Question

vb 拖放文件获取文件名和内容

例如
C:\99.txt里面有内容888
假定有text1.text和text2.text，将99.txt文件拖放到text1.text，那么结果text1.text为99或者为99.txt，text2.text为888 如何实现？

Answer 1

将text的的OLEDragMode和并简OLEDropMode 属性值设为1

Private Sub Text1_OLEDragDrop(Data As DataObject, Effect As Long, Button As Integer, Shift As Integer, X As Single, Y As Single)
If Data.GetFormat(vbCFFiles) = True Then '判断是否为文件对象
then
filename = Data.Files(1) '获得文件名
Text1.Text = filename
Text2.Text=text1.text
End If
End Sub

我是win7 的 好久没用vb了 ，没装vb，不知判蔽漏道运行的怎样，你试试吧，就这个掘烂思路反正

Answer 2

用list控件伪装成TEXT1比较简单
'加一个List控件，属性
List1.OLEDragMode = 1
List1.OLEDropMode = 1

设全局变量 FileName 或者从list中获取文件名，感觉全局变量要方便些。
dim FileName as string
Private Sub list1_OLEDragDrop(Data As DataObject, Effect As Long, Button As Integer, Shift As Integer, X As Single, Y As Single)
For i = 1 To Data.Files.Count - 1
FileName = Data.Files(i) '获得文件名
List1.AddItem FileName '显示文件
Next
End Sub

text2 设置成多行显示
MultiLine属性设置为TRUE
在list change事件中
Open 路径磨伏\携游祥FileName For Input As #1
Do While Not EOF(1) ‘逐行辩搏读取
Input #1, b
Loop
Text2.Text = b