How to break out of case statement in Ruby

Question

I have a method as follows:

def something(param)
  case param
  when 1
    break if <already executed for some similar data>
    #Some code
  when 2
    #Some code
  else
    #Some code
  end
end

If param is 1, I will create a record in DB. In such case, I need to check that the same data are not present already, and if so, I need to break out of when.

When I try break if <condition>, I get this error:

Can't escape from eval with break

As a solution, I can change param's value before the case statement so that it (when 1) never meets the condition:

def something(param)
  param = 0 if <condition>
  case param
  when 1
    .
    .

but I feel it's ugly.

break is used to break out of loops, but I need an equivalent of it here that works like switch statement` break.

a = 1
b = 1

case a
when 1
  return if b == 1
  p "Code Executed"
end

gives me:

LocalJumpError: unexpected return

Possible duplicate: http://stackoverflow.com/questions/8022865/how-to-break-out-of-a-case-while-in-ruby — Lucas Nelson, Mar 17 '16 at 05:42
@LucasNelson Hey, Lucas. Thank you for pointing this post. But can you please see my edit. — Abhi, Mar 17 '16 at 05:46

score 2 · Accepted Answer · answered Mar 17 '16 at 05:45

2

If that is all you have in the method body, then just do return instead of break.

If you cannot do that, then do:

when 1
  unless <already executed for some similar data>
    #Some code
  end
when 2
...

answered Mar 17 '16 at 05:45

sawa

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@Abhi To use `return` you'll have to be inside a method body. In your edit, it's not inside a method, that's why it's throwing `LocalJumpError`. – Babar Al-Amin Mar 17 '16 at 05:53

How to break out of case statement in Ruby

1 Answers1