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Topic Title: how to get current user?
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Created On: 8-May-2008 08:36
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Answer This question was answered by Frank Jensen, on Thursday, May 8, 2008 9:50 AM

Answer:
ups, doorsname() is the function to go for
 8-May-2008 08:36
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chunjia wang

Posts: 39
Joined: 11-Oct-2007

hi , all.
I want to know , now who use doors.
Is there some dxl code ?
Please help,thanks.
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 8-May-2008 09:08
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Frank Jensen

Posts: 24
Joined: 18-Sep-2007

Hi,

I'm not sure if DOORS provides a convinient way but you can always use lmutils.exe to get a list of logged in users.

"C:\Program Files\Telelogic\Doors 7.1\flex\lmutil.exe lmstat -f DOORS"

Cheers,
Frank

-------------------------
Frank Jensen
TRW Automotive
frank.jensenREMOVE_ME@trw.com
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 8-May-2008 09:25
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chunjia wang

Posts: 39
Joined: 11-Oct-2007

hi, thanks.
Maybe , I do not say the detail that I want to do .sorry.
I write a dxl script , it exec some functions .
but I want only some people to use it, others can not ..
I want to know , now who use the doors, and I judge he can use the dxl or not.
thanks.
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 8-May-2008 09:31
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Frank Jensen

Posts: 24
Joined: 18-Sep-2007

oh, sorry.

you can use "username()" to get the currently logged in user.

-------------------------
Frank Jensen
TRW Automotive
frank.jensenREMOVE_ME@trw.com
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 8-May-2008 09:37
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Frank Jensen

Posts: 24
Joined: 18-Sep-2007

Answer Answer
ups, doorsname() is the function to go for

-------------------------
Frank Jensen
TRW Automotive
frank.jensenREMOVE_ME@trw.com
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 8-May-2008 16:18
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Reik Schroeder

Posts: 361
Joined: 28-Jul-2003

Hi chunjia,

I would recommend to use a group to define the users who are allowed to use the script.

Then you may use something like this for checking:

const string szGroupName = "allowed Users";
User u = find ();
if (existsGroup szGroupName) {
Group g = find (szGroupName);
if (member (g, u)) {
// place your code here ......
} else infoBox "You are not allowed to run this script!"
} else infoBox "Group not found ...!\n"


Greetings
Reik

-------------------------
Evosoft GmbH
for Siemens Industry Sector


Berlin, Germany
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