如何让启用线程的窗口为前端活动窗口
CDlg1窗口的按钮事件中开启线程
handle = _beginthreadex(0,0,ThreadFun,(LPVOID)this, 0, 0);
线程处理函数:
unsigned __stdcall ThreadFun(void * parameter)
{
CDlg1 * pDlg1 = (CDlg1 *)parameter;
pDlg1->ShowWindow(FALSE);
CDlg2 dlg;
INT_PTR ret = dlg.DoModal();
if (IDOK == ret)
{
pDlg1->EndDialog(0);
}
else if(IDCANCEL == ret)
{
pDlg1->EndDialog(1);
}
else //这里会出现问题:就是pDlg1这个窗口会跑到其它窗口的后面,不再是活动窗口,有什么办法解决
{
pDlg1->SetFocus();
pDlg1->ShowWindow(TRUE);
}
}