Dutch formula not working in english Excel version - excel-formula

Need help with a dutch formula not working in english Excel.
For some reason the formula below doesn't work with the english version of Excel, can someone help me with this? I'm left with this error #NAME?
=SPATIES.WISSEN(WISSEN.CONTROL((SUBSTITUEREN(A1;TEKEN(160);" "))))
Does not work still results in #NAME?:
=TRIM(CLEAN((SUBSTITUTE(A1;CHAR(160);" "))))

You have got semi colons not commas in the formula
=TRIM(CLEAN((SUBSTITUTE(A1,CHAR(160), " "))))
works fine for me.
From #Scott Craner - it appears that local settings may stop ; being used as a separator for variables in the method call, and , should work.

Related

How to deal with double quotes in conditional formatting's formula

I was having a look at this question, and in order to help, I wrote a conditional formatting rule, based on this formula:
="ABS(B3-A3)>=1,5" // the double-quotes are automatically added by Excel
This, however, seems not to work, although the formula seems to be correct (in the cell at the right, I asked whether this was true and I got a positive result, as you can see (WAAR is Dutch for true)):
I'm not interested in the final solution, I just want to know how I can debug such a situation: apparantly the formula is true, but still the conditional formatting is not applied. This might mean that:
Something is wrong with the language settings.
Something is wrong with the dot/comma decimal separator.
Something else is wrong with the formula (absolute references, relative references, ...)
How can I, step by step, evaluate the formula, entered in a conditional formatting configuration, in order to see what might be wrong?
Thanks in advance
Select the whole range (for the example A2:B5)
Add in conditional formatting formula =ABS($B3-$A3)>=1.5
Result:
Remove the "" at the beginning and end of your formula. Should do the trick.
Sorry, guys, but the truth is really nerve-wrecking:
The answer of Error 1004 is correct. Thanks for the screenshot.
The answer of Jario is correct too.
So what's the deal?
Apparently there seems to be a problem with the decimal separator : when I enter a dot, using Excel regularly, it gets converted into a comma. When I enter a dot in a conditional formatting formula, this conversion is not done, resulting in a wrong formula, where Excel automatically adds double quotes. As the double quotes result in a wrong formula, the conditional formatting won't be applied.
So:
From the moment you see double quotes in a conditional formula, this is an indication that something's wrong, and that your conditional formatting won't work!

I have a median ifs formula but I am getting a error.Any ideas?

I realized that a median ifs formula does not exist but I was able to find a work around on the internet.
=MEDIAN(IF('Source Data'!O2:O230="M",IF('Source Data'!N2:N230="3",'Source Data'!J2:J230)))
I keep getting a #NUM! error but have no idea why. The Columns have the same formats.
Any help would be appreciated!
Try
{=MEDIAN(IF('Source Data'!O2:O230="M",IF('Source Data'!N2:N230=3,'Source Data'!J2:J230)))}
Note:
replace "3" with 3 in your formula;
confirm the formula by press Ctrl+Shift+Enter as it is an array formula.

Adding an iferror correctly

Normally I can add an iferror to my formulas fairly easily. But this formula I am using I seem to be struggling:
Formula:
=SUM((X15-V15)+W15)/Q15
My Iferror:
=IFERROR(SUM((X15-V15)+W15)/Q15),"-")
what am I doing wrong? I have even tried to add extra brackets.
thanks
Sorry, should have posted as an answer...
It looks as if you are closing a bracket after Q15 that you shouldn't: try =IFERROR(SUM((X15-V15)+W15)/Q15,"-")

Need Formula Excel how to delete a specific character in long URL

somebody help me...
In Excel, how the formula to remove all chars after "?ref" In other words only need like this http://www.domain.com/avo/smith-hendry-487895 and so on.
Bellow the sample data:
http://www.domain.com/avo/smith-hendry-482225?ref=cr-sr%7Cnormal%7C23%7C15
http://www.domain.com/avo/merry-77789?ref=cr-sr%7Cnormal%7C15%7C24
http://www.domain.com/avo/margaret-anna-675000?ref=cr-sr%7Cnormal%7C19%7C11
please in Formula, not macro or vbs, thank you so much
You would normally do something like this:
=substring(<source-cell>, 1 , search(<source-cell>,"?ref"))
Not 100% sure about syntax, but the idea should be clear.
EDIT:
Here is a working solution:
=MID(F3,1,FIND("?ref",F3)-1)
It assumes that the source URL is in cell F3.
Try below one
=LEFT(B3, FIND("?ref",B3,1)-1)
Here my B3 cell value is
http://www.domain.com/avo/smith-hendry-482225?ref=cr-sr%7Cnormal%7C23%7C15
and the above formula will give
http://www.domain.com/avo/smith-hendry-482225
Edited Formula to convert the result to string
=HYPERLINK(LEFT(B3, FIND("?",B3,1)-1),LEFT(B3, FIND("?",B3,1)-1))

VBA FormulaR1C1 RC expression is not working when used with division

I am new to vba. I try to put a formula into a range which works well, but the formula is troubling me. I feel the issue might be because of the division.. maybe the slash is a kind of escaping character?
Here the code:
MyWorksheet.Range("J7","J12").FormulaR1C1 = "=R[0]C[-1]/R0C1 *100"
and it gives me this in excel:
=I7/R0C1 *100
So R0C1 is not replaced :-(. How comes? and how can I fix this?
I have goggled a lot but I did not find anything similar.. so hopefully I am not the only one with this issue.
Any help is highly appreciated!
Best regards,
Tobias
Your problem is the zero in your formula - when using absolute reference (i.e. omitting brackets) in R1C1-notation, you count from an imaginary cell one up and one left of A1 - in other words, cell A1 is R1C1 in that notation, B2 is R2C2, etc.
And R0C1 does not exist, thus why it is interpreted as a string in your formula.

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