您可以根据想要上传的文件格式在模块/子例程中定义和写入模式。该模块/子例程必须运行/调用创建模式运行上传前.. (我的样品是用VB6)
注: - Filenametxt是要被上传
IntFileHandleLog = FreeFile
Open App.Path & "schema.ini" For Output As #IntFileHandleLog
Print #IntFileHandleLog, "[" & FileNameTxt & "]"
Print #IntFileHandleLog, "Format=FixedLength"
Print #IntFileHandleLog, "MaxScanRows = 0"
Print #IntFileHandleLog, "CharacterSet = OEM"
Print #IntFileHandleLog, "Col1=" & """" & "Name" & """" & "Text Width 20"
Print #IntFileHandleLog, "Col2=" & """" & "PHONE_NUM" & """" & " Text Width 30"
Print #IntFileHandleLog, "Col3=" & """" & "PHONE_TYPE" & """" & " Text Width 20"
Print #IntFileHandleLog, "col4=" & """" & "UserName" & """" & " Text Width 20"
Print #IntFileHandleLog, "col5=" & """" & "A_STAT" & """" & " Text Width 10"
Print #IntFileHandleLog, "col6=" & """" & "B_STAT" & """" & " Text Width 10"
Print #IntFileHandleLog, "col7=" & """" & "RETRY" & """" & "integer width 2"
Print #IntFileHandleLog, "col8=" & """" & "Birth_Date" & """" & " double width 14"
Print #IntFileHandleLog, "Col9=" & """" & "Joint_Date" & """" & " double width 14"
Close #IntFileHandleLog
文件名这会出现上述*想法不起作用,有没有人有更好的建议? – 2012-04-04 15:16:08