quotation help in VBA

[wildwally]

ok -i've just about pulled the last hair out.

need help with this.

cell contains:

1/2" Thick

I want my code to evaluate to make sure this is whats in the cell.

if rng = "1/2" wall" then do such

I've tried the CHAR(34) route but getting errors that way also and no matter how many """""" i throw at it, its not working.

 

[MrKowz]

Try:

if rng = "1/2"" wall"

 

[Norie]

What are you trying to check for and what is it you arec checking, eg range, single cell?

Are you using code?

 

[Jonmo1]

Try

if rng = "1/2"" wall" then do such

 

[wildwally]

Wow - I've tried nearly every combination to make this work and the one that didn't make the most sense worked. Thats code for you.


Thanks that worked.