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复制文件的源路径是否可以写成读取配置文件?
这是范例!
#include <Process.au3>
dim $Serverip
dim $serverroute
$Serverip = iniread(@scriptdir & "\boot.ini","配置","服务器IP","192.168.0.251")
$serverroute = iniread(@scriptdir & "\boot.ini","配置","批处理路径","\\192.168.0.251
\boot$\boot.bat")
while 1
If Ping($Serverip) Then
_RunDOS("call " & $serverroute)
Exit
EndIf
wend
有没办法让下面\\192.168.123.0.251\XPY.GHO像上面的_RunDOS("call " & $serverroute)一样写成读取配置
文件的方法呢?这个\\192.168.123.0.251\XPY.GHO要可改变才行!
$FileCopy = _CopyWithProgress("\\192.168.123.0.251\XPY.GHO", "E:\SYSBAK\XPY.GHO",2048) |
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