Forum Discussion

Iron Contributor

Feb 18, 2019

Solved

Debug Error on Macro

Does any know why I get Debugged in the following macro. I copied the syntax directly from a video tutorial and have double checked that it appears correct. The the first debug (yellow highlight) ap...

Macros and VBA

Man Fai Chan
Feb 21, 2019
ROUNDDOWN is a built-in function. The use of it is
=ROUNDDOWN(number, num_of_digit)

The macro you wrote is a procedure to repeatedly insert function/formula into the cell in column U. So, it is not related to whether macro recognize ROUNDDOWN

You may try to change the num_of_digit parameter to see if it works. I thought ROUNDDOWN(A1,0) will gives a value round down to the nearest integer.

Wyn Hopkins

MVP

Feb 19, 2019

It's due to the use of "" in your formula which the VBA code sees as the end of the string.

Easiest fix is to replace "" with 0

Greg Bonaparte

Iron Contributor

Feb 19, 2019

Thanks for your reply Wyn.

Just to clarify, Are you saying every line should look like this:

Range("$AZ30").Formula = "=IF(($S30 > 0)*($R30 > 0), $U30=$S30,"")0

with the last line looking like this:

Range("$AZ99").Formula = "=IF(($S99 > 0)*($R99 > 0), $U99=$S99,"")"

Greg Bonaparte
Iron Contributor
Feb 19, 2019
When I attempted to do this
Range("$AZ30").Formula = "=IF(($S30 > 0)*($R30 > 0), $U30=$S30,"")0
every line self populated with this
Range("$AZ99").Formula = "=IF(($S99 > 0)*($R99 > 0), $U99=$S99,"")0"

adding the quote sign automatically.
- Wyn Hopkins
  MVP
  Feb 19, 2019
  Hi Greg Bonaparte
  
  No I mean this sorry
  
  Range("$AZ30").Formula = "=IF(($S30 > 0)*($R30 > 0), $U30=$S30,0)"
  - Greg Bonaparte
    Iron Contributor
    Feb 19, 2019
    Thank you Wyn, The debug error went away. However instead of getting the value in u30 to equal the value is s30, I get a "false" message in az30.