Search Results

Search found 21131 results on 846 pages for 'jaircazarin old account'.

Page 52/846 | < Previous Page | 48 49 50 51 52 53 54 55 56 57 58 59  | Next Page >

  • Why do old programming languages continue to be revised?

    - by SunAvatar
    This question is not, "Why do people still use old programming languages?" I understand that quite well. In fact the two programming languages I know best are C and Scheme, both of which date back to the 70s. Recently I was reading about the changes in C99 and C11 versus C89 (which seems to still be the most-used version of C in practice and the version I learned from K&R). Looking around, it seems like every programming language in heavy use gets a new specification at least once per decade or so. Even Fortran is still getting new revisions, despite the fact that most people using it are still using FORTRAN 77. Contrast this with the approach of, say, the typesetting system TeX. In 1989, with the release of TeX 3.0, Donald Knuth declared that TeX was feature-complete and future releases would contain only bug fixes. Even beyond this, he has stated that upon his death, "all remaining bugs will become features" and absolutely no further updates will be made. Others are free to fork TeX and have done so, but the resulting systems are renamed to indicate that they are different from the official TeX. This is not because Knuth thinks TeX is perfect, but because he understands the value of a stable, predictable system that will do the same thing in fifty years that it does now. Why do most programming language designers not follow the same principle? Of course, when a language is relatively new, it makes sense that it will go through a period of rapid change before settling down. And no one can really object to minor changes that don't do much more than codify existing pseudo-standards or correct unintended readings. But when a language still seems to need improvement after ten or twenty years, why not just fork it or start over, rather than try to change what is already in use? If some people really want to do object-oriented programming in Fortran, why not create "Objective Fortran" for that purpose, and leave Fortran itself alone? I suppose one could say that, regardless of future revisions, C89 is already a standard and nothing stops people from continuing to use it. This is sort of true, but connotations do have consequences. GCC will, in pedantic mode, warn about syntax that is either deprecated or has a subtly different meaning in C99, which means C89 programmers can't just totally ignore the new standard. So there must be some benefit in C99 that is sufficient to impose this overhead on everyone who uses the language. This is a real question, not an invitation to argue. Obviously I do have an opinion on this, but at the moment I'm just trying to understand why this isn't just how things are done already. I suppose the question is: What are the (real or perceived) advantages of updating a language standard, as opposed to creating a new language based on the old?

    Read the article

  • Is there a limit of emails/pictures per Gravatar account?

    - by Steve Taylor
    I'm building a site to connect patients to doctors. Each doctor will have a profile picture. I'm quite happy to manually maintain the profile pictures as there won't be that many doctors nor will they have a need to change their picture very often, if at all. I thought of using Gravatar to host all these profile pictures. The idea is to create a single Gravatar account then keep adding email addresses to it in the form [email protected] and associating each one with a new image. Does anyone know, however, if I will run into any per-account limit? If so, it wouldn't be feasible because I would end up with a bunch of Gravatar accounts instead of just the one.

    Read the article

  • ssh Prompts For Password After Account Unlocked - Despite ssh key?

    - by user1011471
    Here's what happened: I set up ssh key so that user could ssh from A to B without a password. I got user's password wrong in some other context too many times, and user's account got locked out. (IT uses Active Directory here) IT unlocked the account. Concurrent to the unlocking, a script was running, calling something like ssh user@B some-health-check-command every 5 seconds or so -- which seemed to work fine before I caused user to get locked out in step 2. IT reports user reliably gets locked out a short time after each unlock attempt. I thought the ssh key would allow ssh user@B some-command as long as the account is not locked. But it behaves as if, when user gets unlocked, B suddenly asks for a password and since my command repeatedly runs without supplying a password, the account gets locked out after 5 attempts. Account cannot be accessed at this time. Please contact your system administrator. My questions are... Is that what's happening? Or: what's happening? More importantly: How can I reconfigure things such that my script doesn't cause problems? Can I accomplish what I want without having to install Expect? (I don't know if I have permission to do so) Other notes: Not using ssh-agent currently. The ssh command is running on our Jenkins master, a linux box. A and B are Mac OS X. user is managed in Active Directory and normally can sign into all three machines. Other than these things and the ssh key I set up, everything else has the default configuration as far as I know.

    Read the article

  • Associate email account with "Personal Folders" Outlook data file?

    - by TheLQ
    In the process of migrating email servers I've run into an interesting problem: In Outlook 2007 you have the default "Personal Folders" item. This contains the email for the account that was origionally setup with Outlook. My issue is that I have deleted the account associated with that and created an entirely new account. So now I have "Personal Folders" and "[email protected]". However I can't delete "Personal Folders". nor associate "[email protected]" with that PST file. Deleteting it in Outlook (Tools Account Settings Data Files) gave the error "The default data file cannot be removed, because it is your default delivery location. After you have selected a different default delivery location, your current file can be removed." Deleting the PST file itself (outlook.pst) made outlook demand where its default file . would be. So I selected my "[email protected]" PST file and restarted Outlook. Now "Personal Folders" is called "[email protected]", but I still have a duplicate account called this. Which is bad. Worse, my email is associated with the duplicate PST, not the default. How can I associate my email with my default PST or delete the default PST entirely? Luckily I have backu

    Read the article

  • Is it necessary to connect via Cisco VPN using my university account to go online?

    - by stankowait
    Original title: Bug with Xisco VPN - it it necessary to connect via Cisco VPN using my university-account in order to go online for all wireless networks In order to gain wireless access to my university network, I had to download and install the Cisco VPN client. It worked fine under 11.10 and did so for two weeks on 12.04. But since yesterday, I am unable to connect to my wireless network at home. First I have to connect via Cisco VPN using my university account. This is quite annoying and I'm unable to download apps via the software center when using the Cisco VPN client. I really don't know what happened, because it worked fine for two weeks and I did not change a thing.

    Read the article

  • Can the Windows 8 Live SDK use another Microsoft Account other than the current user?

    - by Jerry Nixon
    Using the Windows 8 Live SDK you can have a user give you permission to their Microsoft Account. With this you can get their name and photo and more. But using the Live SDK appears to require the user of the app to use the same Microsoft Account as whoever is signed into the current session of Windows 8. In some scenarios, using a different account is very legitimate. I have simple sign-in working like a charm! This uses the same account. I can't find a way to do use another. Is it possible?

    Read the article

  • My Quicken 401(K) account has changed to Checking. How do I fix this?

    - by user36492
    This is actually the second time this has happened to me, but I don't remember what I did last time (nor can I find the original forum post that helped then). I'm using Quicken Mac 2007. My 401(k) account, previously properly set up, has changed, seemingly irrevocably, to a Checking account. When I click "Edit" and try to change the account type, the 401(k) option is grayed out. I've got years of data stored in this account, so I am really hoping there's a way to salvage this data file!

    Read the article

  • Is it possible to allow the User Account to access the 'Volumes' without asking the Administrator's password?

    - by Tom
    Am the administrator of my Ubuntu system. Recently I added a new user account. But when ever the user tries to access or open the 'Volumes'(Drives where movies, songs and other files are stored) it asks for the Administrator's password. I created the user account to my other family members and I don't want to tell them my password. So is it possible to allow them to access the Volumes without asking Administrator's password ? UPDATE : Ubuntu was installed alongside Windows in my system. I will provide a screenshot of the Volume details -

    Read the article

  • How do I sort account names alphabetically in Thunderbird?

    - by Sri
    I just upgraded to Ubuntu 12.04 and Thunderbird 13.0.1. I had 2 accounts earlier in Thunderbird: [email protected] [email protected] I also had another account on SeaMonkey: [email protected] which I imported into Thunderbird. Now the account order I want is: [email protected] [email protected] [email protected] but it shows: [email protected] [email protected] [email protected] I couldn't find any option in Thunderbird to sort them as I want. I came across a 3rd party extension but I avoid using such extensions. Is there any other way this can be done?

    Read the article

  • How can I disable the guest account on OSX Lion?

    - by Wezly
    I have 'disabled' the guest account on my macbook pro running Lion via System Preferences Users & Groups. However the guest account still seems to appear as an option to login at start up and when switching users. I have never used the guest account for anything, and I have tried a system restore but the guest user has returned as before. How can I get rid of it? Thanks. edit: I also just enabled and disabled the account again - but the guest option still appears at startup offering a safari only restart for a guest user.

    Read the article

  • What is the difference between a "service account" and an "installed application"?

    - by TheBeatlemaniac
    To my understanding, the main difference is that a service account doesn't require a user to log in for authorization, while an installed application does. I am making an Android app (an "installed application"?) that offers an in-app subscription, and doesn't require the user to log in to an account (a "service account"?). To get a Client ID for the Google Play Developer API, I have to declare it as either an installed application or a service account, and am unsure which to go with.

    Read the article

  • Linux Kernel not upgraded (from Ubuntu 12.04 to 12.10) - can't remove old kernels and can't install new apps

    - by Tony Breyal
    Question: How do I remove old kernel images which refuse to be removed? Context: Yesterday I upgraded Ubuntu from 12.04 to 12.10. However, the linux kernel has not upgraded from 3.2 to 3.5 as I would have expected. $ uname -r 3.2.0-32-generic $ uname -a Linux tony-b 3.2.0-32-generic #51-Ubuntu SMP Wed Sep 26 21:33:09 UTC 2012 x86_64 x86_64 x86_64 GNU/Linux $ cat /proc/version Linux version 3.2.0-32-generic (buildd@batsu) (gcc version 4.6.3 (Ubuntu/Linaro 4.6.3-1ubuntu5) ) #51-Ubuntu SMP Wed Sep 26 21:33:09 UTC 2012 Not sure why that happened there. I wanted to install Audacity (v2.0.1-1_amd64) to edit a lecture audio file. When trying this operation through Ubuntu Software Center, it says that to install audacity, four items will need to be removed: linux-image-3.2.0-27-generic linux-image-3.2.0-29-generic linux-image-3.2.0-30-generic linux-image-3.2.0-31-generic So I click "Install Anyway" but it fails with the following output: installArchives() failed: (Reading database ... (Reading database ... 5% (Reading database ... 10% (Reading database ... 15% (Reading database ... 20% (Reading database ... 25% (Reading database ... 30% (Reading database ... 35% (Reading database ... 40% (Reading database ... 45% (Reading database ... 50% (Reading database ... 55% (Reading database ... 60% (Reading database ... 65% (Reading database ... 70% (Reading database ... 75% (Reading database ... 80% (Reading database ... 85% (Reading database ... 90% (Reading database ... 95% (Reading database ... 100% (Reading database ... 259675 files and directories currently installed.) Removing linux-image-3.2.0-27-generic ... Examining /etc/kernel/postrm.d . run-parts: executing /etc/kernel/postrm.d/initramfs-tools 3.2.0-27-generic /boot/vmlinuz-3.2.0-27-generic update-initramfs: Deleting /boot/initrd.img-3.2.0-27-generic run-parts: executing /etc/kernel/postrm.d/zz-update-grub 3.2.0-27-generic /boot/vmlinuz-3.2.0-27-generic Generating grub.cfg ... run-parts: /etc/kernel/postrm.d/zz-update-grub exited with return code 1 Failed to process /etc/kernel/postrm.d at /var/lib/dpkg/info/linux-image-3.2.0-27-generic.postrm line 328. dpkg: error processing linux-image-3.2.0-27-generic (--remove): subprocess installed post-removal script returned error exit status 1 No apport report written because MaxReports is reached already Removing linux-image-3.2.0-29-generic ... Examining /etc/kernel/postrm.d . run-parts: executing /etc/kernel/postrm.d/initramfs-tools 3.2.0-29-generic /boot/vmlinuz-3.2.0-29-generic update-initramfs: Deleting /boot/initrd.img-3.2.0-29-generic run-parts: executing /etc/kernel/postrm.d/zz-update-grub 3.2.0-29-generic /boot/vmlinuz-3.2.0-29-generic Generating grub.cfg ... run-parts: /etc/kernel/postrm.d/zz-update-grub exited with return code 1 Failed to process /etc/kernel/postrm.d at /var/lib/dpkg/info/linux-image-3.2.0-29-generic.postrm line 328. dpkg: error processing linux-image-3.2.0-29-generic (--remove): subprocess installed post-removal script returned error exit status 1 No apport report written because MaxReports is reached already Removing linux-image-3.2.0-30-generic ... Examining /etc/kernel/postrm.d . run-parts: executing /etc/kernel/postrm.d/initramfs-tools 3.2.0-30-generic /boot/vmlinuz-3.2.0-30-generic update-initramfs: Deleting /boot/initrd.img-3.2.0-30-generic run-parts: executing /etc/kernel/postrm.d/zz-update-grub 3.2.0-30-generic /boot/vmlinuz-3.2.0-30-generic Generating grub.cfg ... run-parts: /etc/kernel/postrm.d/zz-update-grub exited with return code 1 Failed to process /etc/kernel/postrm.d at /var/lib/dpkg/info/linux-image-3.2.0-30-generic.postrm line 328. dpkg: error processing linux-image-3.2.0-30-generic (--remove): subprocess installed post-removal script returned error exit status 1 No apport report written because MaxReports is reached already Removing linux-image-3.2.0-31-generic ... Examining /etc/kernel/postrm.d . run-parts: executing /etc/kernel/postrm.d/initramfs-tools 3.2.0-31-generic /boot/vmlinuz-3.2.0-31-generic update-initramfs: Deleting /boot/initrd.img-3.2.0-31-generic run-parts: executing /etc/kernel/postrm.d/zz-update-grub 3.2.0-31-generic /boot/vmlinuz-3.2.0-31-generic Generating grub.cfg ... run-parts: /etc/kernel/postrm.d/zz-update-grub exited with return code 1 Failed to process /etc/kernel/postrm.d at /var/lib/dpkg/info/linux-image-3.2.0-31-generic.postrm line 328. dpkg: error processing linux-image-3.2.0-31-generic (--remove): subprocess installed post-removal script returned error exit status 1 No apport report written because MaxReports is reached already Errors were encountered while processing: linux-image-3.2.0-27-generic linux-image-3.2.0-29-generic linux-image-3.2.0-30-generic linux-image-3.2.0-31-generic Error in function: Setting up grub-pc (2.00-7ubuntu11) ... /usr/sbin/grub-bios-setup: warning: Sector 32 is already in use by the program `FlexNet'; avoiding it. This software may cause boot or other problems in future. Please ask its authors not to store data in the boot track. Installation finished. No error reported. Generating grub.cfg ... dpkg: error processing grub-pc (--configure): subprocess installed post-installation script returned error exit status 1 It seems I need to remove the old linux images somehow. I have tried this through (1) Synaptic, (2) Ubuntu Tweak, and (3) Computer Janitor. The first two fail, whilst Computer Janitor won't even open. The output from Synaptic is: E: linux-image-3.2.0-27-generic: subprocess installed post-removal script returned error exit status 1 E: linux-image-3.2.0-29-generic: subprocess installed post-removal script returned error exit status 1 E: linux-image-3.2.0-30-generic: subprocess installed post-removal script returned error exit status 1 E: linux-image-3.2.0-31-generic: subprocess installed post-removal script returned error exit status 1 How do I remove these old images? Thank you kindly in advance for any help on this matter. P.S. Further information: $ dpkg --list | grep linux-image rH linux-image-3.2.0-27-generic 3.2.0-27.43 amd64 Linux kernel image for version 3.2.0 on 64 bit x86 SMP rH linux-image-3.2.0-29-generic 3.2.0-29.46 amd64 Linux kernel image for version 3.2.0 on 64 bit x86 SMP rH linux-image-3.2.0-30-generic 3.2.0-30.48 amd64 Linux kernel image for version 3.2.0 on 64 bit x86 SMP rH linux-image-3.2.0-31-generic 3.2.0-31.50 amd64 Linux kernel image for version 3.2.0 on 64 bit x86 SMP ii linux-image-3.2.0-32-generic 3.2.0-32.51 amd64 Linux kernel image for version 3.2.0 on 64 bit x86 SMP ii linux-image-3.5.0-17-generic 3.5.0-17.28 amd64 Linux kernel image for version 3.5.0 on 64 bit x86 SMP ii linux-image-extra-3.5.0-17-generic 3.5.0-17.28 amd64 Linux kernel image for version 3.5.0 on 64 bit x86 SMP ii linux-image-generic 3.5.0.17.19 amd64 Generic Linux kernel image But trying to remove using the command line fails too e.g.: $ sudo apt-get purge linux-image-3.2.0-27-generic Reading package lists... Done Building dependency tree Reading state information... Done The following packages will be REMOVED linux-image-3.2.0-27-generic linux-image-3.2.0-29-generic linux-image-3.2.0-30-generic linux-image-3.2.0-31-generic 0 upgraded, 0 newly installed, 4 to remove and 1 not upgraded. 5 not fully installed or removed. After this operation, 597 MB disk space will be freed. Do you want to continue [Y/n]? Y (Reading database ... 259675 files and directories currently installed.) Removing linux-image-3.2.0-27-generic ... Examining /etc/kernel/postrm.d . run-parts: executing /etc/kernel/postrm.d/initramfs-tools 3.2.0-27-generic /boot/vmlinuz-3.2.0-27-generic update-initramfs: Deleting /boot/initrd.img-3.2.0-27-generic run-parts: executing /etc/kernel/postrm.d/zz-update-grub 3.2.0-27-generic /boot/vmlinuz-3.2.0-27-generic Generating grub.cfg ... run-parts: /etc/kernel/postrm.d/zz-update-grub exited with return code 1 Failed to process /etc/kernel/postrm.d at /var/lib/dpkg/info/linux-image-3.2.0-27-generic.postrm line 328. dpkg: error processing linux-image-3.2.0-27-generic (--remove): subprocess installed post-removal script returned error exit status 1 No apport report written because MaxReports has already been reached Removing linux-image-3.2.0-29-generic ... Examining /etc/kernel/postrm.d . run-parts: executing /etc/kernel/postrm.d/initramfs-tools 3.2.0-29-generic /boot/vmlinuz-3.2.0-29-generic update-initramfs: Deleting /boot/initrd.img-3.2.0-29-generic run-parts: executing /etc/kernel/postrm.d/zz-update-grub 3.2.0-29-generic /boot/vmlinuz-3.2.0-29-generic Generating grub.cfg ... run-parts: /etc/kernel/postrm.d/zz-update-grub exited with return code 1 Failed to process /etc/kernel/postrm.d at /var/lib/dpkg/info/linux-image-3.2.0-29-generic.postrm line 328. dpkg: error processing linux-image-3.2.0-29-generic (--remove): subprocess installed post-removal script returned error exit status 1 No apport report written because MaxReports has already been reached Removing linux-image-3.2.0-30-generic ... Examining /etc/kernel/postrm.d . run-parts: executing /etc/kernel/postrm.d/initramfs-tools 3.2.0-30-generic /boot/vmlinuz-3.2.0-30-generic update-initramfs: Deleting /boot/initrd.img-3.2.0-30-generic run-parts: executing /etc/kernel/postrm.d/zz-update-grub 3.2.0-30-generic /boot/vmlinuz-3.2.0-30-generic Generating grub.cfg ... run-parts: /etc/kernel/postrm.d/zz-update-grub exited with return code 1 Failed to process /etc/kernel/postrm.d at /var/lib/dpkg/info/linux-image-3.2.0-30-generic.postrm line 328. dpkg: error processing linux-image-3.2.0-30-generic (--remove): subprocess installed post-removal script returned error exit status 1 No apport report written because MaxReports has already been reached Removing linux-image-3.2.0-31-generic ... Examining /etc/kernel/postrm.d . run-parts: executing /etc/kernel/postrm.d/initramfs-tools 3.2.0-31-generic /boot/vmlinuz-3.2.0-31-generic update-initramfs: Deleting /boot/initrd.img-3.2.0-31-generic run-parts: executing /etc/kernel/postrm.d/zz-update-grub 3.2.0-31-generic /boot/vmlinuz-3.2.0-31-generic Generating grub.cfg ... run-parts: /etc/kernel/postrm.d/zz-update-grub exited with return code 1 Failed to process /etc/kernel/postrm.d at /var/lib/dpkg/info/linux-image-3.2.0-31-generic.postrm line 328. dpkg: error processing linux-image-3.2.0-31-generic (--remove): subprocess installed post-removal script returned error exit status 1 No apport report written because MaxReports has already been reached Errors were encountered while processing: linux-image-3.2.0-27-generic linux-image-3.2.0-29-generic linux-image-3.2.0-30-generic linux-image-3.2.0-31-generic E: Sub-process /usr/bin/dpkg returned an error code (1)

    Read the article

  • an error "has no member named"

    - by helloWorld
    I have this snippet of the code account.cpp #include "account.h" #include <iostream> #include <string> using namespace std; Account::Account(string firstName, string lastName, int id) : strFirstName(firstName), strLastName(lastName), nID(id) {} void Account::printAccount(){ cout << strFirstName; } account.h #include <string> using std::string; class Account{ private: string strLastName; //Client's last name string strFirstName; //Client's first name int nID; //Client's ID number int nLines; //Number of lines related to account double lastBill; public: Account(string firstName, string lastName, int id); void printAccount(); }; company.h #ifndef CELLULAR_COMPANY_H #define CELLULAR_COMPANY_H #include <string> #include <list> #include <iostream> #include "account.h" using namespace std; class Company { private: list<Account> listOfAccounts; public: void addAccount(string firstName, string lastName, int id) { Account newAccount(firstName, lastName, id); listOfAccounts.push_back(newAccount); } void printAccounts(){ for(list<Account>::iterator i = listOfAccounts.begin(); i != listOfAccounts.end(); ++i){ i.printAccount; //here bug } } }; #endif // CELLULAR_COMPANY_H main.cpp #include "cellularcompany.h" int main(){ Company newCompany; newCompany.addAccount("Pavel", "Nedved", 11111); newCompany.printAccounts(); return 0; } can somebody please explain what does my error mean? thanks in advance (I have it in company.h see comment there) I have bug 'struct std::_List_iterator<Account>' has no member named 'printAccount'

    Read the article

  • Spring security - Reach users ID without passing it through every controller

    - by nilsi
    I have a design issue that I don't know how to solve. I'm using Spring 3.2.4 and Spring security 3.1.4. I have a Account table in my database that looks like this: create table Account (id identity, username varchar unique, password varchar not null, firstName varchar not null, lastName varchar not null, university varchar not null, primary key (id)); Until recently my username was just only a username but I changed it to be the email address instead since many users want to login with that instead. I have a header that I include on all my pages which got a link to the users profile like this: <a href="/project/users/<%= request.getUserPrincipal().getName()%>" class="navbar-link"><strong><%= request.getUserPrincipal().getName()%></strong></a> The problem is that <%= request.getUserPrincipal().getName()%> returns the email now, I don't want to link the user's with thier emails. Instead I want to use the id every user have to link to the profile. How do I reach the users id's from every page? I have been thinking of two solutions but I'm not sure: Change the principal to contain the id as well, don't know how to do this and having problem finding good information on the topic. Add a model attribute to all my controllers that contain the whole user but this would be really ugly, like this. Account account = entityManager.find(Account.class, email); model.addAttribute("account", account); There are more way's as well and I have no clue which one is to prefer. I hope it's clear enough and thank you for any help on this. ====== Edit according to answer ======= I edited Account to implement UserDetails, it now looks like this (will fix the auto generated stuff later): @Entity @Table(name="Account") public class Account implements UserDetails { @Id private int id; private String username; private String password; private String firstName; private String lastName; @ManyToOne private University university; public Account() { } public Account(String username, String password, String firstName, String lastName, University university) { this.username = username; this.password = password; this.firstName = firstName; this.lastName = lastName; this.university = university; } public String getUsername() { return username; } public String getPassword() { return password; } public String getFirstName() { return firstName; } public String getLastName() { return lastName; } public void setUsername(String username) { this.username = username; } public void setPassword(String password) { this.password = password; } public void setFirstName(String firstName) { this.firstName = firstName; } public void setLastName(String lastName) { this.lastName = lastName; } public University getUniversity() { return university; } public void setUniversity(University university) { this.university = university; } public int getId() { return id; } public void setId(int id) { this.id = id; } @Override public Collection<? extends GrantedAuthority> getAuthorities() { // TODO Auto-generated method stub return null; } @Override public boolean isAccountNonExpired() { // TODO Auto-generated method stub return false; } @Override public boolean isAccountNonLocked() { // TODO Auto-generated method stub return false; } @Override public boolean isCredentialsNonExpired() { // TODO Auto-generated method stub return false; } @Override public boolean isEnabled() { // TODO Auto-generated method stub return true; } } I also added <%@ taglib prefix="sec" uri="http://www.springframework.org/security/tags" %> To my jsp files and trying to reach the id by <sec:authentication property="principal.id" /> This gives me the following org.springframework.beans.NotReadablePropertyException: Invalid property 'principal.id' of bean class [org.springframework.security.authentication.UsernamePasswordAuthenticationToken]: Bean property 'principal.id' is not readable or has an invalid getter method: Does the return type of the getter match the parameter type of the setter? ====== Edit 2 according to answer ======= I based my application on spring social samples and I never had to change anything until now. This are the files I think are relevant, please tell me if theres something you need to see besides this. AccountRepository.java public interface AccountRepository { void createAccount(Account account) throws UsernameAlreadyInUseException; Account findAccountByUsername(String username); } JdbcAccountRepository.java @Repository public class JdbcAccountRepository implements AccountRepository { private final JdbcTemplate jdbcTemplate; private final PasswordEncoder passwordEncoder; @Inject public JdbcAccountRepository(JdbcTemplate jdbcTemplate, PasswordEncoder passwordEncoder) { this.jdbcTemplate = jdbcTemplate; this.passwordEncoder = passwordEncoder; } @Transactional public void createAccount(Account user) throws UsernameAlreadyInUseException { try { jdbcTemplate.update( "insert into Account (firstName, lastName, username, university, password) values (?, ?, ?, ?, ?)", user.getFirstName(), user.getLastName(), user.getUsername(), user.getUniversity(), passwordEncoder.encode(user.getPassword())); } catch (DuplicateKeyException e) { throw new UsernameAlreadyInUseException(user.getUsername()); } } public Account findAccountByUsername(String username) { return jdbcTemplate.queryForObject("select username, firstName, lastName, university from Account where username = ?", new RowMapper<Account>() { public Account mapRow(ResultSet rs, int rowNum) throws SQLException { return new Account(rs.getString("username"), null, rs.getString("firstName"), rs.getString("lastName"), new University("test")); } }, username); } } security.xml <?xml version="1.0" encoding="UTF-8"?> <beans:beans xmlns="http://www.springframework.org/schema/security" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:beans="http://www.springframework.org/schema/beans" xsi:schemaLocation="http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.1.xsd http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd"> <http pattern="/resources/**" security="none" /> <http pattern="/project/" security="none" /> <http use-expressions="true"> <!-- Authentication policy --> <form-login login-page="/signin" login-processing-url="/signin/authenticate" authentication-failure-url="/signin?error=bad_credentials" /> <logout logout-url="/signout" delete-cookies="JSESSIONID" /> <intercept-url pattern="/addcourse" access="isAuthenticated()" /> <intercept-url pattern="/courses/**/**/edit" access="isAuthenticated()" /> <intercept-url pattern="/users/**/edit" access="isAuthenticated()" /> </http> <authentication-manager alias="authenticationManager"> <authentication-provider> <password-encoder ref="passwordEncoder" /> <jdbc-user-service data-source-ref="dataSource" users-by-username-query="select username, password, true from Account where username = ?" authorities-by-username-query="select username, 'ROLE_USER' from Account where username = ?"/> </authentication-provider> <authentication-provider> <user-service> <user name="admin" password="admin" authorities="ROLE_USER, ROLE_ADMIN" /> </user-service> </authentication-provider> </authentication-manager> </beans:beans> And this is my try of implementing a UserDetailsService public class RepositoryUserDetailsService implements UserDetailsService { private final AccountRepository accountRepository; @Autowired public RepositoryUserDetailsService(AccountRepository repository) { this.accountRepository = repository; } @Override public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException { Account user = accountRepository.findAccountByUsername(username); if (user == null) { throw new UsernameNotFoundException("No user found with username: " + username); } return user; } } Still gives me the same error, do I need to add the UserDetailsService somewhere? This is starting to be something else compared to my initial question, I should maybe start another question. Sorry for my lack of experience in this. I have to read up.

    Read the article

  • Linux: How to rename old mysqld when upgrading MySQL?

    - by Continuation
    I'm upgrading MySQL from MySQL 5.0 to Percona Server 5.1. I'm planning to just use yum remove and yum install to do the upgrade. However, I read in the documentation that it's a good idea to rename the old mysqld to mysqld-5.0. And if the upgrade doesn't work, I could just revert back to the old version. How exactly does this work? If I use yum remove, doesn't that mean the old mysqld is removed? So how do I rename it? Where is mysqld located? How do I find it? Thanks.

    Read the article

  • How to backup old emails locally in Thunderbird and then remove them from IMAP server?

    - by saicode
    I am using Godaddy IMAP email with Thunderbird as my desktop client on Windows 7. The email service has unlimited mailbox size but the local Thunderbird is having troubles due to the large size of the inbox/outbox. I would like to take out old emails from IMAP server and backup them locally. After backing up the old email I would like to delete old email (older than let's say, 2012) from the server. Also I'd like to have them accessible from the local backup if ever needed in the future. This way I might be able to make Thunderbird fast and problem free. Problem is, I am not able to find any instruction to do this in an automated way based on dates etc. I can find some links for Archiving, Compacting and Backup. But unable to find any tutorial about how to backup and archive it locally and delete the original emails from the server.

    Read the article

  • Find an old static IP address on a mac backup?

    - by Joe
    I have an old backup of a mac via time machine, and I was wondering if there is some sort of accessible file that would have the static IP of the old computer it was backed up from. If you are curious, I am in IT and it is a PITA to get a new static IP here. I am just trying to use the old one to save some hassle. I've tried searching through the system log with no success (but I might be doing it wrong- grep with a regexp for an IP). Thanks!

    Read the article

  • Django shows too many warnings when deleting an object

    - by valya
    Hello! I have two models: class Account(models.Model): main_request = models.ForeignKey('JournalistRequest', related_name='main_request') key = models.CharField(_('Key'), max_length=100) class JournalistRequest(models.Model): account = models.ForeignKey(Account, blank=True, null=True) When I try to delete a JournalistRequest, It shows warning with a lot of nesting, like Are you sure you want to delete the selected ?????? ??? objects? All of the following objects and their related items will be deleted: Journalist Request: some request Account: some account Journalist Request: some request Account: some account Journalist Request: some request Account: some account Journalist Request: some request Account: some account Journalist Request: some request All accounts are the same one (ids are same), and all requests are the same one, so I think it becaues of a recursion. But I have no idea how to solve this problem in Django 1.1.1! Can you help me?

    Read the article

  • Factory Girl Association

    - by David Lyod
    I have an association of a Admin - Account in factory girl I now wish to associate a second user with the same account but am unable to do so. I build my Admin-Account association like this u.account { |account| account.association(:account)} This works fine and creates the Account and Admin association. Im looking for a way to setup a second user who's account also points to the record created in the Admin factory association. I currently just build the second user as such @user = Factory.build(:seconduser) @user.account = Account.first @user.save! Which works but seems somewhat hacky .

    Read the article

  • What is the functionality of "sync contacts" in Exchange account in Email application?

    - by santhosh
    Hi i am testing android E-mail application . I have configured an Exchange account where in i could find an option "Sync Contacts from this account" in Account settings. According to my understanding if i check "sync contacts from the account" option , i must be able to access contacts in the exchange account i have configured. But i don't know how to get/access these contacts in android email application. Can any one who have used this functionality or know about it can suggest to me how to make use of "Sync contacts" functionality. Or if you have any idea about, how i can test this functionality, i am very eager to here to you. Kinds & Regards Santhosh Kumar H.E

    Read the article

  • Which is Better? The Start Screen in Windows 8 or the Old Start Menu? [Analysis]

    - by Asian Angel
    There has been quite a bit of controversy surrounding Microsoft’s emphasis on the new Metro UI Start Screen in Windows 8, but when it comes down to it which is really better? The Start Screen in Windows 8 or the old Start Menu? Tech blog 7 Tutorials has done a quick analysis to see which one actually works better (and faster) when launching applications and doing searches. Images courtesy of 7 Tutorials. You can view the results and a comparison table by visiting the blog post linked below. Windows 8 Analysis: Is the Start Screen an Improvement vs. the Start Menu? [7 Tutorials] How to Stress Test the Hard Drives in Your PC or Server How To Customize Your Android Lock Screen with WidgetLocker The Best Free Portable Apps for Your Flash Drive Toolkit

    Read the article

  • Why does bash sometimes think my $HOME isn't the correct directory?

    - by Adam Yanalunas
    Like the title says it seems that bash sometimes misidentifies my $HOME. This cropped up after a seemingly unique series of events that I will now replay in broad strokes. Running OS X 10.6 with normal, local account Work binds my account to Active Directory Much time passes with no issues Set up rvm to manage Ruby installs (this becomes important later) Upgraded to OS X 10.7 a few days ago After successful install, attempted to log in, was presented with "Must reset password" dialog that never allowed a password to be reset. Would simply shake the box after new password was entered. Much googling was done. Much more googling was done. Swearing was had. Logged in as root, created new account, set as admin, deleted /Users/[new account], renamed /Users/[old account] to /Users/[new account] Logged out of root, logged into new account with no issues After OS X asking for a my account password a few times to update Keychain and other system-level stuff it was back to business as usual. Opened Terminal, cd to project folder, tried "rails server" and was presented with: /usr/local/lib/ruby/1.9.1/rubygems/dependency.rb:247:in to_specs': Could not find rails (>= 0) amongst [] (Gem::LoadError) from /usr/local/lib/ruby/1.9.1/rubygems/dependency.rb:256:into_spec' from /usr/local/lib/ruby/1.9.1/rubygems.rb:1210:in gem' from /usr/local/bin/rails:18:in' Ran through a few exercises, decided to rm -rf ~/.rvm and reinstall. Running a --trace on the rvm installer shows it dies on this line: mkdir: /Users/[old account]: Permission denied Scrolling back through the --trace log I see many more mentions of /Users/[old account]. When inspect the install script the offending line is looking at "${HOME}/.rvm" as it tries to run the mkdir. To my confusion I also see mentions of /Users/[new account] in the log. I've tried exporting a new HOME in my .bash_profile to no luck. Can anyone guess why /Users/[old account] would still be kicking around?

    Read the article

< Previous Page | 48 49 50 51 52 53 54 55 56 57 58 59  | Next Page >