More Macro Help - If/else Statements?

[MikeJG]

Is it possible to have an if/else statement in a macro?

 

What I would like to do is have a door label macro that will give me a different result based on whether it is a double door or not.  (There are also a bunch of other applications for this that I'm thinking about).

 

For instance, the way we label doors around here a 3-0 swing door would be "D30".  However, as a double door it would be "2-D16".

 

So the statement would be something like:

 

If is_double_door = true then "2-D" width / 2

Else "D" width

 

If it can be done, what would the syntax look like?

 

Any help would be appreciated.

[Joe_Carrick]

Is it possible to have an if/else statement in a macro?

 

What I would like to do is have a door label macro that will give me a different result based on whether it is a double door or not.  (There are also a bunch of other applications for this that I'm thinking about).

 

For instance, the way we label doors around here a 3-0 swing door would be "D30".  However, as a double door it would be "2-D16".

 

So the statement would be something like:

 

If is_double_door = true then "2-D" width / 2

Else "D" width

 

If it can be done, what would the syntax look like?

 

Any help would be appreciated.

case

  If is_double_door == "true"

    result = "2-D"+(width/2).to_s

  else

    result = "D"+width.to_s

end

result

[MikeJG]

Joe,

 

Thanks again for your help!

 

I'm getting an evaluation error.  See below:

 

[Joe_Carrick]

Mike,try this:

 

case

  when is_double_door == "true"

    result = "2-D"+(width/2).round.to_s

  else

    result = "D"+width.round.to_s

end

result

 

"if" doesn't work within a "case".  The proper syntax is "when"

 

Of course for a 36" single door the above results in D36, not D30.  Likewise a pair of doors totalling 36" would result in 2-D18 not 2-D16.  To get the exact result you indicate would require using divmod(12) as per a previous exercise.

 

like this:

 

case

  when is_double_door == "true"

    w = (width/2).round.divmod(12)

    result = "2-D"+"#{w[0]}#{w[1].round}"

  else

    w = width.round.divmod(12)

    result = "D"+"#{w[0]}#{w[1].round}"

end

result

[MikeJG]

Joe,

 

The result was coming out "D30" regardless.of whether it was a double door or not.  I was determined to try to figure it out for myself, so I did some googling.  It turns out I had to remove the quotes from around "true" on the 2nd line so it reads now:

 

case

when is_double_door == true

w = (width/2).round.divmod(12)

result = "2-D"+"#{w[0]}#{w[1].round}"

else

w = width.round.divmod(12)

result = "D"+"#{w[0]}#{w[1].round}"

end

result

 

Now it works!

 

Thanks again for taking the time to help!