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  • I cannot enter my password when using sudo to install Sophos AV for Linux

    - by dycharlie
    I cannot type my password as shown below. After successfully unlocking root account in Ubuntu 12.04 LTS. saintmichael@ubuntu:~$ sudo usage: sudo [-D level] -h | -K | -k | -V usage: sudo -v [-AknS] [-D level] [-g groupname|#gid] [-p prompt] [-u user name|#uid] usage: sudo -l[l] [-AknS] [-D level] [-g groupname|#gid] [-p prompt] [-U user name] [-u user name|#uid] [-g groupname|#gid] [command] usage: sudo [-AbEHknPS] [-C fd] [-D level] [-g groupname|#gid] [-p prompt] [-u user name|#uid] [-g groupname|#gid] [VAR=value] [-i|-s] [<command>] usage: sudo -e [-AknS] [-C fd] [-D level] [-g groupname|#gid] [-p prompt] [-u user name|#uid] file ... saintmichael@ubuntu:~$ sudo ./sophos-av/install.sh [sudo] password for saintmichael:

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  • How do I reset my password?

    - by doug
    I inherited a machine with Ubuntu desktop installed. It has a password in place and I have no idea what the password may be. I cannot get to the command line to use the methods I have found online. No matter how many times I press "Shift" during the boot process it still goes all the way to the desktop login. I never see grub. I am not sure which version I have but I think may be 9 or 10. Thanks Doug

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  • ban an IP temporarily after x-many incorrect password attempts

    - by sova
    My new web server got hacked (sigh). I have physical access to my machine (in the near future). It seems like the only changes was a new user account and a borked sudoers file. It seems as though the password was discovered by dictionary searching (I didn't pick it). After I fix these problems (or do a full reinstall?) I want to add a mechanism to ban an IP (for maybe 24 hours or some time limit) after getting the password wrong x number of times, but I'm not a unix sysadmin or anything, so I'm not really sure where to get started. The machine is running Lucid Lynx, from an Ubuntu minimal installation. Thanks,I appreciate your help guys. Hopefully this is the right place for this question.

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  • Regarding Lost Administrative Password

    - by Rex Haggard
    I'm working on a Ubuntu 10.04 (Lucid Lynx) system using a Panasonic CF-50 Laptop. My Client has completely forgotten his Administrative Password. He doesn't even remember entering one; however it is there. I've tried the suggestions on the WebSite and I have been unsuccessful in deleting the password so that I can download applets required for running some files. Do you have a solution? I look forward to hearing your response. Thanks for your time and consideration. -- Rex Haggard 1967 North St. Apt. #15 New Orleans, La 70802

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  • Login screen won't accept my password

    - by Raven H
    I recently upgraded to 12.04 from 11.10 and since upgrading have been unable to login to my user profile. The upgrade went okay and I can login to a guest session fine but whenever I try to login to my profile, after entering my password, I just return to the login screen. I've changed my password in Root (passwd 'username')and can log in to tty1 with no issues, it's just in GUI I'm having problems. I'm using a HP dv7 laptop, 32 bit Ubuntu install, Intel® Core™2 Duo CPU P7350 @ 2.00GHz × 2, Nvidia graphics. Any help would be appreciated.

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  • MySQL wants a password but it's empty

    - by gAMBOOKa
    mysql -uroot ERROR 1045 (28000): Access denied for user 'root'@'localhost' (using password: YES) mysql -uroot -p Enter password: <-- leave blank, hit enter without entering anything mysql> <-- i am logged in NOTE: This is a new mysql instance installation So if the password is blank, why won't it log me in without a -p flag? For a little clarification. I am running into this issue when attempting to change the password using a script: We're using a bash script to do that. mysqladmin -u root password abc wouldn't work (access denied) mysqladmin -u root -p password abc cannot be used because it prompts for a password and we need to automate this. mysqladmin -u root -p'' password abc is not working either

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  • Heaps of Trouble?

    - by Paul White NZ
    If you’re not already a regular reader of Brad Schulz’s blog, you’re missing out on some great material.  In his latest entry, he is tasked with optimizing a query run against tables that have no indexes at all.  The problem is, predictably, that performance is not very good.  The catch is that we are not allowed to create any indexes (or even new statistics) as part of our optimization efforts. In this post, I’m going to look at the problem from a slightly different angle, and present an alternative solution to the one Brad found.  Inevitably, there’s going to be some overlap between our entries, and while you don’t necessarily need to read Brad’s post before this one, I do strongly recommend that you read it at some stage; he covers some important points that I won’t cover again here. The Example We’ll use data from the AdventureWorks database, copied to temporary unindexed tables.  A script to create these structures is shown below: CREATE TABLE #Custs ( CustomerID INTEGER NOT NULL, TerritoryID INTEGER NULL, CustomerType NCHAR(1) COLLATE SQL_Latin1_General_CP1_CI_AI NOT NULL, ); GO CREATE TABLE #Prods ( ProductMainID INTEGER NOT NULL, ProductSubID INTEGER NOT NULL, ProductSubSubID INTEGER NOT NULL, Name NVARCHAR(50) COLLATE SQL_Latin1_General_CP1_CI_AI NOT NULL, ); GO CREATE TABLE #OrdHeader ( SalesOrderID INTEGER NOT NULL, OrderDate DATETIME NOT NULL, SalesOrderNumber NVARCHAR(25) COLLATE SQL_Latin1_General_CP1_CI_AI NOT NULL, CustomerID INTEGER NOT NULL, ); GO CREATE TABLE #OrdDetail ( SalesOrderID INTEGER NOT NULL, OrderQty SMALLINT NOT NULL, LineTotal NUMERIC(38,6) NOT NULL, ProductMainID INTEGER NOT NULL, ProductSubID INTEGER NOT NULL, ProductSubSubID INTEGER NOT NULL, ); GO INSERT #Custs ( CustomerID, TerritoryID, CustomerType ) SELECT C.CustomerID, C.TerritoryID, C.CustomerType FROM AdventureWorks.Sales.Customer C WITH (TABLOCK); GO INSERT #Prods ( ProductMainID, ProductSubID, ProductSubSubID, Name ) SELECT P.ProductID, P.ProductID, P.ProductID, P.Name FROM AdventureWorks.Production.Product P WITH (TABLOCK); GO INSERT #OrdHeader ( SalesOrderID, OrderDate, SalesOrderNumber, CustomerID ) SELECT H.SalesOrderID, H.OrderDate, H.SalesOrderNumber, H.CustomerID FROM AdventureWorks.Sales.SalesOrderHeader H WITH (TABLOCK); GO INSERT #OrdDetail ( SalesOrderID, OrderQty, LineTotal, ProductMainID, ProductSubID, ProductSubSubID ) SELECT D.SalesOrderID, D.OrderQty, D.LineTotal, D.ProductID, D.ProductID, D.ProductID FROM AdventureWorks.Sales.SalesOrderDetail D WITH (TABLOCK); The query itself is a simple join of the four tables: SELECT P.ProductMainID AS PID, P.Name, D.OrderQty, H.SalesOrderNumber, H.OrderDate, C.TerritoryID FROM #Prods P JOIN #OrdDetail D ON P.ProductMainID = D.ProductMainID AND P.ProductSubID = D.ProductSubID AND P.ProductSubSubID = D.ProductSubSubID JOIN #OrdHeader H ON D.SalesOrderID = H.SalesOrderID JOIN #Custs C ON H.CustomerID = C.CustomerID ORDER BY P.ProductMainID ASC OPTION (RECOMPILE, MAXDOP 1); Remember that these tables have no indexes at all, and only the single-column sampled statistics SQL Server automatically creates (assuming default settings).  The estimated query plan produced for the test query looks like this (click to enlarge): The Problem The problem here is one of cardinality estimation – the number of rows SQL Server expects to find at each step of the plan.  The lack of indexes and useful statistical information means that SQL Server does not have the information it needs to make a good estimate.  Every join in the plan shown above estimates that it will produce just a single row as output.  Brad covers the factors that lead to the low estimates in his post. In reality, the join between the #Prods and #OrdDetail tables will produce 121,317 rows.  It should not surprise you that this has rather dire consequences for the remainder of the query plan.  In particular, it makes a nonsense of the optimizer’s decision to use Nested Loops to join to the two remaining tables.  Instead of scanning the #OrdHeader and #Custs tables once (as it expected), it has to perform 121,317 full scans of each.  The query takes somewhere in the region of twenty minutes to run to completion on my development machine. A Solution At this point, you may be thinking the same thing I was: if we really are stuck with no indexes, the best we can do is to use hash joins everywhere. We can force the exclusive use of hash joins in several ways, the two most common being join and query hints.  A join hint means writing the query using the INNER HASH JOIN syntax; using a query hint involves adding OPTION (HASH JOIN) at the bottom of the query.  The difference is that using join hints also forces the order of the join, whereas the query hint gives the optimizer freedom to reorder the joins at its discretion. Adding the OPTION (HASH JOIN) hint results in this estimated plan: That produces the correct output in around seven seconds, which is quite an improvement!  As a purely practical matter, and given the rigid rules of the environment we find ourselves in, we might leave things there.  (We can improve the hashing solution a bit – I’ll come back to that later on). Faster Nested Loops It might surprise you to hear that we can beat the performance of the hash join solution shown above using nested loops joins exclusively, and without breaking the rules we have been set. The key to this part is to realize that a condition like (A = B) can be expressed as (A <= B) AND (A >= B).  Armed with this tremendous new insight, we can rewrite the join predicates like so: SELECT P.ProductMainID AS PID, P.Name, D.OrderQty, H.SalesOrderNumber, H.OrderDate, C.TerritoryID FROM #OrdDetail D JOIN #OrdHeader H ON D.SalesOrderID >= H.SalesOrderID AND D.SalesOrderID <= H.SalesOrderID JOIN #Custs C ON H.CustomerID >= C.CustomerID AND H.CustomerID <= C.CustomerID JOIN #Prods P ON P.ProductMainID >= D.ProductMainID AND P.ProductMainID <= D.ProductMainID AND P.ProductSubID = D.ProductSubID AND P.ProductSubSubID = D.ProductSubSubID ORDER BY D.ProductMainID OPTION (RECOMPILE, LOOP JOIN, MAXDOP 1, FORCE ORDER); I’ve also added LOOP JOIN and FORCE ORDER query hints to ensure that only nested loops joins are used, and that the tables are joined in the order they appear.  The new estimated execution plan is: This new query runs in under 2 seconds. Why Is It Faster? The main reason for the improvement is the appearance of the eager Index Spools, which are also known as index-on-the-fly spools.  If you read my Inside The Optimiser series you might be interested to know that the rule responsible is called JoinToIndexOnTheFly. An eager index spool consumes all rows from the table it sits above, and builds a index suitable for the join to seek on.  Taking the index spool above the #Custs table as an example, it reads all the CustomerID and TerritoryID values with a single scan of the table, and builds an index keyed on CustomerID.  The term ‘eager’ means that the spool consumes all of its input rows when it starts up.  The index is built in a work table in tempdb, has no associated statistics, and only exists until the query finishes executing. The result is that each unindexed table is only scanned once, and just for the columns necessary to build the temporary index.  From that point on, every execution of the inner side of the join is answered by a seek on the temporary index – not the base table. A second optimization is that the sort on ProductMainID (required by the ORDER BY clause) is performed early, on just the rows coming from the #OrdDetail table.  The optimizer has a good estimate for the number of rows it needs to sort at that stage – it is just the cardinality of the table itself.  The accuracy of the estimate there is important because it helps determine the memory grant given to the sort operation.  Nested loops join preserves the order of rows on its outer input, so sorting early is safe.  (Hash joins do not preserve order in this way, of course). The extra lazy spool on the #Prods branch is a further optimization that avoids executing the seek on the temporary index if the value being joined (the ‘outer reference’) hasn’t changed from the last row received on the outer input.  It takes advantage of the fact that rows are still sorted on ProductMainID, so if duplicates exist, they will arrive at the join operator one after the other. The optimizer is quite conservative about introducing index spools into a plan, because creating and dropping a temporary index is a relatively expensive operation.  It’s presence in a plan is often an indication that a useful index is missing. I want to stress that I rewrote the query in this way primarily as an educational exercise – I can’t imagine having to do something so horrible to a production system. Improving the Hash Join I promised I would return to the solution that uses hash joins.  You might be puzzled that SQL Server can create three new indexes (and perform all those nested loops iterations) faster than it can perform three hash joins.  The answer, again, is down to the poor information available to the optimizer.  Let’s look at the hash join plan again: Two of the hash joins have single-row estimates on their build inputs.  SQL Server fixes the amount of memory available for the hash table based on this cardinality estimate, so at run time the hash join very quickly runs out of memory. This results in the join spilling hash buckets to disk, and any rows from the probe input that hash to the spilled buckets also get written to disk.  The join process then continues, and may again run out of memory.  This is a recursive process, which may eventually result in SQL Server resorting to a bailout join algorithm, which is guaranteed to complete eventually, but may be very slow.  The data sizes in the example tables are not large enough to force a hash bailout, but it does result in multiple levels of hash recursion.  You can see this for yourself by tracing the Hash Warning event using the Profiler tool. The final sort in the plan also suffers from a similar problem: it receives very little memory and has to perform multiple sort passes, saving intermediate runs to disk (the Sort Warnings Profiler event can be used to confirm this).  Notice also that because hash joins don’t preserve sort order, the sort cannot be pushed down the plan toward the #OrdDetail table, as in the nested loops plan. Ok, so now we understand the problems, what can we do to fix it?  We can address the hash spilling by forcing a different order for the joins: SELECT P.ProductMainID AS PID, P.Name, D.OrderQty, H.SalesOrderNumber, H.OrderDate, C.TerritoryID FROM #Prods P JOIN #Custs C JOIN #OrdHeader H ON H.CustomerID = C.CustomerID JOIN #OrdDetail D ON D.SalesOrderID = H.SalesOrderID ON P.ProductMainID = D.ProductMainID AND P.ProductSubID = D.ProductSubID AND P.ProductSubSubID = D.ProductSubSubID ORDER BY D.ProductMainID OPTION (MAXDOP 1, HASH JOIN, FORCE ORDER); With this plan, each of the inputs to the hash joins has a good estimate, and no hash recursion occurs.  The final sort still suffers from the one-row estimate problem, and we get a single-pass sort warning as it writes rows to disk.  Even so, the query runs to completion in three or four seconds.  That’s around half the time of the previous hashing solution, but still not as fast as the nested loops trickery. Final Thoughts SQL Server’s optimizer makes cost-based decisions, so it is vital to provide it with accurate information.  We can’t really blame the performance problems highlighted here on anything other than the decision to use completely unindexed tables, and not to allow the creation of additional statistics. I should probably stress that the nested loops solution shown above is not one I would normally contemplate in the real world.  It’s there primarily for its educational and entertainment value.  I might perhaps use it to demonstrate to the sceptical that SQL Server itself is crying out for an index. Be sure to read Brad’s original post for more details.  My grateful thanks to him for granting permission to reuse some of his material. Paul White Email: [email protected] Twitter: @PaulWhiteNZ

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  • Ecryptfs: lost passphrase

    - by Sherlock3890
    When i mounted some dir by mount -t ecryptfs private data i entered wrong password. I wrote data in this dir and now i can't mount it. I have no valid password and passphrase (know only the same), but have SIG in /root/.ecryptfs/sig-cache.txt. How i can recover my directory or, at least, "brute it": type many-many passwords like entered when mounting this dir and compare generated sig with existing?

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  • How can * be a safe hashed password?

    - by Exception e
    phpass is a widely used hashing 'framework'. While evaluating phpass' HashPassword I came across this odd method fragment. function HashPassword($password) { // <snip> trying to generate a hash… # Returning '*' on error is safe here, but would _not_ be safe # in a crypt(3)-like function used _both_ for generating new # hashes and for validating passwords against existing hashes. return '*'; } This is the complete phpsalt class: # Portable PHP password hashing framework. # # Version 0.2 / genuine. # # Written by Solar Designer <solar at openwall.com> in 2004-2006 and placed in # the public domain. # # # class PasswordHash { var $itoa64; var $iteration_count_log2; var $portable_hashes; var $random_state; function PasswordHash($iteration_count_log2, $portable_hashes) { $this->itoa64 = './0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'; if ($iteration_count_log2 < 4 || $iteration_count_log2 > 31) $iteration_count_log2 = 8; $this->iteration_count_log2 = $iteration_count_log2; $this->portable_hashes = $portable_hashes; $this->random_state = microtime() . getmypid(); } function get_random_bytes($count) { $output = ''; if (is_readable('/dev/urandom') && ($fh = @fopen('/dev/urandom', 'rb'))) { $output = fread($fh, $count); fclose($fh); } if (strlen($output) < $count) { $output = ''; for ($i = 0; $i < $count; $i += 16) { $this->random_state = md5(microtime() . $this->random_state); $output .= pack('H*', md5($this->random_state)); } $output = substr($output, 0, $count); } return $output; } function encode64($input, $count) { $output = ''; $i = 0; do { $value = ord($input[$i++]); $output .= $this->itoa64[$value & 0x3f]; if ($i < $count) $value |= ord($input[$i]) << 8; $output .= $this->itoa64[($value >> 6) & 0x3f]; if ($i++ >= $count) break; if ($i < $count) $value |= ord($input[$i]) << 16; $output .= $this->itoa64[($value >> 12) & 0x3f]; if ($i++ >= $count) break; $output .= $this->itoa64[($value >> 18) & 0x3f]; } while ($i < $count); return $output; } function gensalt_private($input) { $output = '$P$'; $output .= $this->itoa64[min($this->iteration_count_log2 + ((PHP_VERSION >= '5') ? 5 : 3), 30)]; $output .= $this->encode64($input, 6); return $output; } function crypt_private($password, $setting) { $output = '*0'; if (substr($setting, 0, 2) == $output) $output = '*1'; if (substr($setting, 0, 3) != '$P$') return $output; $count_log2 = strpos($this->itoa64, $setting[3]); if ($count_log2 < 7 || $count_log2 > 30) return $output; $count = 1 << $count_log2; $salt = substr($setting, 4, 8); if (strlen($salt) != 8) return $output; # We're kind of forced to use MD5 here since it's the only # cryptographic primitive available in all versions of PHP # currently in use. To implement our own low-level crypto # in PHP would result in much worse performance and # consequently in lower iteration counts and hashes that are # quicker to crack (by non-PHP code). if (PHP_VERSION >= '5') { $hash = md5($salt . $password, TRUE); do { $hash = md5($hash . $password, TRUE); } while (--$count); } else { $hash = pack('H*', md5($salt . $password)); do { $hash = pack('H*', md5($hash . $password)); } while (--$count); } $output = substr($setting, 0, 12); $output .= $this->encode64($hash, 16); return $output; } function gensalt_extended($input) { $count_log2 = min($this->iteration_count_log2 + 8, 24); # This should be odd to not reveal weak DES keys, and the # maximum valid value is (2**24 - 1) which is odd anyway. $count = (1 << $count_log2) - 1; $output = '_'; $output .= $this->itoa64[$count & 0x3f]; $output .= $this->itoa64[($count >> 6) & 0x3f]; $output .= $this->itoa64[($count >> 12) & 0x3f]; $output .= $this->itoa64[($count >> 18) & 0x3f]; $output .= $this->encode64($input, 3); return $output; } function gensalt_blowfish($input) { # This one needs to use a different order of characters and a # different encoding scheme from the one in encode64() above. # We care because the last character in our encoded string will # only represent 2 bits. While two known implementations of # bcrypt will happily accept and correct a salt string which # has the 4 unused bits set to non-zero, we do not want to take # chances and we also do not want to waste an additional byte # of entropy. $itoa64 = './ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789'; $output = '$2a$'; $output .= chr(ord('0') + $this->iteration_count_log2 / 10); $output .= chr(ord('0') + $this->iteration_count_log2 % 10); $output .= '$'; $i = 0; do { $c1 = ord($input[$i++]); $output .= $itoa64[$c1 >> 2]; $c1 = ($c1 & 0x03) << 4; if ($i >= 16) { $output .= $itoa64[$c1]; break; } $c2 = ord($input[$i++]); $c1 |= $c2 >> 4; $output .= $itoa64[$c1]; $c1 = ($c2 & 0x0f) << 2; $c2 = ord($input[$i++]); $c1 |= $c2 >> 6; $output .= $itoa64[$c1]; $output .= $itoa64[$c2 & 0x3f]; } while (1); return $output; } function HashPassword($password) { $random = ''; if (CRYPT_BLOWFISH == 1 && !$this->portable_hashes) { $random = $this->get_random_bytes(16); $hash = crypt($password, $this->gensalt_blowfish($random)); if (strlen($hash) == 60) return $hash; } if (CRYPT_EXT_DES == 1 && !$this->portable_hashes) { if (strlen($random) < 3) $random = $this->get_random_bytes(3); $hash = crypt($password, $this->gensalt_extended($random)); if (strlen($hash) == 20) return $hash; } if (strlen($random) < 6) $random = $this->get_random_bytes(6); $hash = $this->crypt_private($password, $this->gensalt_private($random)); if (strlen($hash) == 34) return $hash; # Returning '*' on error is safe here, but would _not_ be safe # in a crypt(3)-like function used _both_ for generating new # hashes and for validating passwords against existing hashes. return '*'; } function CheckPassword($password, $stored_hash) { $hash = $this->crypt_private($password, $stored_hash); if ($hash[0] == '*') $hash = crypt($password, $stored_hash); return $hash == $stored_hash; } }

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  • Thumbs Up or Thumbs Down – Intel Debuts Prototype Palm-Reading Tech to Replace Passwords [Poll]

    - by Asian Angel
    This week Intel debuted prototype palm-reading tech that could serve as a replacement for our current password system. Our question for you today is do you think this is the right direction to go for better security or do you feel this is a mistake? Photo courtesy of Jane Rahman. Needless to say password security breaches have been a hot topic as of late, so perhaps a whole new security model is in order. It would definitely eliminate the need to remember a large volume of passwords along with circumventing the problem of poor password creation/selection. At the same time the new technology would still be in the ‘early stages’ of development and may not work as well as people would like. Long-term refinement would definitely improve its performance, but would it really be worth pursuing versus the actual benefits? From the blog post: Intel researcher Sridhar Iyendar demonstrated the technology at Intel’s Developer Forum this week. Waving a hand in front of a “palm vein” detector on a computer, one of Iyendar’s assistants was logged into Windows 7, was able to view his bank account, and then once he moved away the computer locked Windows and went into sleeping mode. How to Get Pro Features in Windows Home Versions with Third Party Tools HTG Explains: Is ReadyBoost Worth Using? HTG Explains: What The Windows Event Viewer Is and How You Can Use It

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  • sudo refuses my password

    - by fredericf
    I've just installed Ubuntu 12.04 and now that I want to install a program through the terminal with a command starting with sudo, the terminal says Sorry, try again. I just don't get it, this is my user password which I check with through the Software Center. What can be then wrong? (Is it possible that the terminal function differently than the rest of the OS in relation to ASCII keyboard?) Please help.

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  • Perl Hash Slice, Replication x Operator, and sub params

    - by user210757
    Ok, I understand perl hash slices, and the "x" operator in Perl, but can someone explain the following code example from here (slightly simplified)? sub test{ my %hash; @hash{@_} = (undef) x @_; } Example Call to sub: test('one', 'two', 'three'); This line is what throws me: @hash{@_} = (undef) x @_; It is creating a hash where the keys are the parameters to the sub and initializing to undef, so: %hash: 'one' = undef, 'two' = undef, 'three' = undef The rvalue of the x operator should be a number; how is it that @_ is interpreted as the length of the sub's parameter array? I would expect you'd at least have to do this: @hash{@_} = (undef) x length(@_);

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  • 3 hash functions to best hash sliding window strings for a bloom filter with minimum collisions

    - by Duaa
    Hi all: I need 3 hash functions to hash strings of a sliding window moving over a text, to be used later to search within a bloom vector. I'm using C# in my programming I read something about rolling hash functions and cyclic polynomials, they are used for sliding window applications. But really, I did not find any codes, they are just descriptions So please, if anyone have any idea about 3 best C# hash functions to use with sliding window strings of fixed size (5-char), that consume less time and have minimum number of collisions, either they are rolling hash functions or others, please help me with some C# codes or links to hash functions names Duaa

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  • No password is complex enough

    - by Blue Warrior NFB
    I have one user in my AD domain who seems to not be able to self-select a password. I may have another one, but they're on a different enough password-expiration schedule that I can't remember who it is right now. I can set a password via ADU&C just fine, but when he tries it via C-A-D he gets the "doesn't meet complexity" message. Figuring he was just doing something like 'pAssword32', I did some troubleshooting of my own and sure enough it doesn't want to take a password that way. He's one of our users that habitually uses a local account and then maps drives using his AD credentials so he doesn't get the your password will expire in 4 days, maybe you should change it prompts, so he's a frequent "my password expired, can you fix it" flyer. I don't want to keep having him set it via ADU&C over my shoulder every N days. I'm just fine setting temp passwords of 48 characters of keyboard-slamming and letting him change it something memorable. My environment is at the Windows 2008 R2 functional level, and I am using fine-grained password policies. In fact, I have two such policies: For normal users (minimum length, remembered passwords) For special utility accounts The password complexities I've tried match both policies for length and char-set selection. The permissions on the User object themselves look normal, SELF does indeed have the "Change Password" right. Is there some other place I should be looking for things that can affect this?

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  • How do i remove a password expiration policy?

    - by jimmygee
    We had a password expiration policy recently removed from our AD but some users continued to get the "..your password will expire in x days. would you like to change it now?" message. So we added a reverse/undo policy to correct the local registry settings Maximum password age = 0 days Minimum password age = 0 days This hasn't worked as new users still seem to encounter the above "change password" message sporadically. We have now removed all custom password policy GPOs and are left with the "Default Domain Policy". Still no good. Can someone point me in the direction to fix this? And an explanation into what i was doing wrong (/how password expiration policies apply) would be useful too. thanks Environment is 2k3 server with mostly XPsp2 clients.

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  • Windows 8 Login Password Out of Sync with Windows Live ID

    - by Israel Lopez
    I'm working with a computer that has setup a local account connected under Windows Live ID. The user can login to Live ID (like hotmail) from another computer with the correct credentials. However from the Windows 8 computer using the correct password it indicates. That password is incorrect. Make sure you're using the password for you Mircrosoft Account. You can always reset it at account.live.com/password/reset. Now, I've used NTPASSWD to reset the password, but it seems that since its not a "Local Account" it wont take the new password or blank one. This account also has a "PIN" the user who also has forgotten it. I also tried to enable/password set the local Administrator account but it does not show up for login. Any ideas?

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  • Password problem while creating domain

    - by Murdock
    Hi, I'm freshman so far in server management stuff but this seems to be clearly against logic. After updating my Windows Server 2008 Standard 32bit, installing DNS server and AD DS I wanted to create domain via using CMD and dcpromo.exe setup. But no matter if I disable demand for comlex password in Password policies or create a password which fully comply with requirements for strong and complex password, still I can't get any further and it says that my password doesn't meet requirements. I'm also asked there to activate password demand by NET USER -passwordreq:yes and when I do so, this password doesn't work any more and I have to remove it from other admin account to be at least able to login with proper Administrator account.

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  • AD User Passwords expiring without any notifications?

    - by scooter133
    We setup password Policies in Active Directory to Expire peoples passwords after so many days. Well it looks like the time has come for the Expiration of the Passwords and people are getting locked out... There has been no warning of user passwords about to expire. They just come in to work and they cannot log in, the phones no longer connect, nothing. Reset the password and all is good. Some of the users are locked out, though most are not, they just cannot log in. On setting the password Expiration, I didn't see anything about nor warning the users of the impending expiration. Seems like it used to warn you 15 days or so before it would expire. Clients range from: WinXP, WinVista, Win7 and Server 2008R2 Remote Desktop Services. How can I make sure my users are warned of the Expiration? Resultant Set of Policy for User that was not prompted: Account Policies/Password Policy Policy Setting Winning GPO Enforce password history 10 passwords remembered Default Domain Policy Maximum password age 270 days Default Domain Policy Minimum password age 0 days Default Domain Policy Minimum password length 4 characters Default Domain Policy Password must meet complexity requirements Disabled Default Domain Policy Store passwords using reversible encryption Disabled Default Domain Policy Account Policies/Account Lockout Policy Policy Setting Winning GPO Account lockout duration 20 minutes Default Domain Policy Account lockout threshold 5 invalid logon attempts Default Domain Policy Reset account lockout counter after 15 minutes Default Domain Policy Local Policies/Audit Policy Policy Setting Winning GPO Audit account logon events Failure Default Domain Policy Audit account management Success, Failure Default Domain Policy Audit directory service access Success, Failure Default Domain Policy Audit logon events Failure Default Domain Policy Audit policy change Success, Failure Default Domain Policy Audit privilege use Failure Default Domain Policy Local Policies/Security Options Interactive Logon Policy Setting Winning GPO Interactive logon: Prompt user to change password before expiration 7 days Default Domain Policy

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  • How do I make Empathy match my keyring with my password?

    - by lisalisa
    I changed my password a few months ago from the password I first used when I installed Ubuntu on my machine. I tried to add a Google Talk account to Empathy, but every time Empathy gives me a message saying the following: Enter password to unlock your login keyring The password you use to log in to your computer no longer matches that of your login keyring. I do not remember my original password and I'm not sure if I should go to Prefrences Passwords and Keys and delete my login password or if there is a way to change the keyring so that it matches up with my current password.

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  • Use Outlook password for website verification

    - by Jack Lockyer
    I am currently building an internal employee dashboard for our global company (it is hosted on an external website for logistical reasons) I'd like (need) to password protect the page as we will be displaying sensitive information, my question is, is it possible to integrate with Outlook passwords? We have over 350 staff all of whom use outlook on a daily basis, I'd love for the website to check whether the visitor is logged into Outlook and if they're not, prompt them to log in. Is it possible?? If it is I'll get is developed straight away.

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  • Windows XP: Consequences of setting a password for an account

    - by sleske
    I do not quite understand how Windows (specifically Windows XP) handles accounts with/without passwords. As far as I can see, on a fresh Windows XP install I have one default account which has admin rights does not have a password will auto-login (without password prompt) when the computer boots What happens if I set a password for this account? Will it still auto-login? Or will it always prompt for the PW? And generally, what consequences does it have if I set a password? I noted that Scheduled Tasks apparently cannot run under an account w/o password (creating a scheduled task will prompt for the account PW). Is there anything that will not work with a password set? Why is it even possible to have accounts without a password? I have some Unix/Linux background, but the concepts appear a little different under Windows.

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  • Broke my sudoers password, how do I reset it without using sudo?

    - by Eric Dand
    I thought it would be a good idea to finally take the password off my little netbook since it has never actually been of any use, and has mostly just slowed down . But when I went to change my password, there wasn't even an option to make it blank, and any attempt to make it a few easy characters was met with "Password too weak". So I did what any good geek would do and popped open the terminal, read the manual entry for passwd and quickly used the -d option to remove the password from my account. It all went well for a couple days (I even managed to also make my keychain password blank) until I tried to update the thing. My sudoers password is not blank, and it's not my old password. I have no idea what it is. How do I reset it (or even better, make it blank) without the use of the sudo command?

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  • problem to change my Xenserver password

    - by Michlaou
    I try to change my root password on my Xenserver 6.0. I follow these steps: enter boot: menu.c32 selecet xe-serial and press tab add "single" before the 2nd triple hyphens and i press enter. I have that: mboot.c32 /boot/xen.gz com1=115200,8n1 console=com1, vga mem=1024G dom0_max_vcpus4 dom0_mem=752M lowmem_emergency_pool=1M crashkernel=64M@32M single --- /boot/vmlinuz-2.6-xen root=LABEL=root-rodraxar ro console=tty0 xencons=hvc console=hvc0 --- /boot/initrd-2.6-xen.img I have commande on the screen and it's stop at: ext3-fs: monted filesystem with ordered data mode. Can you help me?

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