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Syntax help in a formula

Greg Bonaparte
Contributor

Hello Community

 

Can anyone help me with my syntax below? This gets "too few arguments" error

 

=IF((($R30 > 0.00001), ROUNDDOWN(($Y$17-$Y$18)/$Q30,0,"")))

12 Replies
It should be like this:
=IF($R30>0.00001,
ROUNDDOWN(($Y$17-$Y$18)/$Q30,0),
“”)

Thank you very much. This took away the error. However the new formula continues to place a value in the cell when the r30 cell is empty. Why does it think an empty cell is greater than 0.00001?

Or better yet, is it possible to say "process this function only if r has any value" as opposed to saying "process if r is greater than 0.00001?

Modify the logical_test argument of IF to this:
ISNUMBER($R30)

Thank you, but when I try \

=ISNUMBER($R30, ROUNDDOWN(($Y$17-$Y$18)/$Q30,0),“”)

Im back to "too few arguments" error.

Solution
This is the complete formula:
=IF(ISNUMBER($R30),
ROUNDDOWN(($Y$17-$Y$18)/$Q30,0),
“”)

Thank you, this works. One final question. This leaves a "#NAME?" the cell. Not critical but I prefer either "0" or blank. I tried

 

"=IF(ISNUMBER($R30),
ROUNDDOWN(($Y$17-$Y$18)/$Q30,0),
“0”)

 

without success.

If you prefer to return 0, if the result of the logical_test argument is FALSE, then replace the value_if_false argument with 0, like this:
=IF(ISNUMBER($R30),
ROUNDDOWN(($Y$17-$Y$18)/$Q30,0),
0)
Note that a #NAME? error is returned, if the formula includes a name that is not defined in the Name Manager.

Most probably it returns #NAME? since you copy the formula from that page as it is, on Web page apostrophes at the end of the formula are rounded apostrophes, CHAR(147). You need to change them on standard ones, CHAR(34). Or copy paste from code pane here:

=IF(ISNUMBER($R30),
  ROUNDDOWN(($Y$17-$Y$18)/$Q30,0),
  "")

it doesn't transform apostrophes.

Thank you Twifoo, all is perfect now

Thank you

Welcome!
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