以下是引用bccn0906在2014-1-17 10:21:28的发言:
filename=GETFILE('txt')
IF FILE(filename)
DECLARE INTEGER ShellExecute IN shell32.DLL INTEGER HWND,STRING, STRING, STRING, STRING, INTEGER
ShellExecute(0,"open",filename,"","",1)
ENDIF
filename=GETFILE('c:\1.txt')
IF FILE(filename)
DECLARE INTEGER ShellExecute IN shell32.DLL INTEGER HWND,STRING, STRING, STRING, STRING, INTEGER
ShellExecute(0,"open",filename,"","",1)
ENDIF
运行后怎么出来的是windows打开对话框?