spss conditional based on macro text

Question

I'm trying to build a simple variable in spss like below but can't figure it out from documentation. In sas, you simply need to put qoutes around your macro to resolve. Can't figure out how in SPSS.

define !reg (name= !tokens(1)).

compute !concat(!name,'_f') = 0.

if type =("!name")  compute !concat(!name,'_f') = 1.

execute.

!enddefine. 

!reg name=joe.

eli-k · Accepted Answer · 2025-03-26 04:25:02Z

0

OK so you want the value of !name to be in quotes in the resulting syntax, but putting the quotes there in type =("!name") disrupts the macro. So the solution is

!quote(!name)

Now I'm not sure what you are doing with the macro here, but I suggest a couple of corrections to the full line of code - no need for parentheses around the quote, no need for compute after the if:

if type = !quote(!name) !concat(!name,'_f') = 1.

answered Mar 26 at 4:25

eli-k

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2 Comments

Mark Hedman Apr 9 at 14:13

I'm trying to build a simply binary using a macro call. I'd like to first create a variable called joe_f = 0. Then using a macro do something like this :

Mark Hedman Apr 9 at 14:22

Thanks, the !quote addition achieved what I wanted, the new syntax is : define !reg (name= !tokens(1)). compute !concat(!name,'_f') = 0. if (name=!quote(!name)) !concat(!name,'_f') = 1. execute. !enddefine. !reg name=joe.

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