Conditionally apply blue font color to table rows using VBA

Dear Experts:

The following macro (courtesy of the MS Word 'Sage' Graham Skan from EE) applies a blue font color on all table rows where the text entry in the 8th column of the tables is 'blue'

I got a a new requirement:

It is always the last column where this text entry is found. Hence instead of hard coding the 8th column, Line 6 and Line 7 should be tweaked to accomodate the new requirement.

Sub BlueRow()
    Dim tbl As Table
    Dim rw As row
    For Each tbl In ActiveDocument.Tables
        For Each rw In tbl.Rows
            If Len(rw.Cells(8).Range.Text) = 6 Then
                If InStr(1, rw.Cells(8).Range.Text, "blue", vbTextCompare) Then
                    rw.Range.Font.Color = wdColorBlue
                End If
            End If
        Next rw
    Next tbl
End Sub

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Help is very much appreciated. Thank you very much in advance.

Regards, Andreas
Andreas HermleTeam leaderAsked:
Who is Participating?
 
Ryan ChongCommented:
try:

rw.Cells.Count

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hence:

Sub BlueRow()
    Dim tbl As Table
    Dim rw As Row
    For Each tbl In ActiveDocument.Tables
        For Each rw In tbl.Rows
            If Len(rw.Cells(rw.Cells.Count).Range.Text) = 6 Then
                If InStr(1, rw.Cells(rw.Cells.Count).Range.Text, "blue", vbTextCompare) Then
                    rw.Range.Font.Color = wdColorBlue
                End If
            End If
        Next rw
    Next tbl
End Sub

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0
 
Ryan ChongCommented:
not too sure about the condition:

If Len(rw.Cells(8).Range.Text) = 6 Then

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which you may simply remove it.

to become:

Sub BlueRow()
    Dim tbl As Table
    Dim rw As Row
    For Each tbl In ActiveDocument.Tables
        For Each rw In tbl.Rows
            If InStr(1, rw.Cells(rw.Cells.Count).Range.Text, "blue", vbTextCompare) Then
                rw.Range.Font.Color = wdColorBlue
            End If
        Next rw
    Next tbl
End Sub

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0
 
Andreas HermleTeam leaderAuthor Commented:
Great Ryan, this did the trick, thank you very much for your swift help.

Regards, Andreas
0
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