我想启动另外一个程序:int intReValue;
char szAppPath[]="C:\Program Files\SEL5040";
char szAppName[]="SelReport.exe";
char szOperation[]="Open";intReValue=ShellExecute(@@,szTemp,szAppName,NULL,szAppName,SW_MAXIMIZE);//@@处应该怎么写??
//
char szAppPath[]="C:\Program Files\SEL5040";
char szAppName[]="SelReport.exe";
char szOperation[]="Open";intReValue=ShellExecute(@@,szTemp,szAppName,NULL,szAppName,SW_MAXIMIZE);//@@处应该怎么写??
//
[in] Handle to a parent window. This window receives any message boxes that an application produces, such as error reporting.
没有父窗口,置为NULL也行。
如果只单纯的打开一个exe文件
用WinExec就行了
[in] Handle to a parent window. This window receives any message boxes that an application produces, such as error reporting.
NULL试试~
或
ShellExecute(NULL,"open","notepad.exe",
"c:\\MyLog.log","",SW_SHOW );
As you can see, I haven't passed the full path of the programs.