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• Restore "lost" user after Active Directory removal?

  - by Zulgrib

  Is it possible to restore lost users after Active Directory unistallation ? (I forgot to switch users to local users) The computer run Windows Server 2008 R2 Entreprise, and all the registry linked to the user i want to restore seems to still be there, user's folder is still on the harddrive, and useraccount2 still show the user (But flagged as unknown user) Some folders still have rigts set to this lost user, and even the local default Admin account cannot open/delete the folder. (But the real problem here is to find how to recover users account, the folder can be deleted an other way) All users i want te restore was originaly local users, converted to domain users after Active Directory installation. I think that if i can change user's sid (choosing the sid manually) i'll be able to easily recover rights on folders Regards

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• Drupal 7: One-time user account

  - by Noob

  I'm going to create a survey in Drupal 7 with the webform module, installed on a debian system which may be adapted in every way. The users (personally known, approx. 120) doing that survey will walk into a room and complete the survey in browsers on different computers. After that, they'll leave the room and other persons will enter, complete the survey on the same computers and so on. Each user may enter only one submission. The process needs to be anonymous, i. e. I mustn't have any idea of who did wich submission. My current solution is to generate random one-time-passwords and hand out one password per user (without noting who got which password). Within the survey there will be a password field where the one-time-password is entered. The value is checked by webform to be unique. I'll get the data via csv or Excel and verify the passwords manually in excel by comparing them to the list of valid passwords. The problem is: I don't like the idea of manually generating the password list, copying it to excel and doing a manual check. That's a good idea for one-time-use, but we're going to repeat the survey every once in a while. I'd rather generate one-time-logins (like user0001/fdlkjewf, user0002/dfrefnnr, ...) for each survey, hand them out to the users and let drupal/debian/whatever check whether a submission is valid or not. Do you have any idea how to batch-generate about 120 users with one-time-passwords in Drupal 7 and verify that each user may submit the form only once? Do you even have a better idea how to accomplish the task within the intranet? Thank you for your help.

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• Sorting Algorithms

  - by MarkPearl

  General Every time I go back to university I find myself wading through sorting algorithms and their implementation in C++. Up to now I haven’t really appreciated their true value. However as I discovered this last week with Dictionaries in C# – having a knowledge of some basic programming principles can greatly improve the performance of a system and make one think twice about how to tackle a problem. I’m going to cover briefly in this post the following: Selection Sort Insertion Sort Shellsort Quicksort Mergesort Heapsort (not complete) Selection Sort Array based selection sort is a simple approach to sorting an unsorted array. Simply put, it repeats two basic steps to achieve a sorted collection. It starts with a collection of data and repeatedly parses it, each time sorting out one element and reducing the size of the next iteration of parsed data by one. So the first iteration would go something like this… Go through the entire array of data and find the lowest value Place the value at the front of the array The second iteration would go something like this… Go through the array from position two (position one has already been sorted with the smallest value) and find the next lowest value in the array. Place the value at the second position in the array This process would be completed until the entire array had been sorted. A positive about selection sort is that it does not make many item movements. In fact, in a worst case scenario every items is only moved once. Selection sort is however a comparison intensive sort. If you had 10 items in a collection, just to parse the collection you would have 10+9+8+7+6+5+4+3+2=54 comparisons to sort regardless of how sorted the collection was to start with. If you think about it, if you applied selection sort to a collection already sorted, you would still perform relatively the same number of iterations as if it was not sorted at all. Many of the following algorithms try and reduce the number of comparisons if the list is already sorted – leaving one with a best case and worst case scenario for comparisons. Likewise different approaches have different levels of item movement. Depending on what is more expensive, one may give priority to one approach compared to another based on what is more expensive, a comparison or a item move. Insertion Sort Insertion sort tries to reduce the number of key comparisons it performs compared to selection sort by not “doing anything” if things are sorted. Assume you had an collection of numbers in the following order… 10 18 25 30 23 17 45 35 There are 8 elements in the list. If we were to start at the front of the list – 10 18 25 & 30 are already sorted. Element 5 (23) however is smaller than element 4 (30) and so needs to be repositioned. We do this by copying the value at element 5 to a temporary holder, and then begin shifting the elements before it up one. So… Element 5 would be copied to a temporary holder 10 18 25 30 23 17 45 35 – T 23 Element 4 would shift to Element 5 10 18 25 30 30 17 45 35 – T 23 Element 3 would shift to Element 4 10 18 25 25 30 17 45 35 – T 23 Element 2 (18) is smaller than the temporary holder so we put the temporary holder value into Element 3. 10 18 23 25 30 17 45 35 – T 23   We now have a sorted list up to element 6. And so we would repeat the same process by moving element 6 to a temporary value and then shifting everything up by one from element 2 to element 5. As you can see, one major setback for this technique is the shifting values up one – this is because up to now we have been considering the collection to be an array. If however the collection was a linked list, we would not need to shift values up, but merely remove the link from the unsorted value and “reinsert” it in a sorted position. Which would reduce the number of transactions performed on the collection. So.. Insertion sort seems to perform better than selection sort – however an implementation is slightly more complicated. This is typical with most sorting algorithms – generally, greater performance leads to greater complexity. Also, insertion sort performs better if a collection of data is already sorted. If for instance you were handed a sorted collection of size n, then only n number of comparisons would need to be performed to verify that it is sorted. It’s important to note that insertion sort (array based) performs a number item moves – every time an item is “out of place” several items before it get shifted up. Shellsort – Diminishing Increment Sort So up to now we have covered Selection Sort & Insertion Sort. Selection Sort makes many comparisons and insertion sort (with an array) has the potential of making many item movements. Shellsort is an approach that takes the normal insertion sort and tries to reduce the number of item movements. In Shellsort, elements in a collection are viewed as sub-collections of a particular size. Each sub-collection is sorted so that the elements that are far apart move closer to their final position. Suppose we had a collection of 15 elements… 10 20 15 45 36 48 7 60 18 50 2 19 43 30 55 First we may view the collection as 7 sub-collections and sort each sublist, lets say at intervals of 7 10 60 55 – 20 18 – 15 50 – 45 2 – 36 19 – 48 43 – 7 30 10 55 60 – 18 20 – 15 50 – 2 45 – 19 36 – 43 48 – 7 30 (Sorted) We then sort each sublist at a smaller inter – lets say 4 10 55 60 18 – 20 15 50 2 – 45 19 36 43 – 48 7 30 10 18 55 60 – 2 15 20 50 – 19 36 43 45 – 7 30 48 (Sorted) We then sort elements at a distance of 1 (i.e. we apply a normal insertion sort) 10 18 55 60 2 15 20 50 19 36 43 45 7 30 48 2 7 10 15 18 19 20 30 36 43 45 48 50 55 (Sorted) The important thing with shellsort is deciding on the increment sequence of each sub-collection. From what I can tell, there isn’t any definitive method and depending on the order of your elements, different increment sequences may perform better than others. There are however certain increment sequences that you may want to avoid. An even based increment sequence (e.g. 2 4 8 16 32 …) should typically be avoided because it does not allow for even elements to be compared with odd elements until the final sort phase – which in a way would negate many of the benefits of using sub-collections. The performance on the number of comparisons and item movements of Shellsort is hard to determine, however it is considered to be considerably better than the normal insertion sort. Quicksort Quicksort uses a divide and conquer approach to sort a collection of items. The collection is divided into two sub-collections – and the two sub-collections are sorted and combined into one list in such a way that the combined list is sorted. The algorithm is in general pseudo code below… Divide the collection into two sub-collections Quicksort the lower sub-collection Quicksort the upper sub-collection Combine the lower & upper sub-collection together As hinted at above, quicksort uses recursion in its implementation. The real trick with quicksort is to get the lower and upper sub-collections to be of equal size. The size of a sub-collection is determined by what value the pivot is. Once a pivot is determined, one would partition to sub-collections and then repeat the process on each sub collection until you reach the base case. With quicksort, the work is done when dividing the sub-collections into lower & upper collections. The actual combining of the lower & upper sub-collections at the end is relatively simple since every element in the lower sub-collection is smaller than the smallest element in the upper sub-collection. Mergesort With quicksort, the average-case complexity was O(nlog2n) however the worst case complexity was still O(N*N). Mergesort improves on quicksort by always having a complexity of O(nlog2n) regardless of the best or worst case. So how does it do this? Mergesort makes use of the divide and conquer approach to partition a collection into two sub-collections. It then sorts each sub-collection and combines the sorted sub-collections into one sorted collection. The general algorithm for mergesort is as follows… Divide the collection into two sub-collections Mergesort the first sub-collection Mergesort the second sub-collection Merge the first sub-collection and the second sub-collection As you can see.. it still pretty much looks like quicksort – so lets see where it differs… Firstly, mergesort differs from quicksort in how it partitions the sub-collections. Instead of having a pivot – merge sort partitions each sub-collection based on size so that the first and second sub-collection of relatively the same size. This dividing keeps getting repeated until the sub-collections are the size of a single element. If a sub-collection is one element in size – it is now sorted! So the trick is how do we put all these sub-collections together so that they maintain their sorted order. Sorted sub-collections are merged into a sorted collection by comparing the elements of the sub-collection and then adjusting the sorted collection. Lets have a look at a few examples… Assume 2 sub-collections with 1 element each 10 & 20 Compare the first element of the first sub-collection with the first element of the second sub-collection. Take the smallest of the two and place it as the first element in the sorted collection. In this scenario 10 is smaller than 20 so 10 is taken from sub-collection 1 leaving that sub-collection empty, which means by default the next smallest element is in sub-collection 2 (20). So the sorted collection would be 10 20 Lets assume 2 sub-collections with 2 elements each 10 20 & 15 19 So… again we would Compare 10 with 15 – 10 is the winner so we add it to our sorted collection (10) leaving us with 20 & 15 19 Compare 20 with 15 – 15 is the winner so we add it to our sorted collection (10 15) leaving us with 20 & 19 Compare 20 with 19 – 19 is the winner so we add it to our sorted collection (10 15 19) leaving us with 20 & _ 20 is by default the winner so our sorted collection is 10 15 19 20. Make sense? Heapsort (still needs to be completed) So by now I am tired of sorting algorithms and trying to remember why they were so important. I think every year I go through this stuff I wonder to myself why are we made to learn about selection sort and insertion sort if they are so bad – why didn’t we just skip to Mergesort & Quicksort. I guess the only explanation I have for this is that sometimes you learn things so that you can implement them in future – and other times you learn things so that you know it isn’t the best way of implementing things and that you don’t need to implement it in future. Anyhow… luckily this is going to be the last one of my sorts for today. The first step in heapsort is to convert a collection of data into a heap. After the data is converted into a heap, sorting begins… So what is the definition of a heap? If we have to convert a collection of data into a heap, how do we know when it is a heap and when it is not? The definition of a heap is as follows: A heap is a list in which each element contains a key, such that the key in the element at position k in the list is at least as large as the key in the element at position 2k +1 (if it exists) and 2k + 2 (if it exists). Does that make sense? At first glance I’m thinking what the heck??? But then after re-reading my notes I see that we are doing something different – up to now we have really looked at data as an array or sequential collection of data that we need to sort – a heap represents data in a slightly different way – although the data is stored in a sequential collection, for a sequential collection of data to be in a valid heap – it is “semi sorted”. Let me try and explain a bit further with an example… Example 1 of Potential Heap Data Assume we had a collection of numbers as follows 1[1] 2[2] 3[3] 4[4] 5[5] 6[6] For this to be a valid heap element with value of 1 at position [1] needs to be greater or equal to the element at position [3] (2k +1) and position [4] (2k +2). So in the above example, the collection of numbers is not in a valid heap. Example 2 of Potential Heap Data Lets look at another collection of numbers as follows 6[1] 5[2] 4[3] 3[4] 2[5] 1[6] Is this a valid heap? Well… element with the value 6 at position 1 must be greater or equal to the element at position [3] and position [4]. Is 6 > 4 and 6 > 3? Yes it is. Lets look at element 5 as position 2. It must be greater than the values at [4] & [5]. Is 5 > 3 and 5 > 2? Yes it is. If you continued to examine this second collection of data you would find that it is in a valid heap based on the definition of a heap.

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• django create user and log them in

  - by Scott Willman

  In a view I'm trying to create a new user and then log them in but result in a new url on success. def create(request): if request.method == "POST": # do user creation # user.save() auth_user = authenticate(username=user.username,password=user.password) if auth_user is not None: login(request, auth_user) return HttpResponseRedirect('/user/account/') return render_to_response('create_form.html') So, how do I maintain the user object using the HttpResponseRedirect or validate the logged in user in an unassociated view?

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• cPanel Virtfs won't umount

  - by JPerkSter

  Anyone have any experience with virtfs on cPanel servers? I can't seem to get them to unmount, as they say they are already unmounted: [root@Server ~]# cat /proc/mounts | grep user /dev/root /home/virtfs/user/lib ext3 rw,errors=continue,data=ordered 0 0 /dev/root /home/virtfs/user/opt ext3 rw,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/lib ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/sbin ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/share ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/bin ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/man ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/X11R6 ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/kerberos ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/libexec ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/local/bin ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/local/share ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/local/Zend ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/local/IonCube ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/include ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda3 /home/virtfs/user/usr/local/lib ext3 rw,nodev,errors=continue,data=ordered 0 0 /dev/sda2 /home/virtfs/user/var/spool ext3 rw,nodev,noatime,nodiratime,errors=continue,data=ordered 0 0 /dev/sda2 /home/virtfs/user/var/lib ext3 rw,nodev,noatime,nodiratime,errors=continue,data=ordered 0 0 /dev/sda2 /home/virtfs/user/var/cpanel ext3 rw,nodev,noatime,nodiratime,errors=continue,data=ordered 0 0 /dev/sda2 /home/virtfs/user/var/run ext3 rw,nodev,noatime,nodiratime,errors=continue,data=ordered 0 0 /dev/sda2 /home/virtfs/user/var/log ext3 rw,nodev,noatime,nodiratime,errors=continue,data=ordered 0 0 /dev/sda6 /home/virtfs/user/tmp ext3 rw,nosuid,nodev,noexec,noatime,errors=continue,data=ordered 0 0 /dev/root /home/virtfs/user/bin ext3 rw,errors=continue,data=ordered 0 0 [root@Server ~]# for i in cat /proc/mounts |grep virtfs |grep user |awk '{print$2}'; do umount $i; done umount: /home/virtfs/user/lib: not mounted umount: /home/virtfs/user/opt: not mounted umount: /home/virtfs/user/usr/lib: not mounted umount: /home/virtfs/user/usr/sbin: not mounted umount: /home/virtfs/user/usr/share: not mounted umount: /home/virtfs/user/usr/bin: not mounted umount: /home/virtfs/user/usr/man: not mounted umount: /home/virtfs/user/usr/X11R6: not mounted umount: /home/virtfs/user/usr/kerberos: not mounted umount: /home/virtfs/user/usr/libexec: not mounted umount: /home/virtfs/user/usr/local/bin: not mounted umount: /home/virtfs/user/usr/local/share: not mounted umount: /home/virtfs/user/usr/local/Zend: not mounted umount: /home/virtfs/user/usr/local/IonCube: not mounted umount: /home/virtfs/user/usr/include: not mounted umount: /home/virtfs/user/usr/local/lib: not mounted umount: /home/virtfs/user/var/spool: not mounted umount: /home/virtfs/user/var/lib: not mounted umount: /home/virtfs/user/var/cpanel: not mounted umount: /home/virtfs/user/var/run: not mounted umount: /home/virtfs/user/var/log: not mounted umount: /home/virtfs/user/tmp: not mounted umount: /home/virtfs/user/bin: not mounted umount: /home/virtfs/user/dev: not mounted umount: /home/virtfs/user/proc: not mounted

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• Learn About Oracle’s Strategy for a Simple, Modern User Experience at OpenWorld 2012

  - by Applications User Experience

  By Kathy Miedema, Oracle Applications User Experience If you’re interested in what the best possible user experience looks like, you’ll want to hear what Oracle’s Applications User Experience team is planning for OpenWorld 2012, Sept. 30-Oct. 4 in San Francisco. This year, we will talk Fusion, Fusion, Fusion. We were among the first to show Oracle Fusion Applications in the last couple of years, and we’ll be showing it again this year so you can see what Oracle is planning for the next generation of enterprise applications. Attend our sessions to learn more about the user experience strategy in which Oracle is investing. Simplicity is the driving force behind the demos that we are unveiling now, which you can see at OpenWorld. We want to create opportunities for productivity and efficiency, and deliver enterprise data across devices to help you do your work in the way best suited to your job and needs, said Jeremy Ashley, Vice President, Oracle Applications User Experience. You can see the new look for Fusion Applications at a general session led by Ashley at 3:30 p.m. on Wednesday, Oct. 3. You’ll also have the chance to learn more about tailoring in Oracle Fusion Applications, and gain a new understanding of the investment in the user experience behind Fusion Applications at our sessions (see session information below). Inside the Oracle Applications User Experience team’s on-site lab at Oracle OpenWorld 2011. Head to the demogrounds to see new demos from the Applications User Experience team, including the new look for Fusion Applications and what we’re building for mobile platforms. Take a spin on our eye tracker, a very cool tool that we use to research the usability of a particular design. Visit the Usable Apps OpenWorld page to find out where our demopods will be located. We are also recruiting participants for our on-site lab, in which we gather feedback on new user experience designs, and taking reservations for a charter bus that will bring you to Oracle headquarters for a lab tour Thursday, Oct. 4, or Friday, Oct. 5. Tours leave at 10 a.m. and 1:45 p.m. from the Moscone Center in San Francisco. You’ll see more of our newest designs at the lab tour, and some of our research tools in action. Can’t participate in a customer feedback session or take a lab tour this time around? Visit Usable Apps to participate or book a tour another time. For more information on any OpenWorld sessions, check the content catalog – also available at www.oracle.com/openworld. For information on Applications User Experience (Apps UX) sessions and activities, go to the Usable Apps OpenWorld page. APPS UX OPENWORLD SESSIONS Oracle’s Roadmap to a Simple, Modern User Experience Presenter: Jeremy Ashley, Vice President Applications User Experience, Oracle; with Debra Lilley, Fujitsu Consulting; Basheer Khan, Innowave; and Edward Roske, InterRelSession ID: CON9467Date: Wednesday, Oct. 3 Time: 3:30 - 4:30 p.m.Location: Moscone West - 3002/3004 Jeremy Ashley Oracle Fusion Applications: Transforming Insight into Action Presenters: Killian Evers and Kristin Desmond, OracleSession ID: CON8718Date: Thursday, Oct. 4Time: 11:15 a.m. - 12:15 p.m.Location: Moscone West - 2008 “FRIENDS OF UX” OPENWORLD SESSIONS Sessions by the Oracle Usability Advisory Board (OUAB) members: Advances in Oracle Enterprise Governance, Risk, and Compliance Manager  Presenters: Koen Delaure, KPMG Advisory NV, and Oracle Usability Advisory Board member; Russell Stohr, Oracle Session ID: CON9389Date: Tuesday, Oct. 2Time: 1:15 - 2:15 p.m.Location: Palace Hotel - Concert Optimize Oracle E-Busines Suite Procure-to-Pay: Cut Inefficiences/Fraud with Oracle GRC Apps Presenters: Koen Delaure, KPMG Advisory NV, and Solveig Wagner, Seadrill Management AS, both Oracle Usability Advisory Board members; and Swarnali Bag, OracleSession ID: CON9401Date: Monday, Oct. 1Time: 12:15 - 1:15 p.m.Location: Intercontinental - Sutter Showcase of JD Edwards EnterpriseOne Mobility Presenters: Jon Wells, Westmoreland Coal Co., Oracle Usability Advisory Board member; Rob Mills and Liz Davson, Town of Oakville; Keith Sholes and Louise Farner, Oracle Session ID: CON9123Date: Tuesday, Oct. 2Time: 1:15 - 2:15 p.m.Location: InterContinental - Grand Ballroom B Sessions by the Fusion User Experience Adovcates (FXA) Usability and Features of Oracle Fusion Applications, Built upon Oracle Fusion Middleware Presenters: Debra Lilley, Fujitsu Consulting and Oracle Usability Advisory Board member; John King, King Training ResourcesSession ID: UGF10371Date: Sunday, Sept. 30Time: 11 a.m. - 11:45 a.m. Location: Moscone West – 2010 Ten Things to Love About Oracle Fusion Project Portfolio Management  Presenter: Floyd Teter, EiS TechnologiesSession ID: CON6021Date: Tuesday, Oct. 2Time: 10:15 - 11:15 a.m.Location: Moscone West – 2003

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• os x 10.4 Old, deleted user mail account problems

  - by Chris

  Hello- A while back I tried to add a user 'david' as a mail user on my OS X 10.4 server using dscl (I only had terminal access at the time, no ability to use workgroup manager). I could never get this account to work properly, so I deleted it. dscl . -list /Users no longer shows 'david' as an entry. I have since gained access via Workgroup Manager, and I am trying to re-create the 'david' account. Workgroup manager creates the account fine, along with an email account, which I can then log into via IMAP ('login david password' returns 'OK user logged in'). However, this mail account does not have an inbox, and I can not create one thru a mail client, IMAP or cyradm (they all say 'system I/O error'). When I re-delete this user, I can't find any record of him in any of the mail spool locations. Creating a user with any other name works fine (Inbox, mail access, everything). Any ideas on how I can get this user up and running again? -Chris P.S. - to create this user in the first place, I used dscl . create, then dscl . append /Users/david "some XML I found on the 'net" to add email privileges, if this helps...

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• finding the user - without knowing the innards of IIS

  - by LosManos

  Place of crime is WinSrv2008 with IIS7. My IIS apppool user is trying to create a folder but fails. How do I find out which User it is? Let's say I don't know much about IIS7 and Aspnet but need to trace whatever is happening through tools. So I fire up Sysinternals/ProcessMonitor to find out what is happening. I find Access denied on a folder just as I suspected. But which user? I add the User column to the output or ProcessMonitor and it says IIS Apppool\Defaultapppool in capitals. Well... that isn't a user is it? If I go to IIS and its Apppools and Advanced settings and Process model and Identity I can see clues about which user it is but that is only because I know IIS. What if it had been Apache or LightHttpd or whatever? How do I see the user to give the appropriate rights to?

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• Script to move specific user folders in Windows 7

  - by Evan M.

  Hi there. When I install Windows Vista/7, I move some of my user folders onto a new partition (i.e. Documents, Musics, Pictures, etc.). This does not include moving the whole User directory, just some of the data folders. %AppData% remains in it's default location (%SystemDrive%\Users). I'm getting tired of manually moving each of these folder's by changing their location under the properties dialog. Does anyone know of a way that I can script this to apply to the folders that I wish?

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• How to get transparent user interface background in Windows 7

  - by blasteralfred

  I use Windows 7 Home Premium. Recently, when I came through Photoshop in OSX, I was interested by its user interface. It has got a "100% transparent" user interface background, whereas it's usually gray in Windows. I googled for some glass solutions, but all of them provide a "transparent window" or a semi-transparent one, which makes the complete window transparent. These solutions do not work for me, as I just need the UI (gray) background transparent but require all the other elements (such as nav-bars, control buttons, etc.) to have no transparency. Below is a screenshot of what I'm saying: Is this a feature of Photoshop only? How can I achieve the same in Windows 7? Thanks in advance...:)

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• User Profile cannot be loaded - Windows 7

  - by Ryan

  After uninstalling an HP Vector Mouse driver, then rebooting, when Windows tries to auto log me in, i get an error message saying tyhe following: The User Profile Service failed the Login. User Profile cannot be loaded. Due to the fact that it is the only account on this PC, I cannot even go into another account. I rebooted the machine several times, before going into Safe Mode with Networking. For some reason, I cannot create a new account whilst in safe mode (I think it is to do with UAC, nothing with UAC is clickable). Thus, I am stuck. I cannot get into my account, nor can I create a new one to copy files over to. Any ideas? Thanks in advance! Ryan EDIT: System Restore was, for some reason turned off. Thus, I cannot restore to a working point.

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• Why is the reported user "root" when a "normal user" executes "ps ux" on OS X? Is this normal behavi

  - by snies

  I am running OS X 10.6.1 . When i am logged in as a normal user of group staff and do a ps ux it lists my ps ux command as being run by root: snies 181 0.0 0.3 2774328 12500 ?? S 6:00PM 0:20.96 /System/Library... root 1673 0.0 0.0 2434788 508 s001 R+ 8:16AM 0:00.00 ps ux snies 177 0.0 0.0 2457208 984 ?? Ss 6:00PM 0:00.52 /sbin/launchd snies 1638 0.0 0.0 2435468 1064 s001 S 8:13AM 0:00.03 -bash Is this normal behaviour? And if so why? Please note that the user is not an Administrator account and is not able to sudo.

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• Can't access Administrator account on Windows XP after adding local user account

  - by bwerks

  I have an installation of windows XP, and it's not part of a domain. Previously, it just had only the administrator account, and upon creating a different user account, all access was lost to the administrator account. When the machine starts up, only the new local account is offered for login, which seemed strange. I've checked that the administrator account was not disabled, nor are any rights missing from the local security policy. Furthermore, the administrator account is accessible via remote desktop, where an opportunity is given to type the desired account. REALLY strange. Upon deletion of the new local user account, the administrator account appeared again. Can anyone tell me what's going on?

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• Why is the reported user "root" when a "normal user" executes "ps ux" on OSX? Is this normal behavio

  - by snies

  I am running OS X 10.6.1 . When i am logged in as a normal user of group staff and do a ps ux it lists my ps ux command as being run by root: snies 181 0.0 0.3 2774328 12500 ?? S 6:00PM 0:20.96 /System/Library... root 1673 0.0 0.0 2434788 508 s001 R+ 8:16AM 0:00.00 ps ux snies 177 0.0 0.0 2457208 984 ?? Ss 6:00PM 0:00.52 /sbin/launchd snies 1638 0.0 0.0 2435468 1064 s001 S 8:13AM 0:00.03 -bash Is this normal behaviour? And if so why? Please note that the user is not an Administrator account and is not able to sudo.

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• Copying OS X account settings to new user account

  - by Donna Crain

  Hi I hope someone can help me... I accidentally renamed my own user name on my mac running OS X 10.4.11 and tried renaming back to what it was but now I have two accounts (the one I want to use which has all my preferences, applications, mail boxes, etc etc) is it possible to copy my settings from my original into this new user name? When I open my computer it defaults to this new account but everything is setting up from new and I just need to get access to all my work documents, mail settings etc. If there is anyone who could help out, I would be very grateful. Thank you in advance.

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• Switch user from locked screensaver?

  - by desert69

  I'm having the same problem as this, but the answer from there doesn't work for me. Neither in my iMac upgraded from Lion to Mountain Lion (currently at 10.8), nor from a Mac Mini with Lion Server (10.7.4). I have "Show fast user switching menu as" enabled at "Users & Groups" Login Options, but when I have to login from screensaver, there's no option to switch user. How can I enable such an option? EDIT: I think I forgot to mention that we believe (just guessing) that it has to do with network accounts. We use network accounts here, and have that issue. My boss says he could solve this issue at his house's Mac, which don't use network accounts. Does it make a difference?

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• Temp profile used when User logs in

  - by TJ

  One of my users logged into his computer, Windows XP, last night only to be meet with an error message that it could not load his profile and he will be logged in using a temporary profile. Typically when this happens I shut the machine down and restart and the correct profile will load when they log in again. Not this time. In the user profile options under computer-properties-Advanced-user profiles it show that there are three profiles with his name. Two are the exact same size with the same modified date (5/5/10) and the other is what I would expect size wise for a new profile with a modified date of today. What are my options to restore his profile?

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• User account shows two Downloads folders

  - by Chris Lieb

  I have my user account on my D drive and junction'd to the C:\users folder. I accidentally moved my profile Downloads folder (C:\users\me\Downloads) and then moved it back to its path on the D drive (C:\me\Downloads). After doing this, the directory tree for my user profile lists two Downloads directories, one located at C:\users\me and one at D:\me. I tried deleting the directory from the D drive, then restoring it from the Recycle Bin to the proper location on the C drive (actually the D drive, accessed through the junction), but it gave me the two Downloads directories again. Is there some way to fix this so that the only listing is for the C:\users\me\Downloads directory, like it was to begin with?

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• Remove Live ID authentication from user account

  - by slugster

  I've just run in to a really annoying issue with Windows 8.1 - it seems I cannot remove the need to use Live ID credentials from an account without completely deleting that account. I know the process to do it - use the Disconnect link from the Accounts-Your account screen. The trouble comes when you get to the Switch to a local account screen, it will not let you enter the current account for the user name, instead you must enter a new one thus creating a new user account. Can I revert back to using just a local login without having to recreate the account? It seems quite retarded that I have to recreate the account, as deep down the only change required is which credential provider is used to authenticate the login. (Note that this Live ID linkage was created by using the Windows Store, not as a result of an upgrade from 8 to 8.1).

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• non-privileged normal user passing environment variables to /bin/login [closed]

  - by AAAAAAAA

  Suppose that in FreeBSD (or linux maybe) there is a non-privileged normal user (non-superuser). And there is a telnet standalone (I know that telnet is usually run under inetd) running under (owned by) this user. (Suppose that there was no original, root-owned telnet running.) This telnet server is programmed so that it does not check ld_* environment variables before passing it to /bin/login owned by root that has setuid set up. The question would be: 1. Will this telnet work? 2. If it does work, will it even be able to pass environment variables to /bin/login?

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• Cannot schedule task to run under domain user account other than current user

  - by Filburt

  On our Win 2008 machines I can't schedule tasks for domain users because the domain name does not resolve to network name but the AD dc name. The "network name" looks like ABCDEFGE-HIJKLM and the "dc" / "name" would look like ABCDEFGE-HIJKLMN. When selecting the domain user account the account qualifier will look like ABCDEFGE-HIJKLMN\task.user. This results in an "invalid account" error. When however keeping the currently logged in user it will display ABCDEFGE-HIJKLM\current.user. Does this behaviour result from the presumable "illegal" domain name? Is there a workaround for this? update I could of course log in as the desired domain account and create the task but since this account is a account used for running services I want to avoid creating a user profile on the machine.

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• How to create limited user accounts in Linux

  - by James Willson

  I want to create a user account for each of the key programs installed on my debian server. For example, for the following programs: Tomcat Nginx Supervisor PostgreSQL This seems to be recommended based on my reading online. However, I want to restrict these user accounts as much as possible, so that they dont have a shell login, dont have access to the other programs and are as limited as possible but still functional. Would anyone mind telling me how this could be achieved? My reading so far suggests this: echo "/usr/sbin/nologin" /etc/shells useradd -s /usr/sbin/nologin tomcat But I think there may be a more complete way of doing it. EDIT: I'm using debian squeeze

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• Limit user to one CPU core in windows 7

  - by Dean Ward

  I let my family members use my pc over rdp to play their flash-based games as their laptops overheat if they use them directly. I have it setup so I can use the pc at the same time as them. The pc has a quad core cpu and I would like to be able to assign one of those cores to the user logged in via RDP so that the other 3 cores are left alone. Is this possible? They login via a specific user account setup for the purpose. Thanks for any advice!

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• Link to user directory displayed with wrong name in start menu

  - by wierob

  In the start menu the link to my user directory is displayed with a wrong name e.g. foo When I click on the link in the start menu the explorer opens my correct user directory but the addressbar still names it foo. However, when I open a cmd from that directory the location is correctly shown as C:\Users\myUserName. Furthermore there is no C:\Users\foo directory. How can I fix this (i.e. ths link in the startmenu should be named myUserName)?

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• Certain SFTP user cannot connect

  - by trobrock

  I have my Ubuntu Server set up so users with the group of sftponly can connect with sftp, but have a shell of /bin/false, and they connect to their home directories. This is working fine with three of the user accounts I have. But I added a new user account today the same way that I added the others and it will not successfully connect. sftp -vvv user@hostname debug1: Next authentication method: password user@hostname's password: debug3: packet_send2: adding 48 (len 73 padlen 7 extra_pad 64) debug2: we sent a password packet, wait for reply debug1: Authentication succeeded (password). debug2: fd 5 setting O_NONBLOCK debug3: fd 6 is O_NONBLOCK debug1: channel 0: new [client-session] debug3: ssh_session2_open: channel_new: 0 debug2: channel 0: send open debug1: Requesting [email protected] debug1: Entering interactive session. debug1: channel 0: free: client-session, nchannels 1 debug3: channel 0: status: The following connections are open: #0 client-session (t3 r-1 i0/0 o0/0 fd 5/6 cfd -1) debug3: channel 0: close_fds r 5 w 6 e 7 c -1 debug1: fd 0 clearing O_NONBLOCK debug3: fd 1 is not O_NONBLOCK Connection to hostname closed by remote host. Transferred: sent 2176, received 1848 bytes, in 0.0 seconds Bytes per second: sent 127453.3, received 108241.6 debug1: Exit status -1 Connection closed For a successful user: sftp -vvv good_user@hostname debug1: Next authentication method: password good_user@hostname's password: debug3: packet_send2: adding 48 (len 63 padlen 17 extra_pad 64) debug2: we sent a password packet, wait for reply debug1: Authentication succeeded (password). debug2: fd 5 setting O_NONBLOCK debug3: fd 6 is O_NONBLOCK debug1: channel 0: new [client-session] debug3: ssh_session2_open: channel_new: 0 debug2: channel 0: send open debug1: Requesting [email protected] debug1: Entering interactive session. debug2: callback start debug2: client_session2_setup: id 0 debug1: Sending subsystem: sftp debug2: channel 0: request subsystem confirm 1 debug2: fd 3 setting TCP_NODELAY debug2: callback done debug2: channel 0: open confirm rwindow 0 rmax 32768 debug2: channel 0: rcvd adjust 2097152 debug2: channel_input_status_confirm: type 99 id 0 debug2: subsystem request accepted on channel 0 debug2: Remote version: 3 debug2: Server supports extension "[email protected]" revision 1 debug2: Server supports extension "[email protected]" revision 2 debug2: Server supports extension "[email protected]" revision 2 debug3: Sent message fd 3 T:16 I:1 debug3: SSH_FXP_REALPATH . -> / sftp> I cannot figure out why one user will work and the other wont, I have restart the ssh service after adding the user. I have even removed the user and added them again to be sure I am adding it correctly.

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