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  • Creating a dynamic proxy generator with c# – Part 2 – Interceptor Design

    - by SeanMcAlinden
    Creating a dynamic proxy generator – Part 1 – Creating the Assembly builder, Module builder and caching mechanism For the latest code go to http://rapidioc.codeplex.com/ Before getting too involved in generating the proxy, I thought it would be worth while going through the intended design, this is important as the next step is to start creating the constructors for the proxy. Each proxy derives from a specified type The proxy has a corresponding constructor for each of the base type constructors The proxy has overrides for all methods and properties marked as Virtual on the base type For each overridden method, there is also a private method whose sole job is to call the base method. For each overridden method, a delegate is created whose sole job is to call the private method that calls the base method. The following class diagram shows the main classes and interfaces involved in the interception process. I’ll go through each of them to explain their place in the overall proxy.   IProxy Interface The proxy implements the IProxy interface for the sole purpose of adding custom interceptors. This allows the created proxy interface to be cast as an IProxy and then simply add Interceptors by calling it’s AddInterceptor method. This is done internally within the proxy building process so the consumer of the API doesn’t need knowledge of this. IInterceptor Interface The IInterceptor interface has one method: Handle. The handle method accepts a IMethodInvocation parameter which contains methods and data for handling method interception. Multiple classes that implement this interface can be added to the proxy. Each method override in the proxy calls the handle method rather than simply calling the base method. How the proxy fully works will be explained in the next section MethodInvocation. IMethodInvocation Interface & MethodInvocation class The MethodInvocation will contain one main method and multiple helper properties. Continue Method The method Continue() has two functions hidden away from the consumer. When Continue is called, if there are multiple Interceptors, the next Interceptors Handle method is called. If all Interceptors Handle methods have been called, the Continue method then calls the base class method. Properties The MethodInvocation will contain multiple helper properties including at least the following: Method Name (Read Only) Method Arguments (Read and Write) Method Argument Types (Read Only) Method Result (Read and Write) – this property remains null if the method return type is void Target Object (Read Only) Return Type (Read Only) DefaultInterceptor class The DefaultInterceptor class is a simple class that implements the IInterceptor interface. Here is the code: DefaultInterceptor namespace Rapid.DynamicProxy.Interception {     /// <summary>     /// Default interceptor for the proxy.     /// </summary>     /// <typeparam name="TBase">The base type.</typeparam>     public class DefaultInterceptor<TBase> : IInterceptor<TBase> where TBase : class     {         /// <summary>         /// Handles the specified method invocation.         /// </summary>         /// <param name="methodInvocation">The method invocation.</param>         public void Handle(IMethodInvocation<TBase> methodInvocation)         {             methodInvocation.Continue();         }     } } This is automatically created in the proxy and is the first interceptor that each method override calls. It’s sole function is to ensure that if no interceptors have been added, the base method is still called. Custom Interceptor Example A consumer of the Rapid.DynamicProxy API could create an interceptor for logging when the FirstName property of the User class is set. Just for illustration, I have also wrapped a transaction around the methodInvocation.Coninue() method. This means that any overriden methods within the user class will run within a transaction scope. MyInterceptor public class MyInterceptor : IInterceptor<User<int, IRepository>> {     public void Handle(IMethodInvocation<User<int, IRepository>> methodInvocation)     {         if (methodInvocation.Name == "set_FirstName")         {             Logger.Log("First name seting to: " + methodInvocation.Arguments[0]);         }         using (TransactionScope scope = new TransactionScope())         {             methodInvocation.Continue();         }         if (methodInvocation.Name == "set_FirstName")         {             Logger.Log("First name has been set to: " + methodInvocation.Arguments[0]);         }     } } Overridden Method Example To show a taster of what the overridden methods on the proxy would look like, the setter method for the property FirstName used in the above example would look something similar to the following (this is not real code but will look similar): set_FirstName public override void set_FirstName(string value) {     set_FirstNameBaseMethodDelegate callBase =         new set_FirstNameBaseMethodDelegate(this.set_FirstNameProxyGetBaseMethod);     object[] arguments = new object[] { value };     IMethodInvocation<User<IRepository>> methodInvocation =         new MethodInvocation<User<IRepository>>(this, callBase, "set_FirstName", arguments, interceptors);          this.Interceptors[0].Handle(methodInvocation); } As you can see, a delegate instance is created which calls to a private method on the class, the private method calls the base method and would look like the following: calls base setter private void set_FirstNameProxyGetBaseMethod(string value) {     base.set_FirstName(value); } The delegate is invoked when methodInvocation.Continue() is called within an interceptor. The set_FirstName parameters are loaded into an object array. The current instance, delegate, method name and method arguments are passed into the methodInvocation constructor (there will be more data not illustrated here passed in when created including method info, return types, argument types etc.) The DefaultInterceptor’s Handle method is called with the methodInvocation instance as it’s parameter. Obviously methods can have return values, ref and out parameters etc. in these cases the generated method override body will be slightly different from above. I’ll go into more detail on these aspects as we build them. Conclusion I hope this has been useful, I can’t guarantee that the proxy will look exactly like the above, but at the moment, this is pretty much what I intend to do. Always worth downloading the code at http://rapidioc.codeplex.com/ to see the latest. There will also be some tests that you can debug through to help see what’s going on. Cheers, Sean.

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  • PAM Winbind Expired Password

    - by kernelpanic
    We've got Winbind/Kerberos setup on RHEL for AD authentication. Working fine however I noticed that when a password has expired, we get a warning but shell access is still granted. What's the proper way of handling this? Can we tell PAM to close the session once it sees the password has expired? Example: login as: ad-user [email protected]'s password: Warning: password has expired. [ad-user@server ~]$ Contents of /etc/pam.d/system-auth: auth required pam_env.so auth sufficient pam_unix.so nullok try_first_pass auth requisite pam_succeed_if.so uid >= 500 quiet auth sufficient pam_krb5.so use_first_pass auth sufficient pam_winbind.so use_first_pass auth required pam_deny.so account [default=2 success=ignore] pam_succeed_if.so quiet uid >= 10000000 account sufficient pam_succeed_if.so user ingroup AD_Admins debug account requisite pam_succeed_if.so user ingroup AD_Developers debug account required pam_access.so account required pam_unix.so broken_shadow account sufficient pam_localuser.so account sufficient pam_succeed_if.so uid < 500 quiet account [default=bad success=ok user_unknown=ignore] pam_krb5.so account [default=bad success=ok user_unknown=ignore] pam_winbind.so account required pam_permit.so password requisite pam_cracklib.so try_first_pass retry=3 password sufficient pam_unix.so md5 shadow nullok try_first_pass use_authtok password sufficient pam_krb5.so use_authtok password sufficient pam_winbind.so use_authtok password required pam_deny.so session [default=2 success=ignore] pam_succeed_if.so quiet uid >= 10000000 session sufficient pam_succeed_if.so user ingroup AD_Admins debug session requisite pam_succeed_if.so user ingroup AD_Developers debug session optional pam_mkhomedir.so umask=0077 skel=/etc/skel session optional pam_keyinit.so revoke session required pam_limits.so session optional pam_mkhomedir.so session [success=1 default=ignore] pam_succeed_if.so service in crond quiet use_uid session required pam_unix.so session optional pam_krb5.so

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  • duplicity fail: not promping for password: "Running 'sftp user@host' failed"

    - by Thr4wn
    I have two linode VPS accounts and I want to back up one onto the other (the reasons are mainly for fun and to practice server administration.) the short version Duplicity isn't even asking for my password, but immediately says "invalid SSH password" (but I can ssh into the other server). why? the long version When I run duplicity /home/me scp://[email protected]//root/backup I get Invalid SSH password Running 'sftp [email protected]' failed (attempt #1) Invalid SSH password Running 'sftp [email protected]' failed (attempt #2) Invalid SSH password Running 'sftp [email protected]' failed (attempt #3) And it says Invalid SSH password immediately with no opportunity for me to actually type the password. When I type duplicity full -v9 --num-retries 4 /home/me scp://[email protected]//root/backup I get Main action: full Running 'sftp [email protected]' (attempt #1) State = sftp, Before = 'Connecting to 97.107.129.67... [email protected]'s' State = sftp, Before = '' Invalid SSH password Running 'sftp [email protected]' failed (attempt #1) I can ssh into [email protected] fine, and in fact have the ip in known_hosts before I tried any of this. serer 1 (from which I'm running the duplicity command) is Linode's default Ubuntu 8 setup with only a handful of programs installed via apt-get. server 2 (represented by x.x.x.x) is literally only Linode's default Ubuntu 8 setup I previously tried using SystemImager -- would that have changed settings in a destructive way? (I have removed and rebooted since then) Isn't Duplicity supposed to prompt for password? Am I using it wrong? are there common mistakes/dependencies I need to know about? Is there any way that x.x.x.x could be setup that could make this not work (I used Linode's default Ubuntu 8 setup and barely )?

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  • C++/boost generator module, feedback/critic please

    - by aaa
    hello. I wrote this generator, and I think to submit to boost people. Can you give me some feedback about it it basically allows to collapse multidimensional loops to flat multi-index queue. Loop can be boost lambda expressions. Main reason for doing this is to make parallel loops easier and separate algorithm from controlling structure (my fieldwork is computational chemistry where deep loops are common) 1 #ifndef _GENERATOR_HPP_ 2 #define _GENERATOR_HPP_ 3 4 #include <boost/array.hpp> 5 #include <boost/lambda/lambda.hpp> 6 #include <boost/noncopyable.hpp> 7 8 #include <boost/mpl/bool.hpp> 9 #include <boost/mpl/int.hpp> 10 #include <boost/mpl/for_each.hpp> 11 #include <boost/mpl/range_c.hpp> 12 #include <boost/mpl/vector.hpp> 13 #include <boost/mpl/transform.hpp> 14 #include <boost/mpl/erase.hpp> 15 16 #include <boost/fusion/include/vector.hpp> 17 #include <boost/fusion/include/for_each.hpp> 18 #include <boost/fusion/include/at_c.hpp> 19 #include <boost/fusion/mpl.hpp> 20 #include <boost/fusion/include/as_vector.hpp> 21 22 #include <memory> 23 24 /** 25 for loop generator which can use lambda expressions. 26 27 For example: 28 @code 29 using namespace generator; 30 using namespace boost::lambda; 31 make_for(N, N, range(bind(std::max<int>, _1, _2), N), range(_2, _3+1)); 32 // equivalent to pseudocode 33 // for l=0,N: for k=0,N: for j=max(l,k),N: for i=k,j 34 @endcode 35 36 If range is given as upper bound only, 37 lower bound is assumed to be default constructed 38 Lambda placeholders may only reference first three indices. 39 */ 40 41 namespace generator { 42 namespace detail { 43 44 using boost::lambda::constant_type; 45 using boost::lambda::constant; 46 47 /// lambda expression identity 48 template<class E, class enable = void> 49 struct lambda { 50 typedef E type; 51 }; 52 53 /// transform/construct constant lambda expression from non-lambda 54 template<class E> 55 struct lambda<E, typename boost::disable_if< 56 boost::lambda::is_lambda_functor<E> >::type> 57 { 58 struct constant : boost::lambda::constant_type<E>::type { 59 typedef typename boost::lambda::constant_type<E>::type base_type; 60 constant() : base_type(boost::lambda::constant(E())) {} 61 constant(const E &e) : base_type(boost::lambda::constant(e)) {} 62 }; 63 typedef constant type; 64 }; 65 66 /// range functor 67 template<class L, class U> 68 struct range_ { 69 typedef boost::array<int,4> index_type; 70 range_(U upper) : bounds_(typename lambda<L>::type(), upper) {} 71 range_(L lower, U upper) : bounds_(lower, upper) {} 72 73 template< typename T, size_t N> 74 T lower(const boost::array<T,N> &index) { 75 return bound<0>(index); 76 } 77 78 template< typename T, size_t N> 79 T upper(const boost::array<T,N> &index) { 80 return bound<1>(index); 81 } 82 83 private: 84 template<bool b, typename T> 85 T bound(const boost::array<T,1> &index) { 86 return (boost::fusion::at_c<b>(bounds_))(index[0]); 87 } 88 89 template<bool b, typename T> 90 T bound(const boost::array<T,2> &index) { 91 return (boost::fusion::at_c<b>(bounds_))(index[0], index[1]); 92 } 93 94 template<bool b, typename T, size_t N> 95 T bound(const boost::array<T,N> &index) { 96 using boost::fusion::at_c; 97 return (at_c<b>(bounds_))(index[0], index[1], index[2]); 98 } 99 100 boost::fusion::vector<typename lambda<L>::type, 101 typename lambda<U>::type> bounds_; 102 }; 103 104 template<typename T, size_t N> 105 struct for_base { 106 typedef boost::array<T,N> value_type; 107 virtual ~for_base() {} 108 virtual value_type next() = 0; 109 }; 110 111 /// N-index generator 112 template<typename T, size_t N, class R, class I> 113 struct for_ : for_base<T,N> { 114 typedef typename for_base<T,N>::value_type value_type; 115 typedef R range_tuple; 116 for_(const range_tuple &r) : r_(r), state_(true) { 117 boost::fusion::for_each(r_, initialize(index)); 118 } 119 /// @return new generator 120 for_* new_() { return new for_(r_); } 121 /// @return next index value and increment 122 value_type next() { 123 value_type next; 124 using namespace boost::lambda; 125 typename value_type::iterator n = next.begin(); 126 typename value_type::iterator i = index.begin(); 127 boost::mpl::for_each<I>(*(var(n))++ = var(i)[_1]); 128 129 state_ = advance<N>(r_, index); 130 return next; 131 } 132 /// @return false if out of bounds, true otherwise 133 operator bool() { return state_; } 134 135 private: 136 /// initialize indices 137 struct initialize { 138 value_type &index_; 139 mutable size_t i_; 140 initialize(value_type &index) : index_(index), i_(0) {} 141 template<class R_> void operator()(R_& r) const { 142 index_[i_++] = r.lower(index_); 143 } 144 }; 145 146 /// advance index[0:M) 147 template<size_t M> 148 struct advance { 149 /// stop recursion 150 struct stop { 151 stop(R r, value_type &index) {} 152 }; 153 /// advance index 154 /// @param r range tuple 155 /// @param index index array 156 advance(R &r, value_type &index) : index_(index), i_(0) { 157 namespace fusion = boost::fusion; 158 index[M-1] += 1; // increment index 159 fusion::for_each(r, *this); // update indices 160 state_ = index[M-1] >= fusion::at_c<M-1>(r).upper(index); 161 if (state_) { // out of bounds 162 typename boost::mpl::if_c<(M > 1), 163 advance<M-1>, stop>::type(r, index); 164 } 165 } 166 /// apply lower bound of range to index 167 template<typename R_> void operator()(R_& r) const { 168 if (i_ >= M) index_[i_] = r.lower(index_); 169 ++i_; 170 } 171 /// @return false if out of bounds, true otherwise 172 operator bool() { return state_; } 173 private: 174 value_type &index_; ///< index array reference 175 mutable size_t i_; ///< running index 176 bool state_; ///< out of bounds state 177 }; 178 179 value_type index; 180 range_tuple r_; 181 bool state_; 182 }; 183 184 185 /// polymorphic generator template base 186 template<typename T,size_t N> 187 struct For : boost::noncopyable { 188 typedef boost::array<T,N> value_type; 189 /// @return next index value and increment 190 value_type next() { return for_->next(); } 191 /// @return false if out of bounds, true otherwise 192 operator bool() const { return for_; } 193 protected: 194 /// reset smart pointer 195 void reset(for_base<T,N> *f) { for_.reset(f); } 196 std::auto_ptr<for_base<T,N> > for_; 197 }; 198 199 /// range [T,R) type 200 template<typename T, typename R> 201 struct range_type { 202 typedef range_<T,R> type; 203 }; 204 205 /// range identity specialization 206 template<typename T, class L, class U> 207 struct range_type<T, range_<L,U> > { 208 typedef range_<L,U> type; 209 }; 210 211 namespace fusion = boost::fusion; 212 namespace mpl = boost::mpl; 213 214 template<typename T, size_t N, class R1, class R2, class R3, class R4> 215 struct range_tuple { 216 // full range vector 217 typedef typename mpl::vector<R1,R2,R3,R4> v; 218 typedef typename mpl::end<v>::type end; 219 typedef typename mpl::advance_c<typename mpl::begin<v>::type, N>::type pos; 220 // [0:N) range vector 221 typedef typename mpl::erase<v, pos, end>::type t; 222 // transform into proper range fusion::vector 223 typedef typename fusion::result_of::as_vector< 224 typename mpl::transform<t,range_type<T, mpl::_1> >::type 225 >::type type; 226 }; 227 228 229 template<typename T, size_t N, 230 class R1, class R2, class R3, class R4, 231 class O> 232 struct for_type { 233 typedef typename range_tuple<T,N,R1,R2,R3,R4>::type range_tuple; 234 typedef for_<T, N, range_tuple, O> type; 235 }; 236 237 } // namespace detail 238 239 240 /// default index order, [0:N) 241 template<size_t N> 242 struct order { 243 typedef boost::mpl::range_c<size_t,0, N> type; 244 }; 245 246 /// N-loop generator, 0 < N <= 5 247 /// @tparam T index type 248 /// @tparam N number of indices/loops 249 /// @tparam R1,... range types 250 /// @tparam O index order 251 template<typename T, size_t N, 252 class R1, class R2 = void, class R3 = void, class R4 = void, 253 class O = typename order<N>::type> 254 struct for_ : detail::for_type<T, N, R1, R2, R3, R4, O>::type { 255 typedef typename detail::for_type<T, N, R1, R2, R3, R4, O>::type base_type; 256 typedef typename base_type::range_tuple range_tuple; 257 for_(const range_tuple &range) : base_type(range) {} 258 }; 259 260 /// loop range [L:U) 261 /// @tparam L lower bound type 262 /// @tparam U upper bound type 263 /// @return range 264 template<class L, class U> 265 detail::range_<L,U> range(L lower, U upper) { 266 return detail::range_<L,U>(lower, upper); 267 } 268 269 /// make 4-loop generator with specified index ordering 270 template<typename T, class R1, class R2, class R3, class R4, class O> 271 for_<T, 4, R1, R2, R3, R4, O> 272 make_for(R1 r1, R2 r2, R3 r3, R4 r4, const O&) { 273 typedef for_<T, 4, R1, R2, R3, R4, O> F; 274 return F(F::range_tuple(r1, r2, r3, r4)); 275 } 276 277 /// polymorphic generator template forward declaration 278 template<typename T,size_t N> 279 struct For; 280 281 /// polymorphic 4-loop generator 282 template<typename T> 283 struct For<T,4> : detail::For<T,4> { 284 /// generator with default index ordering 285 template<class R1, class R2, class R3, class R4> 286 For(R1 r1, R2 r2, R3 r3, R4 r4) { 287 this->reset(make_for<T>(r1, r2, r3, r4).new_()); 288 } 289 /// generator with specified index ordering 290 template<class R1, class R2, class R3, class R4, class O> 291 For(R1 r1, R2 r2, R3 r3, R4 r4, O o) { 292 this->reset(make_for<T>(r1, r2, r3, r4, o).new_()); 293 } 294 }; 295 296 } 297 298 299 #endif /* _GENERATOR_HPP_ */

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  • How is the "change password at next logon" requirement supposed to work with RDP using Network Level Authentication?

    - by NReilingh
    We have a Windows server (2008 R2) with the "Remote Desktop Services" feature installed and no Active Directory domain. Remote desktop is set up to "Allow connections only from computers running Remote Desktop with Network Level Authentication (more secure)". This means that before the remote screen is displayed, the connection is authenticated in a "Windows Security: Enter your credentials" window. The only two role services installed on this server is the RD Session Host and Licensing. When the "User must change password at next logon" checkbox is selected in the properties for a local user on this server, the following displays on a client computer after attempting to connect using the credentials that were last valid: On some other servers using RDP for admin access (but without the Remote Desktop Services role installed), the behavior is different -- the session begins and the user is given a change password prompt on the remote screen. What do I need to do to replicate this behavior on the Remote Desktop Services server?

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  • How to reset Administrator password Windows Server 2003 installed on Vmware?

    - by Cucumber
    I want to reset administrator password on Windows Server 2003. OS installed on VMware server VMware Server version 2.0.1. And problem is this: when i try to boot from live cd, after boot disk with Windows not detected. I tried to use Windows Admin Hack - Linux Boot; Hirens.BootCD.10.4; ophcrack-xp-livecd-2.3.1.iso None of these programs did not see the hard drive! Any ideas? Thank! ADD: I want to reset LOCAL admin password, not domain. And this computer are not domain controller.

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  • OpenLDAP Password Expiration with pwdReset=TRUE?

    - by jsight
    I have configured the ppolicy overlay for OpenLDAP to enable password policies. These things work: Password lockouts on too many failed attempts Password Change required once pwdReset=TRUE added to user entry Password Expirations If the account is locked out due to intrusion attempts (too many bad passwords) or time (expiration time hit), the account must be reset by an administrator. However, when the administrator sets pwdReset=TRUE in the profile, this seems to also override the expiration policy. So, the password that the administrator sent out (which should be a temporary password) ends up being valid permanently. Is there a way in OpenLDAP to have a password that must be changed, but also MUST expire?

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  • OpenLDAP Password Expiration with pwdReset=TRUE?

    - by jsight
    I have configured the ppolicy overlay for OpenLDAP to enable password policies. These things work: Password lockouts on too many failed attempts Password Change required once pwdReset=TRUE added to user entry Password Expirations If the account is locked out due to intrusion attempts (too many bad passwords) or time (expiration time hit), the account must be reset by an administrator. However, when the administrator sets pwdReset=TRUE in the profile, this seems to also override the expiration policy. So, the password that the administrator sent out (which should be a temporary password) ends up being valid permanently. Is there a way in OpenLDAP to have a password that must be changed, but also MUST expire?

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  • Blackberry Gmail password change

    - by Highstead
    I've updated my gmail password and so i must update my blackberry password. I tried updating the email password to which i got the following message. Invalid email address or password. Please verify your email address and password. The information you provided is incorrect. If the error persists contact gmail.com (Your email provider). Please try again. I tried again, with what i know the password to be, with password show on. I've also deleted the account and tried to create it. I've tried going to the "Last account activity: XXXX details" menu and signing out all devices. I'm continually getting the above error, but the account activities don't seem to show any sign of a mobile attempt to access my mail account. Has anyone had this issue before and how did you sign it. Thanks in advance.

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  • Update saved password for basic authentication using a script

    - by Kalamane
    I have a website that uses basic authentication as described on this webpage. Each of the computers I manage have the password saved in their browser. There is only one username and password for this. After someone logs in to the site this way, they are presented with their individual username and password prompt as part of the web page. The purpose of the initial username/password is to discourage non-technical employees that aren't supposed to be using the page from even viewing it. So far, when we've had to change this password, I've manually gone to each computer and updated the saved password. I'm writing a startup script to configure other aspects of these systems so that I can maintain them easier. I'd like to be able to update the saved password via this script. The operating system running on these machines is Windows XP SP3 and the browsers they're using to access this site are IE8 and IE9. How can I update the saved basic authentication information for a website via a script?

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  • How to allow password protected start-stop-daemon functionality?

    - by Mahmoud Abdelkader
    I would like to use Ubuntu's start-stop-daemon to start my application, but the application protects some sensitive information, so I have a mechanism where the application prompts for a password that's then used to generate a hashkey, which is used as the secret key for a symmetric encryption (AES) to encrypt and decrypt things from a database. I'd like to daemonize this application and have it run from start-stop-daemon, so that sudo service appname stop and sudo service appname start would work, but, I'm not sure how to go about doing this with the added complexity of a password prompt. Is there something that supports this or do I have to program it from scratch? I figured I should ask first before re-inventing the wheel. Thanks in advance.

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  • SSH only works after intentionally failed password

    - by pyraz
    So, I'm having a rather weird problem. I have a server, that when I try to SSH into, immediately closes the connection if I type in the correct password on the first attempt. However, if I purposefully enter a wrong password on the first attempt, and then enter a correct password at the second or third prompt, it successfully logs me into the computer. Similarly, when I try to use public key authentication, I get an immediate closed connection. If, however, I enter a wrong password for my key file, followed by another wrong password once it reverts to password authentication, I can successfully log in as long as I provide the correct password at the second or third prompt. The machine is running Red Hat Enterprise Linux Server release 6.2 (Santiago), and is using LDAP and PAM for authentication. Any ideas on where to start debugging this one? Let me know what config files I need to provide and I'll be happy to do so.

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  • iPhone 3G backup encryption? I've never entered a password?

    - by Lewis
    I can't unclick or access my backup iPhone encrypted file. For the life of me I can not remember ever entering a password for the encrypted iPhone backups. I've tried every password I've used or use and nothing is working. I'm not getting anywhere with long searches online. Can anyone here help? iPhone 3.1.2 iTunes 9.1.1 Mac OSX 10.5.8 Please help, how do I get my iPhone backed up from my 'locked' file I've never locked?

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  • How to Use a PIN Instead of a Password in Windows 8

    - by Taylor Gibb
    Entering your full password on a touch screen device can really become a pain in the neck, luckily for us we can link a short 4 digit PIN to our user account and log in with that instead. Note: PIN codes are nowhere near as safe as using an alphanumeric password, however, they do still have a purpose when you don’t want to enter your 15 character password on a touch screen device. Creating a PIN Press the Win + I keyboard combination to bring up the Settings Charm, then click on the Change PC settings link. This will open up the Modern UI PC Settings app, where you can click on the Users section. On the right hand side you will see a Create a PIN button, click on it. Now you will need to verify that you are the owner of this user account by entering your password. Then you can choose a PIN, remember that it can only contain digits. Now when you get to the login screen you will have the option to use a PIN. How To Boot Your Android Phone or Tablet Into Safe Mode HTG Explains: Does Your Android Phone Need an Antivirus? How To Use USB Drives With the Nexus 7 and Other Android Devices

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  • SSH asks for password

    - by user1435470
    I have already : Installed the server Generated the pub/pri keys with -P "" Copied the id_rsa.pub to authorized_keys ssh localhost answered "yes", copied to known_hosts tried ssh localhost still asks for password Output: hduser@hduser1-desktop:~$ ssh -v localhost OpenSSH_5.3p1 Debian-3ubuntu7, OpenSSL 0.9.8k 25 Mar 2009 debug1: Reading configuration data /etc/ssh/ssh_config debug1: Applying options for * debug1: Connecting to localhost [127.0.0.1] port 22. debug1: Connection established. debug1: identity file /home/hduser/.ssh/identity type -1 debug1: identity file /home/hduser/.ssh/id_rsa type 1 debug1: Checking blacklist file /usr/share/ssh/blacklist.RSA-2048 debug1: Checking blacklist file /etc/ssh/blacklist.RSA-2048 debug1: identity file /home/hduser/.ssh/id_dsa type -1 debug1: Remote protocol version 2.0, remote software version OpenSSH_5.3p1 Debian- 3ubuntu7 debug1: match: OpenSSH_5.3p1 Debian-3ubuntu7 pat OpenSSH* debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_5.3p1 Debian-3ubuntu7 debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug1: kex: server->client aes128-ctr hmac-md5 none debug1: kex: client->server aes128-ctr hmac-md5 none debug1: SSH2_MSG_KEX_DH_GEX_REQUEST(1024<1024<8192) sent debug1: expecting SSH2_MSG_KEX_DH_GEX_GROUP debug1: SSH2_MSG_KEX_DH_GEX_INIT sent debug1: expecting SSH2_MSG_KEX_DH_GEX_REPLY debug1: Host 'localhost' is known and matches the RSA host key. debug1: Found key in /home/hduser/.ssh/known_hosts:3 debug1: ssh_rsa_verify: signature correct debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: SSH2_MSG_NEWKEYS received debug1: SSH2_MSG_SERVICE_REQUEST sent debug1: SSH2_MSG_SERVICE_ACCEPT received debug1: Authentications that can continue: publickey,password debug1: Next authentication method: publickey debug1: Offering public key: /home/hduser/.ssh/id_rsa debug1: Authentications that can continue: publickey,password debug1: Trying private key: /home/hduser/.ssh/identity debug1: Trying private key: /home/hduser/.ssh/id_dsa debug1: Next authentication method: password Any suggestions ? Cheers

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  • Password Cracking in 2010 and Beyond

    - by mttr
    I have looked a bit into cryptography and related matters during the last couple of days and am pretty confused by now. I have a question about password strength and am hoping that someone can clear up my confusion by sharing how they think through the following questions. I am becoming obsessed about these things, but need to spend my time otherwise :-) Let's assume we have an eight-digit password that consists of upper and lower-case alphabetic characters, numbers and common symbols. This means we have 8^96 ~= 7.2 quadrillion different possible passwords. As I understand there are at least two approaches to breaking this password. One is to try a brute-force attack where we try to guess each possible combination of characters. How many passwords can modern processors (in 2010, Core i7 Extreme for eg) guess per second (how many instructions does a single password guess take and why)? My guess would be that it takes a modern processor in the order of years to break such a password. Another approach would consist of obtaining a hash of my password as stored by operating systems and then search for collisions. Depending on the type of hash used, we might get the password a lot quicker than by the bruteforce attack. A number of questions about this: Is the assertion in the above sentence correct? How do I think about the time it takes to find collisions for MD4, MD5, etc. hashes? Where does my Snow Leopard store my password hash and what hashing algorithm does it use? And finally, regardless of the strength of file encryption using AES-128/256, the weak link is still my en/decryption password used. Even if breaking the ciphered text would take longer than the lifetime of the universe, a brute-force attack on my de/encryption password (guess password, then try to decrypt file, try next password...), might succeed a lot earlier than the end of the universe. Is that correct? I would be very grateful, if people could have mercy on me and help me think through these probably simple questions, so that I can get back to work.

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  • Can't work out security

    - by user215351
    I installed Ubuntu on puter I am the only user I alone use. I was trying a to find out how to repair hardware faults. Surprised to find I was not the owner and that there is a password that locks me out. I only set one password during set up so what is this mysterious password. As far as I'm concerned it is overdone on security, Im sick of authenticating every 3 seconds. I need a simpler system

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  • Fixing /etc/shadow with md5 passwords to sha512 passwords

    - by dr jimbob
    I recently upgraded an ubuntu server with many users to a recent version from a version from 2008. The server used to use md5 password hashes (e.g., the shadow passwords began with $1$) and now is configured to use sha512. I'd prefer to keep using sha512, but would like the old users to be able to partially login once with their old password and then be forced to update their password (even if its the same password) generating a sha512. Right now, the old md5-based passwords in /etc/shadow won't let the user login at all (and just appear to be incorrect passwords). This seems like plenty of people should have had to do this before; yet I can't see how to do it, looking in the common places like /etc/pam.d/common-password nad /etc/login.defs. Also users will be logging in via ssh; and I do not have everyone's contact info (email or otherwise); and some login fairly rarely. Any help? (Googling doesn't seem to give any good solutions).

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  • Can't Enter Password in Recovery Mode

    - by Mike Lentini
    When I go into recovery mode to get to the command line, I enter "telinit 3" to drop out of root. This works on my desktop, but on my laptop it asks for my username and password, then I hit enter and the first letter of my password appears and it doesn't log me in. It then proceeds to ask me for my username and password again, and the issue continues. Am I doing something wrong? Is this a known issue with a solution? EDIT: Worked around this by going into /etc/default/grub and setting it to boot in text mode. Still would like a solution for this though.

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  • Can not enter password for sudo [duplicate]

    - by Michael
    This question already has an answer here: add repository to ubuntu from terminal with pgp key 3 answers I have used Ubuntu for several years, and I can not enter password for sudo, and this is when i wanna add a key to public.gpg for itunes10 it dos not work, and the password normal works wite sudo but not in the terminal when i enter: sudo wget -q "http:// deb.playonlinux.com/public.gpg" -o- | sudo apt-get add - and it says this 'sorry try again', and i just had installed itunes10, and have to add a key whit wget to public.gpg, and i tried to enter in the terminal: sudo apt-get update and the password works fine but not whit using sudo wget, and can some one please help.

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  • How to disable password prompt for default keyring?

    - by user110431
    I'm using Ubuntu 12.10 64Bit. How to disable ubuntu enter password for keyring default to unlock prompt ? Every time I open Chrome or ubuntu software center, this annoying window jump out. I have being searching online for a long time, most of the answers are like delete ~/.gnome2/keyring XX , but I don't have such directory or go to password and keys , disable some option, but this window is empty in my case, very strange , even I add a new password keyring, it is still empty. None of these methods works in my case. I will be very appreciate if you can help

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  • installation password

    - by tdot
    I just updated to ubuntu 11.10 and am getting "Your authentication attempt was unsuccessful." I am the only user and am the Administrator, during login on my keyring i set to logon automatically, but it still askes me for my password. So i put in my old one and it works. But when i go to install apps it asks again and nothing works! Also if i go to user accounts and try to unlock or change my password it won't let me! It says under password "None" so what's going on?! I'm new to Ubuntu! Help!

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  • WinRAR extracting file before checking password? [closed]

    - by opatachibueze
    I tried extracting an encrypted rar file today, and I discovered that I had to wait the same amount of time I'll wait before a file is extracted (extraction reaches 99% completion) for WinRAR to conclude it's the wrong password (winrar message: "CRC failed wrong password or corrupt file?") . My guess is that this file is somewhere on the Computer just before the detection, - it has to be and then gets deleted after it's verified that the password is not the same? Is there anyway I can forcefully get this file from the PC? Thanks.

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  • Encryption Password help!

    - by Carlos L.
    Ok so let me summarize this up. I encrypted my Home to protect against hackers of course when I first installed Ubuntu. It loaded up the Terminal and was attempting to show me my encryption password incase it ever needed to be used. So I thought "Ehh what the heck, I can find it out later..." So I closed Terminal and went on with the (amazing!) Ubuntu life. But now I am having to install Java JDK 7.0.0.4 onto my computer to ya know, play games and such. But it is asking for my password for the encrypted Home folder but it never gave it to me... HELP!!! Does anyone remember the command for Terminal to give you you're randomly generated Encryption password pop up on the famous purple window? Please give legitimate answer and fast please!

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