ClipboardFormat 值 44 和 50 的含义

Question

Excel 的Application对象有一个ClipboardFormats属性。从文档：

以数值数组的形式返回剪贴板上当前的格式。

但是，当我运行以下代码时：

var app = new ActiveXObject('Excel.Application');
app.Visible = true;
var results = new VBArray(app.ClipboardFormats).toArray();
app.Quit();
window.alert(results.join(','));

我回来了：

0, 44, 50

0对应于XlClipboardFormat.xlClipboardFormatText，但其他值在XlClipboardFormat枚举中没有匹配的枚举成员。

什么44和50意味着作为剪贴板格式的值？

Answer 1

不知道这对走得更远有多大帮助，但是以这个来源的例子为例，我想出了以下测试代码：

Option Explicit

Private Declare Function OpenClipboard Lib "user32" (ByVal hwnd As Long) As Long
Private Declare Function CloseClipboard Lib "user32" () As Long
Private Declare Function GetClipboardData Lib "user32" (ByVal wFormat As Long) As Long
Private Declare Function EmptyClipboard Lib "user32" () As Long
Private Declare Function EnumClipboardFormats Lib "user32" _
                         (ByVal wFormat As Long) As Long
Private Declare Function GetClipboardFormatName Lib "user32" _
                         Alias "GetClipboardFormatNameA" (ByVal wFormat As Long, _
                                                          ByVal lpString As String, _
                                                          ByVal nMaxCount As Long) As Long

Private Sub test()
    Dim results As Variant
    Dim fmtName As String
    Dim fmt As Long

    Range("A1").Copy
    results = Application.ClipboardFormats
    Debug.Print "For cell A1 (plain) = " & Join(results, ",")
    ClipboardFormats

    Range("A2").Copy
    results = Application.ClipboardFormats
    Debug.Print "For cell A2 (bold ) = " & Join(results, ",")
    ClipboardFormats

End Sub

Private Sub ClipboardFormats()
    Dim fmt As Long
    Dim fmtName As String
    Dim iClipBoardFormatNumber As Long

    OpenClipboard 0&
    If iClipBoardFormatNumber = 0 Then
        fmt = EnumClipboardFormats(0)
        Do While fmt <> 0
            fmtName = Space(255)
            GetClipboardFormatName fmt, fmtName, 255
            fmtName = Trim(fmtName)
            If fmtName <> vbNullString Then
                fmtName = Left(fmtName, Len(fmtName) - 1)
                Debug.Print "fmtName (" & fmt & ") = " & fmtName
            End If
            fmt = EnumClipboardFormats(fmt)
        Loop
    End If

    EmptyClipboard
    CloseClipboard
End Sub

输出很有趣，但不一定有启发性：

For cell A1 (plain) = 0,2,4,5,6,7,8,9,11,12,14,17,19,22,23,31,32,33,44,45,50,58,63
fmtName (49161) = DataObject
fmtName (50023) = Biff12
fmtName (50004) = Biff8
fmtName (50006) = Biff5
fmtName (49910) = XML Spreadsheet
fmtName (49349) = HTML Format
fmtName (49566) = CSV
fmtName (49273) = Rich Text Format
fmtName (49163) = Embed Source
fmtName (49156) = Native
fmtName (49155) = OwnerLink
fmtName (49166) = Object Descriptor
fmtName (49165) = Link Source
fmtName (49167) = Link Source Descriptor
fmtName (50003) = Link
fmtName (49154) = ObjectLink
fmtName (49171) = Ole Private Data
For cell A2 (bold ) = 0,2,4,5,6,7,8,9,11,12,14,17,19,22,23,31,32,33,44,45,50,58,63
fmtName (49161) = DataObject
fmtName (50023) = Biff12
fmtName (50004) = Biff8
fmtName (50006) = Biff5
fmtName (49910) = XML Spreadsheet
fmtName (49349) = HTML Format
fmtName (49566) = CSV
fmtName (49273) = Rich Text Format
fmtName (49163) = Embed Source
fmtName (49156) = Native
fmtName (49155) = OwnerLink
fmtName (49166) = Object Descriptor
fmtName (49165) = Link Source
fmtName (49167) = Link Source Descriptor
fmtName (50003) = Link
fmtName (49154) = ObjectLink
fmtName (49171) = Ole Private Data