Oh yeah, you're right. First it evaluated true even if $ was 0 .. But a quick "if four=$ | 4=$" solved the variant type problem ;\)

Oh well, in reality I had a very long time this instead

 Code:
if $=4 | $=four & $>0


then .. for a shorter period this:

 Code:
if instr(four4,$)




Edited by Jochen (2011-05-09 04:06 PM)
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