如何从 AX 2009 读取 AD 用户?还有什么安全问题?
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2 回答
2
我刚刚在我的博客上发布了一个旧作业,用于从 X++ 访问 AD 用户,代码如下:
static void JAEE_IterateActiveDirectoryUsers(Args _args)
{
str computer = new xSession().clientComputerName();
xAxaptaUserManager mgr = new xAxaptaUserManager();
xAxaptaUserDetails usr;
container doms;
int d, u;
str dom, login, name, sid, email;
;
// iterate AD domains
doms = mgr.enumerateDomains(computer);
for (d = 1; d <= conlen(doms); d++)
{
dom = conpeek(doms, d);
setprefix(dom);
// iterate AD domain users
usr = mgr.enumerateDomainUsers(dom);
for (u = 0; u < usr.getUserCount(); u++)
{
if (usr.isUserEnabled(u) && !usr.isUserExternal(u))
{
// get information from AD
login = usr.getUserLogin(u);
name = usr.getUserName(u);
sid = usr.getUserSid(u);
email = usr.getUserMail(u);
// stuff happens here, you can compare AD data with AX User info
info(strfmt("%1 - %2 - %3 - %4 - %5", dom, login, name, email, sid));
}
}
}
}
于 2013-04-02T09:27:37.273 回答
0
看看下面,你会从那里得到很多关于 AX 2009 的 AD 和安全结构的帮助 http://technet.microsoft.com/en-us/library/aa497043%28v=ax.50%29。 aspx
于 2013-03-30T10:50:13.950 回答