如题。
可以直接获取到像素吗?
用picture对象返回的值单位是twip~
------解决方案--------------------------------------------------------
1.
Dim sFile,str
sFile = "a.bmp"
Set oPicture = LoadPicture(sFile)
str = "文件:" & vbTab & sFile & vbCrLf
str = str & "宽度:" & vbTab & Fix(oPicture.Width / 26.458) & vbCrLf
str = str & "高度:" & vbTab & Fix(oPicture.Height / 26.458)
Set oPicture = Nothing
WScript.Echo str
2.
Dim arrFile
Dim oFile,oDir,oShell
arrFile = MyGetFile()
Set oShell = CreateObject("Shell.Application")
Set oDir = oShell.NameSpace(arrFile(1) + "\")
Set oFile = oDir.ParseName(arrFile(0))
WScript.Echo oDir.GetDetailsOf(oFile,-1)
Set oFile = Nothing
Set oDir = Nothing
Set oShell = Nothing
'***********************************************************************************
'获得要操作的文件,返回一个包含文件名和路径的数组
'***********************************************************************************
Function MyGetFile()
On Error Resume Next
Dim strFile,objFso,objFile
If WScript.Arguments.Count < 1 Then
Set objDialog = CreateObject("UserAccounts.CommonDialog")
objDialog.Filter = "bmp 文件|*.bmp|jpg 文件|*.jpg|ico 文件|*.ico|所有 文件|*.*"
objDialog.ShowOpen
strFile = objDialog.FileName
Set objDialog = Nothing
Else
strFile = WScript.Arguments(0)
end if
Set objFso = CreateObject("Scripting.FileSystemObject")
Set objFile = objFso.GetFile(strFile)
If Err Then
If Err.Number = 5 Then WScript.Quit
WScript.Echo Err.Description
Err.Clear
WScript.Quit
Else
MyGetFile = Array(objFile.Name,objFile.ParentFolder)
End If
Set objFile = Nothing
Set objFso = Nothing
End Function
3.
Dim sFile,str
Dim oStream
Dim bWidth,bHeight
sFile = "a.bmp"
Set oStream = CreateObject("Adodb.Stream")
With oStream
.Type = 1
.Open
.LoadFromFile sFile
End With
'如果是BMP格式，则Position=18，读4字节；
'如果是gif格式，则为Position=6，读2字节；
'如果是png格式，则Position=16，读4字节
oStream.Position = 18
bWidth = oStream.Read(4)
bHeight = oStream.Read(4)
oStream.Close
str = "文件:" & vbTab & sFile & vbCrLf
str = str & "宽度:" & vbTab & Bin2Num(bWidth) & vbCrLf
str = str & "高度:" & vbTab & Bin2Num(bHeight)
Set oStream = Nothing
WScript.Echo str
'二进制流转换为数值
Function Bin2Num(binStr)
Dim i,numLen
numLen = Lenb(binStr)
For i = numLen To 1 Step -1
Bin2Num = Bin2Num * 256 + Ascb(Midb(binStr,i,1))
Next
End Function
很久以前写的了。