# Making a name like "Copy of...", "Copy 2 of..."

#1

Here’s a little ruby exercise - i’m a bit brain damaged today and can’t
think of a nice way to do this.

I want a method where you pass a name as a string and it returns ‘copy
of x’ or ‘copy 2 of x’ eg

new_name(“My Lesson”)
=> “Copy of My Lesson”

new_name(“Copy of My Lesson”)
=> “Copy 2 of My Lesson”

new_name(“Copy 10 of My Lesson”)
=> “Copy 11 of My Lesson”

I can only come up with horrible looking solutions but i feel there’s a
nice way to do it. Anyone?

cheers
max

#2

Max W. wrote:

=> “Copy 2 of My Lesson”

new_name(“Copy 10 of My Lesson”)
=> “Copy 11 of My Lesson”

I can only come up with horrible looking solutions but i feel there’s a
nice way to do it. Anyone?

cheers
max

Use a closure to keep track of how many copies you’ve made:

def copy(string)
x = 0
p = lambda do
x += 1
if x == 1
“Copy of #{string}”
else
“Copy #{x} of #{string}”
end
end
p
end

c = copy(“My Lesson”)

puts c.call
puts c.call
puts c.call

#3

On Thu, May 28, 2009 at 8:04 AM, Max W.

s = “My Lesson”
=> “My Lesson”

copied_name(_)
=> “Copy of My Lesson”

copied_name(_)
=> “Copy 1 of My Lesson”

copied_name(_)
=> “Copy 2 of My Lesson”

Well, taking that approach how about:

def copied_name(source_name)
case source_name
when /^Copy of (.)\$/i
“Copy 1 of #{\$1}”
when /^Copy (\d+) of (.
)\$/i
num = \$1.to_i
“Copy #{num+1} of #{\$2}”
else
“Copy of #{source_name}”
end
end

s = “My Lesson” # => “My Lesson”
s = copied_name(s) # => “Copy of My Lesson”
s = copied_name(s) # => “Copy 1 of My Lesson”
s = copied_name(s) # => “Copy 2 of My Lesson”

However, since you used the i option on your regex maybe you wanted to
preserve the case of the “Copy of part” which this doesn’t

s = “copy of My Lesson” # => “copy of My Lesson”
s = copied_name(s) # => “Copy 1 of My Lesson”
s = copied_name(s) # => “Copy 2 of My Lesson”
s = copied_name(s) # => “Copy 3 of My Lesson”

If so then this does:

def copied_name(source_name) # !> method redefined; discarding old
copied_name
case source_name
when /^(Copy )(of .)\$/i
“#{\$1}1 #{\$2}”
when /^(Copy )(\d+)( of .
)\$/i
num = \$2.to_i
“#{\$1}#{num+1}#{\$3}”
else
“Copy of #{source_name}”
end
end

s = “My Lesson” # => “My Lesson”
s = copied_name(s) # => “Copy of My Lesson”
s = copied_name(s) # => “Copy 1 of My Lesson”
s = copied_name(s) # => “Copy 2 of My Lesson”

s = “copy of My Lesson” # => “copy of My Lesson”
s = copied_name(s) # => “copy 1 of My Lesson”
s = copied_name(s) # => “copy 2 of My Lesson”
s = copied_name(s) # => “copy 3 of My Lesson”

Rick DeNatale

#4

def copied_name(source_name)
case source_name
when /^Copy of (.)\$/i
“Copy 1 of #{\$1}”
when /^Copy (\d+) of (.
)\$/i
num = \$1.to_i
“Copy #{num+1} of #{\$2}”
else
“Copy of #{source_name}”
end
end

I’ll go with this one, thanks Rick!

max

#5

Tim H. wrote:

Max W. wrote:

=> “Copy 2 of My Lesson”

new_name(“Copy 10 of My Lesson”)
=> “Copy 11 of My Lesson”

I can only come up with horrible looking solutions but i feel there’s a
nice way to do it. Anyone?

cheers
max

Use a closure to keep track of how many copies you’ve made:

def copy(string)
x = 0
p = lambda do
x += 1
if x == 1
“Copy of #{string}”
else
“Copy #{x} of #{string}”
end
end
p
end

c = copy(“My Lesson”)

puts c.call
puts c.call
puts c.call

That method returns a proc though, rather than a string. Wouldn’t that
mean that every time i want to call it i have to assign it to a variable
and then call that variable? Or am i being dumb? here’s my
hacky-feeling solution:

def copied_name(source_name)
if source_name.match /^Copy of/i
new_name = source_name.gsub!(/^Copy of/i, “Copy 1 of”)
elsif source_name.match /^Copy \d+ of/i
num = source_name.scan(/\d+/).first.to_i
new_name = source_name.gsub!(/^Copy #{num} of/i, “Copy #{num+1}
of”)
else
new_name = "Copy of " + source_name
end
end

s = “My Lesson”
=> “My Lesson”

copied_name(_)
=> “Copy of My Lesson”

copied_name(_)
=> “Copy 1 of My Lesson”

copied_name(_)
=> “Copy 2 of My Lesson”

#6

A bit more compact:

def copied_name(source_name)
if source_name =~ /\ACopy (\d+ )?of (.*)\z/i
“Copy #{(\$1||1).to_i+1} of #{\$2}”
else
“Copy of #{source_name}”
end
end

You can simplify (\$1||1) to just \$1, if you don’t mind starting at Copy
1 of… instead of Copy 2 of…