Question

If the nested if-statement doesn’t fulfill a condition, how do I continue on to the outer if? Eg. I have this (very impractical) example:

a = 2

if( a > 1 ):
    if( a == 3 ):
        print "yes"
elif( a == 2 ):
    print "yes"

I want a == 2 to be checked next. How would I do this?

(I have multiple conditions in that nested if that I need to check so I’d rather not have a huge string of and/or statements in that one outer-if. I also have more than one elif statement so I don’t want to mash all the elifs together under that nested one.)

A deeper example:

b = 8

if( a > 1 ):
    if( b == 3 ):
        doSomething()
    elif( b == 4 ):
        doSomethingElse()
    elif( b == 5 ):
        more()
elif( -1 <= a <= 1 ):
    asd()
elif( a < -1 ):
    if( b == 7 ):
        asdfasdf()
    elif( b == 8 ):
        asasdf()

Answer 1

How do you expect this to be clear to a computer?

if a > 1:
  # ANYTHING
elif a == 2:
  # ANYTHING

by definition will never execute anything in the second block.

I believe your intend might be more along this lines:

def action():
  if a > 1:
    if b == 2:
      doSomething()
      return
    if b == 3:
      doSomethingElse()
      return
 if a == 2: # Note: NO elif!
   doDefaultForA2()
   return
 showError("Unknown command")
 return

There are multiple ways to achieve this behaviour. I like the function+return pattern, because it structures code nicely. Another way is to use a handled flag:

unhandled = True
if a > 1:
  if b == 2:
    doSomething()
    unhandled = False
  if unhandled and b == 3:
    doSomethingElse()
    unhandled = False
if unhandled and a == 2: # Note: NO elif!
  doDefaultForA2()
  unhandled = False
if unhandled:
  showError("Unknown command")

(You can obviously put in a number of shortcuts here by using elif. But using explicit “if unhandled” makes the code more verbose about the logic and easier to add new options.)