Logging out a user

Hi,
I'm new to WebLogic, and I'm trying to figure out how to log out a user. I am currently using Authenticate's logout method (weblogic.security.auth.Authenticate.logout(weblogic.security.Security.getCurrentSubject()),
and it seems to work, but the user who was just logged out can't log in in the same browser until another user has logged in and out. I've tried destroying all the cookies and ivalidating the current session, but to no avail. Any help would be appreciated. Thanks,
Dmitriy

Actually, most importantly in our logout procedure we must set loggedIn to false in a database table.
We've talked about having the clients send a "heartbeat" to the model residing on the OraAS. If we do not get a heartbeat for a set number of times we could "time out" a client. If we were to go this route, we'll have to register the clients (probably by IP) and spawn a new thread for this service.
Does anyone feel this is not the route to go? Is there anything a bit more KISS to implement. Obviously this adds a bit more complication and more points of failure to our current system.
-brian

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Log out the user

How can I log out some user on the VPN 5000 system With the VPN 5000 Manager? Is any possble for doing this? Thanks.
Mike

Hi there,
To log a user out of the Portal I decided to go the route of: UMFactory.getAuthenticator().logout(req, res)
Code is below.
Thanks for your replies,
- Vik.
<code>
IAuthentication Authen = UMFactory.getAuthenticator();
HttpServletRequest req = request.getServletRequest();
HttpServletResponse res = request.getServletResponse(true);
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Logging out all users

To perform a back-up I want to log out all users. Currently I use the 'unloadapp' ESSCMD, but this does not seem to log out users who are currently retrieving data. As a result my back-up fails.Does the 'logoutallusers' command log of all users (active, inactive)?Regards,JG

Howdy!We do this via script for each application using the following essmsh syntax:"alter application appname disable connects;"This works great. We then re-establish any connections that we killed with the following statement:"alter application appname enable connects;"Hope this helps!JR

How to log out all users from Application to switch partitions?

Hi,
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alter system kill request on application FCSTA4;
alter system kill request on application HISTA;
alter system kill request on application ABANSWER;
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Quite often we seem to get the issue that some user can't be logged out because they are running something intensive and this causes the partition switch to fail.
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that means that the app was purchased under that seperate apple id. u will need to delete the app and redownload it using the yahoo one if u want to be able to update it.

Account log out without user consent

Hi all,
There are 4 separate user accounts under my MacBook Pro running Lion. Today I logged into one account and then used switch user to switch to another account.
When I wanted to switch back to the previous account, it appeared to be logged out. I didn't do it, and was not sure about what caused it.
May I ask whether anyone has the same issue or knows the causes and solutions?
Thanks a lot.

>>At the Log in Screen I guess there is no way to log out a user unless you log in and log off.
That's true, other than restart, which gives you the option, while warning that they may lose unsaved changes.
>>roam wrote :
You can log off a user without first logging into their account, as shown in the following...
It works at least in Mac OS 10.6. I don't know about Mavericks.
It does work, and testing here, on the next login, windows & apps are just as they were left for the other user, even just-edited & unsaved files... no promises though.

Can administrator log out a user from application

Hi
My application has functionality to display all the logged in users of the application. This screen ia available to only administrators of the application. Now can the administrator logout any of the logged in users from his own screen? If yes then how?
Could you please give me some pointers?
Thanks & Regards
Inder Jeet Singh

there are several ways an administrator can do that. but that depends on the way you manage your sessions.
how are you getting the list of logged users? if it is from the database based on some field in some table then you can give the functionality to administrator to set that value to the logged out status(or whatever you are using).
but that depends on the way your application is keeping trak of the logged users.

Log out other user - what if I do not know who they are?

I work in a hostel and use a computer that has two user accounts. When I plug my iShuffle on my user account (that is only used by me) in the following warning comes up:
"Another user on this computer is using the iPod software, so iTunes is unable to communicate with this iPod. Please ask the other user to log out?"
I do not know who the other user is, so cannot log them out. Is there a way of fixing this without re & de-installing the software?
Windows XP

-Open the Task Manager (press CTRL-SHIFT-ESC).
-Click the Processes tab.
-Find the iPodService.exe, select it, and End Process it.
-Repeat for iTunes.exe and iTunesHelper.exe, if they are there.
-Close the Task Manager.
-Start iTunes again.

Log Out Behavior - Users appear to stay logged in

I cannot get a user to logout using the Log Out User
behavior. I created a restricted page which displays the user's
name once logged in, however when the user logs out and a different
user id is used, the restricted page returns the first user's id. I
cannot get the first user cleared. Thought it might be cookies but
not sure. I'm using DW CS4's built in behaviors (no hand coding),
MySQL 5.1.11, and PHP 5.2.9-1. Any ideas

David I'll try to clarify.
I've built 1 test MySql database, and 3 test pages
(login.php, login_failure.php, and authorized.php).
database: 6 fields (recordID, userFirstName, userLastName,
userID, userPW, userAccess). userAccess values were 1 and 0, now
are admin and user. admin is the allowed value in this case.
Records in MySql InnoDB:
recordID: 1 (Primary Key)
userFirstName: User1
userLastName: User1 Last
userID: user1 (Indexed)
userPW: user
userAccess: admin
recordID: 2 (Primary Key)
userFirstName: User2
userLastName: User2 Last
userID: user2 (Indexed)
userPW: user
userAccess: admin
login.php: Has one behavior, Log In User with Restrict access
based on User, password and access level which is derived from
userAccess field in database. If successful, user goes to
authorized.php, if not, user goes to login_failure.php.
login_failure.php: No SB, simply presents user with message
that they are not authorized for access and a link to return to
login.php.
authorized.php: 4 SBs. Restrict Access to Page(), Recordset
(Recordset1), Dynamic Text (Recordset1.userFirstName), and Log Out
User. This page allows userAccess value "admin" from database and
presents user with text Welcome (Recordset1.userFirstName). There
is also the Log Out User link to log out.
Steps to recreate:
Set both userIDs userAccess to "admin" to allow access to
authorized.php
Login to login.php using userID "user1" and password "user"
Take me to authorized.php, with "Welcome User1" displayed
Click Logout
Login to login.php using userID "user2" and password "user"
Take me to authorized.php, with "Welcome User1" displayed
(note: supposed to be Welcome User2)
Click Logout
Change userAccess for userID "user1" to something other than
"admin"
Login to login.php using userID "user1" and password "user"
Takes me to login_failure.php as expected
Return to login.php via link
Login to login.php using userID "user2" and password "user"
(which is now the only auth user)
Take me to authorized.php, with "Welcome User1" displayed
(note: supposed to be Welcome User2)
Click Logout
The security part works as seen above. I just don't
understand why the authorization.php page always shows the first
record in the DB. That's why I was assuming that the Log Out User
function was not working.
Steve

Clients get stuck when logging out current user

I'm using ARD 3.1 to control 9 different Macs (from G4's with 10.3.9 to Dual G5's with 10.4.8, all with 3.1 clients, all with several user accounts) in the same company in Germany. I'm connected to the company through an SSH tunnel all the time.
Some of these Macs regulary go to "ARD Not Active" after I log out any current user from the Apple menu on that computer (i. e. if I want to log out the admin and switch back to some normal user).
From my end it then looks as if this computer would be down, but actually the computer is still on and seems to be in some kind of deep sleep that can only be ended by hitting some keys on the keyboard in Germany or sometimes even only by restarting the computer.
This happpens frequently on 3 of these 9 computers. 2 of them have 10.4.8 and 1 has 10.4.6. The same thing also happened all the time with ARD 3.0.
Any idea what this might be and how I could fix this? It's often very annoying because I have to wait for the next day before I can continue to work on these computers.
Thanks and greetings
azdesertman
Mac OS X (10.4.8)

Part of your problem sounds like the energy saver settings are putting the machine to sleep. You might be able to wake them simply by executing a "Wake" command on the computers.
You could also set the energy saver settings to not allow the machines to go to sleep.

Logging out portal user programitically? (JSP)

I am trying to log out users using a JSP. Here is a snippet of
code calling the wwsec_app_priv package. I am not having any
luck with this. Any suggestions?
Thanks,
Tom Henricksen
CallableStatement logout = null;
logout = con.prepareCall ("begin if wwctx_api.is_logged_on
then wwsec_app_priv.logout(p_done_url=>?); end if; end;");
logout.setString(1,logout_url);
logout.execute();
logout.close();

The syntax you have works within our current web page (PSP
page), but we have never tried it directly with PL/SQL.
One thing you might want to look at is that "normally" (from a
web page) the URL is called via the portal30 DAD:
/pls/portal30/PORTAL30.wwsec_app_priv.logout?p_done_url=
http%3A%2F%wherever.com
Could it be that this works if PORTAL30 is running the package,
and not the actual user? Just a wag. --Kate

Inactivity Log Out Multiple users

As a family we share one IMac. All five of us are pretty active. I have selected "Log out after 30 min inactivity" however this is not occurring. None of the accounts logout automatically, and when all 5 accounts are logged on, the system is real slow. I also manage the users accounts and have passworded the system preferences, just so I control the setup. Any suggestion on why the users are not logging out after 30 min?

Where are you setting that preference? I've never seen it.
Also, as noted, I don't imagine it will allow logout if there are any unsaved changes to anything being worked on. Safari has a preference to ask if you want to close multiple tabs.
Message was edited by: Barney-15E

Logging out another user?

my cousin tried plugging his mp3 player in and using my Itunes, but it didnt work for him. Now whenever I plug my Ipod in, it says it can't run my Ipod unless I log out the other user. How do I do this?
pc Windows XP
pc Windows XP

You can restart your PC or follow these directions:
You can also try this
Close iTunes.
ctl-alt-del > Task Manager > Processes tab
Click the ImageName column to sort alphabetically
Check the box for showing processes for all users
End these processes
iPodService.exe
iTunesHelper.exe
Close TaskManager.
Start > Run > "C:\Program Files\iTunes\iTunesHelper.exe"
The quotes are important, make sure they are there. You can cut/paste from this post.

MBP won't log out of user account

Hello,-
My MBP won't log out of my user account. It just sits at a white screen with the apple logo. I have to reboot it (several times) to get back
to my user accounts.
I have done a disk repair and it worked fine for 2 days then back to issue.
Any suggestions?

If you don't already have a current backup, back up all data before doing anything else. This procedure is a diagnostic test. It changes nothing, for better or worse, and therefore will not, in itself, solve your problem. The backup is necessary on general principle, not because of anything suggested in this comment.
Third-party system modifications are a common cause of usability problems. By a “system modification,” I mean software that affects the operation of other software — potentially for the worse. The procedure will help identify which such modifications you've installed, as well as some other aspects of the configuration that may be related to the problem.
Don’t be alarmed by the seeming complexity of these instructions — they’re easy to carry out. Here's a brief summary: In each of two steps, you copy a line of text from this web page into a window in another application. You wait about a minute. Then you paste some other text, which will have been copied automatically, back into a reply on this page. The sequence is copy; paste; paste again. That's all there is to it. Details follow.
While investigating the problem, you may have started the computer in "safe" mode. If possible, these steps should be taken while booted in “normal” mode, not in safe mode. If you’re now running in safe mode, reboot as usual before continuing. If you can only boot in safe mode, you can still use this procedure, but not all of it will work. Be sure to mention that in your reply, if you haven't already done so.
Below are instructions to enter UNIX shell commands. The commands are safe and do nothing but produce human-readable output, but they must be entered exactly as given in order to work. If you question the safety of the procedure suggested here — which you should — search this site for other discussions in which it’s been followed without any report of ill effects. I am not asking you to trust me. If you can't satisfy yourself that these instructions are safe, don't follow them.
The commands will line-wrap or scroll in your browser, but each one is really just a single long line, all of which must be selected. You can accomplish this easily by triple-clicking anywhere in the line. The whole line will highlight, and you can then copy it.
Note: If you have more than one user account, Step 2 must be taken as an administrator. Ordinarily that would be the user created automatically when you booted the system for the first time. Step 1 should be taken as the user who has the problem, if different. Most personal Macs have only one user, and in that case this paragraph doesn’t apply.
Launch the Terminal application in any of the following ways:
☞ Enter the first few letters of its name into a Spotlight search. Select it in the results (it should be at the top.)
☞ In the Finder, select Go ▹ Utilities from the menu bar, or press the key combination shift-command-U. The application is in the folder that opens.
☞ Open LaunchPad. Click Utilities, then Terminal in the icon grid.
When you launch Terminal, a text window will open with a line already in it, ending either in a dollar sign (“$”) or a percent sign (“%”). If you get the percent sign, enter “sh” and press return. You should then get a new line ending in a dollar sign.
Step 1
Triple-click anywhere in the line of text below on this page to select it:
PB=/usr/libexec/PlistBuddy; PR () { [[ "$o" ]] && printf '\n%s:\n\n%s\n\n' "$1" "$o"; }; PF () { o=$($PB -c Print "$2" | awk -F'= ' \/$3'/{print $2}'); PR "$1"; }; { o=$(( $(vm_stat | awk '/Pageo/{sub("\\.",""); print $2}')/256 )); [[ $o -gt 1024 ]] && printf "\nPageouts: %s MiB\n\n" $o; o=$(kextstat -kl | grep -v com\\.apple | cut -c53- | cut -d\< -f1); PR "Loaded extrinsic kernel extensions"; o=$(launchctl list | sed 1d | awk '!/0x|com\.apple|org\.(x|openbsd)|\.[0-9]+$/{print $3}'); PR "Loaded extrinsic user agents"; o=$(launchctl getenv DYLD_INSERT_LIBRARIES); PR "Inserted libraries"; o=$(crontab -l); PR "User cron tasks"; o=$(cat /e*/lau*); PR "Global launchd configuration"; o=$(cat ~/.lau*); PR "Per-user launchd configuration"; PF "Global login items" /L*/P*/loginw* Path; PF "Per-user login items" L*/P*/*loginit* Name; PF "Safari extensions" L*/Saf*/*/E*.plist Bundle | sed 's/\..*$//;s/-[1-9]$//'; o=$(find ~ $TMPDIR.. $ -flags +sappnd,schg,uappnd,uchg -o ! -user $UID -o ! -perm -600 $ | wc -l); [[ $o -eq 0 ]] || printf "\nRestricted user files: %s\n\n" $o; cd; o=$(find -L /S*/L*/E* {,/}L*/{A*d,Compon,Ex,In,Keyb,Mail/Bu,P*P,Qu,Scripti,Servi,Spo}* -type d -name Contents -prune | while read d; do ID=$($PB -c 'Print :CFBundleIdentifier' "$d/Info.plist"); ID=${ID:-No bundle ID}; egrep -qv "^com\.apple\.[^x]|Accusys|ArcMSR|ATTO|HDPro|HighPoint|driver\.stex|hp-fax|JMicron|microsoft\.MDI|print|SoftRAID" <<< $ID && printf '%s\n\t(%s)\n' "${d%/Contents}" "$ID"; done); PR "Extrinsic loadable bundles"; o=$(find /u*/{,*/}lib -type f -exec sh -c 'file -b "$1" | grep -qw shared && ! codesign -v "$1"' {} {} \; -print); PR "Unsigned shared libraries"; for d in {,/}L*/{La,Priv,Sta}* L*/Fonts; do o=$(ls -A "$d"); PR "$d"; done; } 2> /dev/null | pbcopy; echo $'\nStep 1 done'
Copy the selected text to the Clipboard by pressing the key combination command-C. Then click anywhere in the Terminal window and paste (command-V). I've tested these instructions only with the Safari web browser. If you use another browser, you may have to press the return key after pasting.
The command may take up to a few minutes to run, depending on how many files you have and the speed of the computer. Wait for the line "Step 1 done" to appear below what you entered. The output of the command will be automatically copied to the Clipboard. If the command produced no output, the Clipboard will be empty. Paste into a reply to this message. No typing is involved in this step.
Step 2
Remember that you must be logged in as an administrator for this step. Do as in Step 1 with this line:
PR () { [[ "$o" ]] && printf '\n%s:\n\n%s\n\n' "$1" "$o"; }; { o=$(sudo launchctl list | sed 1d | awk '!/0x|com\.(apple|openssh|vix\.cron)|org\.(amav|apac|calendarse|cups|dove|isc|ntp|post[fg]|x)/{print $3}'); PR "Loaded extrinsic daemons"; o=$(sudo defaults read com.apple.loginwindow LoginHook); PR "Login hook"; o=$(sudo crontab -l); PR "Root cron tasks"; o=$(syslog -k Sender kernel -k Message CReq 'error|GPU |hfs: Ru|I/O e|find tok|n Cause: -|NVDA\(|pagin|timed? ?o' | tail -n25 | awk '/:/{$4=""; print}'); PR "Log check"; } 2> /dev/null | pbcopy; echo $'\nStep 2 done'
This time you'll be prompted for your login password, which you do have to type. Nothing will be displayed when you type it. Type it carefully and then press return. You may get a one-time warning to be careful. Heed that warning, but don't post it. If you see a message that your username "is not in the sudoers file," then you're not logged in as an administrator.
You can then quit Terminal.
To prevent confusion, I'll repeat: When you type your password in the Terminal window, you won't see what you're typing.
Note: If you don’t have a login password, set one before taking Step 2. If that’s not possible, skip the step.
Important: If any personal information, such as your name or email address, appears in the output of these commands, anonymize it before posting. Usually that won't be necessary.
Remember, Steps 1 and 2 are all copy-and-paste — type only your password. Also remember to post the output.

When logging out of user account iMac will not start back up at login screen.

To prevent guests from using my iMac I frequently "log out" and created a guest account. However when I log out and shut down, my iMac turns on
right back to my desktop not even returning to the "log-in" screen. I tried fixing this under preferences but this did not yield any help as it's still doing
this? I don't understand I can't recall ever having this problem before. Please help! Thanks

Try rebooting into Safe Mode, this will clear some caches. It is possible one (or more) is corrupt. To reboot in Safe Model when you hear the startup tone hold down the Shift Key until you see a progress bar appearing.

SecurityAgent seems to be thrashing when ParentalControls logs out a user with limited access time.

I have a 2007 iMac running OS X Lion, all latest updates and patches applied.
My daughter has an account on the system, which I have limited to only being able to be used for a few hours a day.
The problem arises when she leaves herself logged in, but walks away, and the time runs out. When I sit down at the console, the first thing that happens is that the cursor disappears, and the systems starts to go into a logout. Makes sesne so far, I guess. However, the system then goes into what I can only presume is some sort of loop. The cube animation starts, and rotates back and forth several times between two screens, both equally hessian wrapped, with no distinguising features.
I can log into the system via ssh, and I can confirm that the system is spiking out on SecurityAgent:
Processes: 661 total, 5 running, 29 stuck, 627 sleeping, 2082 threads
19:59:08 Load Avg: 1.68, 3.07, 5.24 CPU usage: 43.38% user, 23.52% sys, 33.8% idle SharedLibs: 3620K resident, 1948K data, 0B linkedit.
MemRegions: 104350 total, 1295M resident, 189M private, 2974M shared.
PhysMem: 1166M wired, 3312M active, 1655M inactive, 6133M used, 9560K free.
VM: 1611G vsize, 1118M framework vsize, 34169473(356) pageins, 5796949(92) pageouts.
Networks: packets: 100372238/47G in, 162918821/146G out. Disks: 23914908/3252G read, 27697160/2570G written.
PID    COMMAND      %CPU      TIME     #TH #WQ #PORT #MREG RPRVT RSHRD RSIZE VPRVT VSIZE PGRP PPID STATE    UID FAULTS
27013 SecurityAgen 98.0      37:41.27 2/1 1    88    84    508K   3904K 2664K 31M    2410M 27013 1     running 92   3028
28561 top          25.9      02:27.05 1/1 0    32    43    3752K 216K   4176K 20M    2381M 28561 28505 running 0    1924014+
1      launchd      4.6       50:43.63 3    0    3159- 343   2804K+ 2836K 2128K+ 15M+   2405M 1     0     stuck    0    553252+
0      kernel_task 1.5       08:01:03 71/2 0    2     976   14M    0B     595M- 15M    5487M- 0     0     running 0    23008
197    WindowServer 0.7       01:57:50 5    1    6653 6448 22M+   885M   486M+ 137M   40G    197   1     sleeping 88   25934192+
18548 Transmission 0.5       01:59:18 7    1    175   372   10M    7352K 16M-   47M    2550M 18548 282   sleeping 501 10297632
14     opendirector 0.4       02:27.27 13   13   617+ 108+ 6072K+ 1872K 6860K+ 32M+   2417M+ 14    1     stuck    0    407028+
25117 mcxalr       0.2       00:40.33 4/1 3    43    41    200K   252K   1780K 30M    2391M 25117 1     running 54   1580
26441 SecurityAgen 0.2       00:01.55 2    1    121   245   2856K 19M    6820K 35M    2503M 26441 1     stuck    92   27427
28001 ManagedClien 0.1       00:00.11 4    3    98+   86    1196K+ 4900K 3748K+ 31M    2413M 28001 27998 sleeping 0    3718+
294    UserEventAge 0.1       00:20.22 3    1    212   140   976K+ 3964K 2632K+ 34M    2422M 294   282   sleeping 501 294201+
28006 ManagedClien 0.0       00:00.11 4    3    97    86    1184K+ 4904K 3700K+ 32M    2414M 28006 28003 sleeping 0    3725+
28573 mdworker     0.0       00:00.41 4/1 2    55    101   8648K 6052K 13M    38M    2422M 28573 28443 running 89   4766+
27831 loginwindow 0.0       00:00.50 4    2    153+ 108+ 1692K+ 4804K 3252K+ 35M+   2425M+ 27831 197   sleeping 0    4875+
15     configd      0.0       00:33.57 6    1    2316 230   2392K 1788K 3524K+ 27M    2420M 15    1     sleeping 0    320735+
28318 loginwindow 0.0       00:00.54 3    1    151   101   924K+ 4712K 3120K+ 33M+   2423M 28318 197   sleeping 0    4778+
Now, obviously, because I can ssh in, I can start a clean shutdown and restart, but I'd like a way to do this that doesn't involve me bringing down the system, when the whole point of multiple users and parental controls is that a system can be shared cleanly.
Thoughts?
Thanks,
Will

See if this blog post that our support team created helps you with the connection information. This can create the desktop Access application linked up to the Access 2013 web app tables you have.
http://blogs.technet.com/b/the_microsoft_access_support_team_blog/archive/2014/03/24/how-to-make-external-connections-to-an-access-web-app-new.aspx
Also, you might just try clicking the Create Reports button in the Access web app client interface and Access will create the reporting database for you with links to the web app tables. On the File menu (when using the Access web app in client) make sure
to click the option to Allow Any Location under the Manage button. After that, just click the Create Reports button on the File menu (right above the Manage button) and wait a few seconds. Access will create a new desktop database linked up to all of the Access
web app tables. You can then create as many reports or other desktop objects as you like.
Hope that helps,
Jeff Conrad - Access Junkie - MVP Alumnus
Senior Content Developer - Office Content Development Team - Microsoft Corporation
Author - Microsoft Access 2013 Inside Out
Author - Microsoft Access 2010 Inside Out
Co-author - Microsoft Office Access 2007 Inside Out
Access 2007/2010/2013 Info: http://www.AccessJunkie.com
This posting is provided "AS IS" with no warranties, and confers no rights.
Use of included script samples are subject to the terms specified at
http://www.microsoft.com/en-us/legal/Copyright/Default.aspx

Logging out a user

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