Callback after every `each` iteration when `next` is called - ruby

Question

I want to run some code after every iteration of each. Is there a way to do this without repeating the code? I tried this:
(1..10).each do |n|
continue = Proc.new {
puts "ended #{n}"
next
}
continue.call if n == 2
puts n
continue.call
end
but it didn't work.
In my actual code, I have lots of next calls. That's why it's unpractical to call a method every time I call next.

Answer 1

Approach 1
Define the contents of the loop in a method that you call from the loop. You can put an ensure block in the method. That way, your method can use return anywhere you want to move onto the next iteration, but you still guarantee you execute the ensure code:
def doit(x)
return if x == 2
puts "I like the number #{x}"
ensure
puts "LOOP"
end
[1,2,3,4].each{|x| doit(x)}
results in
I like the number 1
LOOP
LOOP
I like the number 3
LOOP
I like the number 4
LOOP
Approach 2
Similar to approach 1, but allows you reuse the "callback" code for different concerns. It also keeps you using next instead of return. This is to define a method that yields and then does other stuff:
def ensure_LOOP(x)
yield
puts "LOOP"
end
[1,2,3,4].each do |x|
ensure_LOOP(x) do
next if x == 2
puts "I really like the number #{x}"
end
end
Results in
I really like the number 1
LOOP
LOOP
I really like the number 3
LOOP
I really like the number 4
LOOP
And
[1,2,3,4].each do |x|
ensure_LOOP(x) do
next unless x == 2
puts "I don't like the number #{x}"
end
end
results in
LOOP
I don't like the number 2
LOOP
LOOP
LOOP

Answer 2

As I understand the question, you don't want anything executed after continue.call if n==2. If that's correct, you could use the control expression next with an argument.
def m(n)
puts "Only #{n} more days!"
end
(1..6).each do |n|
next m(n) if n==3
puts n
m(n)
end
1
Only 1 more days!
2
Only 2 more days!
Only 3 more days!
4
Only 4 more days!
5
Only 5 more days!
6
Only 6 more days!

Answer 3

Just call it within the each loop?
(1..10).each do |n|
puts n
puts "ended #{n}"
end

Answer 4

The code you provided actually does run, and outputs the following:
1
ended 1
ended 2
2
ended 2
3
ended 3
4
ended 4
5
ended 5
6
ended 6
7
ended 7
8
ended 8
9
ended 9
10
ended 10
As you can see, the Proc gets called twice for the number 2, as your if condition passes in that case and calls the Proc.
Stepping back, defining a method outside the context of the iteration is probably a better choice. It will make the code easier to read and avoid redefining the Proc each time the each block executes.
That said, technically what you have here seems to run okay.