我正在努力使用VBA宏,它应该为文本的一部分着色。
宏看起来像
Sub Note()
Dim c As Range
Dim val As String
Set c = ActiveCell
val = InputBox("Add note", "Note text")
If IsEmpty(c.Value) = True Then
c.Value = Format(Now(), "DD MMM YY Hh:Nn") & ": " & val
Else
c.Value = c.Value & Chr(10) & Format(Now(), "DD MMM YY Hh:Nn") & ": " & val
End If
End Sub
我希望实现Now()将为红色,其余文本将为绿色。
我试着玩.Font.Color = vbRed等但没有运气
我也看this answer,但这不是我想要的
答案 0 :(得分:3)
试试这样:
Option Explicit
Sub Note()
Dim c As Range
Dim val As String: val = "vit"
Dim lngLen As Long
Set c = ActiveCell
c.Value = Format(Now(), "DD MMM YY Hh:Nn") & ": " & val
lngLen = Len(Format(Now(), "DD MMM YY Hh:Nn"))
c.Characters(Start:=1, Length:=lngLen).Font.Color = vbRed
End Sub
我删除了输入框,但您可以轻松退回。它可能提供你想要的东西。很多,它要求Now()格式的长度,并按照你在问题中提到的问题的逻辑,将公式中的前N个符号用红色着色。
答案 1 :(得分:2)
你链接了一个答案,但你没有使用那里的内容,为什么?
试试这个:
Sub Note()
Dim c As Range
Dim val As String
Dim StartChar As Integer, _
LenColor As Integer
Set c = ActiveCell
val = InputBox("Add note", "Note text")
With c
.Font.Color = RGB(0, 0, 0)
If IsEmpty(.Value) = True Then
StartChar = 1
LenColor = Len("DD MMM YY Hh:Nn")
.Value = Format(Now(), "DD MMM YY Hh:Nn") & ": " & val
.Characters(Start:=StartChar, Length:=LenColor).Font.Color = RGB(255, 0, 0)
Else
StartChar = Len(.Value) + 1
LenColor = Len("DD MMM YY Hh:Nn")
.Value = .Value & Chr(10) & Format(Now(), "DD MMM YY Hh:Nn") & ": " & val
.Characters(Start:=StartChar, Length:=LenColor).Font.Color = RGB(255, 0, 0)
End If
End With 'c
End Sub
答案 2 :(得分:1)
试试这个:
Sub Note()
Dim c As Range
Dim val As String
Dim lngPos As Integer
Set c = ActiveCell
val = InputBox("Add note", "Note text")
c.Value = ""
If IsEmpty(c.Value) = True Then
c.Value = Format(Now(), "DD MMM YY Hh:Nn") & " - " & val
lngPos = InStr(ActiveCell.Value, " - ")
With ActiveCell.Font
.ColorIndex = 4
End With
With ActiveCell.Characters(Start:=1, Length:=lngPos - 1).Font
.ColorIndex = 3 'or .Color = RGB(255, 0, 0)
End With
Else
c.Value = c.Value & Chr(10) & Format(Now(), "DD MMM YY Hh:Nn") & " - " & val
lngPos = InStr(ActiveCell.Value, " - ")
With ActiveCell.Font
.ColorIndex = 4
End With
With ActiveCell.Characters(Start:=1, Length:=lngPos - 1).Font
.ColorIndex = 3 'or .Color = RGB(255, 0, 0)
End With
End If
End Sub