return match[0] end nil

end

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Many times, you can simply rely on basic conversion methods, for example, in ruby, if you had a string like “05” and you only needed 5, you would just do “05” .to_i

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