问题
I have written the function below to test if a cell has conditional formatting activated based upon the cell fill.
Function cfTest(inputCell)
If inputCell.DisplayFormat.Interior.Color <> 16777215 Then
cfTest = True
Else
cfTest = False
End If
End Function
It does not work however. Saying that, this method does.
Sub myCFtest()
Dim R As Integer
R = 2
Do
If Range("I" & R).DisplayFormat.Interior.Color <> 16777215 Then
Range("K" & R).Value = True
Else
Range("K" & R).Value = False
End If
R = R + 1
Loop Until R = 20
End Sub
Can anyone explain to me why the function will not work?
Cheers.
EDIT: Updated function but not working for conditional formatting
Function cfTest(inputCell)
If inputCell.Interior.ColorIndex <> -4142 Then
cfTest = True
Else
cfTest = False
End If
End Function
回答1:
Here is a working demo if the desired result. Column E looks at column D and displays the value TRUE if it is conditionally formatted by cell fill color. i.e. click on the name 'Bob', and conditionally formatting highlights the cell via the code below
=IF(AND(CELL("row")=ROW(D1),CELL("col")=COLUMN(D1)),TRUE)

Click on another name, and the same result occurs.

However, when I click off the names onto another cell, I last name selected remains highlighted, giving the impression of a button still depressed.

The VBA code behind is as follows.
This sits within the Sheet1 code:
Private Sub Worksheet_SelectionChange(ByVal Target As Range)
If Target.Column = 4 And Target.Row <= Application.WorksheetFunction.CountA(Range("D:D")) Then
Range("D:D").Calculate
Call cfTest
End If
End Sub
And this is the method itself:
Sub cfTest()
Range("E:E").ClearContents
If ActiveCell.DisplayFormat.Interior.color <> 16777215 Then
ActiveCell.Offset(0, 1) = True
End If
End Sub
The application I ended up building off this example had much more too it, but going back to the posted question, the cfTest() method allowed me to test if a cell was conditionally formatted based upon cell fill.
回答2:
Here are two related functions that implement mathematical conditions. This is slightly less complicated than the Chip Pearson version, and also less complete, but I think this should cover most cases, and this shouldn't be too difficult to extend.
Function isConditionallyFormatted(rng As Range) As Boolean
Dim f As FormatCondition
On Error Resume Next
isConditionallyFormatted = False
For Each f In rng.FormatConditions
isConditionallyFormatted = checkFormula(rng.Value, f.operator, f.Formula1)
isConditionallyFormatted = checkFormula(rng.Value, f.operator, f.Formula2)
Next
End Function
Function checkFormula(rng As Variant, operator As Variant, condition As Variant)
On Error GoTo errHandler:
Dim formula As String
condition = Right(condition, Len(condition) - 1)
Select Case operator
Case xlEqual: formula = rng & "=" & condition
Case xlGreater: formula = rng & ">" & condition
Case xlGreaterEqual: formula = rng & ">=" & condition
Case xlLess: formula = rng & "<" & condition
Case xlLessEqual: formula = rng & "<=" & condition
Case xlExpression: formula = condition
End Select
checkFormula = Evaluate(formula)
Exit Function
errHandler:
Debug.Print Err.Number & " : " & Err.Description
End Function
This will work for some common operators, but there are two other operators (xlBetween and xlNotBetween) and there are other types of condition that would have to be caught as well, and the logic for some of those would be a little more complicated than this. Some of them, however (like databars), inherently convey that there is a condition, so no processing would be necessary.
Here is a link to the full documentation:
http://msdn.microsoft.com/en-us/ff835850(v=office.15)
回答3:
I'm not sure as to the why of this but maybe it'll help. VB doesn't seem to allow access to a cells color when that color is based on conditional formatting.
For example..
'cell A1 colored yellow through conditional formatting
MsgBox Range("A1").Interior.ColorIndex
'returns the incorrect result of -4142 regardless of cell color
'cell B1 colored yellow via the fill option on the ribbon
MsgBox Range("B1").Interior.ColorIndex
'returns the correct result of 6
That being said, is there a reason you couldn't just test the cell for whatever formatting rules you have in effect. That would eliminate the need for a UDF.
=IF(A1<50,False,True)
回答4:
I would perform a prior check for the color index your condition is for using this:
Function cfTest_color_chk(inputCell As Range)
cfTest_color_chk = inputCell.Interior.ColorIndex
End Function
Then your function
Function cfTest(inputCell As Range)
If inputCell.Interior.ColorIndex <> -4142 Then
cfTest = True
Else
cfTest = False
End If
End Function
Another solution to make things rock solid is to combine both function so that cfTest takes cfTest_color_chk as a parameter and cfTest_color_chk will return the value of the color to match...
Hope this helps
Pascal
来源:https://stackoverflow.com/questions/22366145/vba-function-to-test-if-cell-is-conditionally-formatted-in-excel