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  • Login failed for user 'sa' because the account is currently locked out. The system administrator can

    - by cabhilash
    Login failed for user 'sa' because the account is currently locked out. The system administrator can unlock it. (Microsoft SQL Server, Error: 18486) SQL server has local password policies. If policy is enabled which locks down the account after X number of failed attempts then the account is automatically locked down.This error with 'sa' account is very common. sa is default administartor login available with SQL server. So there are chances that an ousider has tried to bruteforce your system. (This can cause even if a legitimate tries to access the account with wrong password.Sometimes a user would have changed the password without informing others. So the other users would try to lo) You can unlock the account with the following options (use another admin account or connect via windows authentication) Alter account & unlock ALTER LOGIN sa WITH PASSWORD='password' UNLOCK Use another account Almost everyone is aware of the sa account. This can be the potential security risk. Even if you provide strong password hackers can lock the account by providing the wrong password. ( You can provide extra security by installing firewall or changing the default port but these measures are not always practical). As a best practice you can disable the sa account and use another account with same privileges.ALTER LOGIN sa DISABLE You can edit the lock-ot options using gpedit.msc( in command prompt type gpedit.msc and press enter). Navigate to Account Lokout policy as shown in the figure The Following options are available Account lockout threshold This security setting determines the number of failed logon attempts that causes a user account to be locked out. A locked-out account cannot be used until it is reset by an administrator or until the lockout duration for the account has expired. You can set a value between 0 and 999 failed logon attempts. If you set the value to 0, the account will never be locked out. Failed password attempts against workstations or member servers that have been locked using either CTRL+ALT+DELETE or password-protected screen savers count as failed logon attempts. Account lockout duration This security setting determines the number of minutes a locked-out account remains locked out before automatically becoming unlocked. The available range is from 0 minutes through 99,999 minutes. If you set the account lockout duration to 0, the account will be locked out until an administrator explicitly unlocks it. If an account lockout threshold is defined, the account lockout duration must be greater than or equal to the reset time. Default: None, because this policy setting only has meaning when an Account lockout threshold is specified. Reset account lockout counter after This security setting determines the number of minutes that must elapse after a failed logon attempt before the failed logon attempt counter is reset to 0 bad logon attempts. The available range is 1 minute to 99,999 minutes. If an account lockout threshold is defined, this reset time must be less than or equal to the Account lockout duration. Default: None, because this policy setting only has meaning when an Account lockout threshold is specified.When creating SQL user you can set CHECK_POLICY=on which will enforce the windows password policy on the account. The following policies will be applied Define the Enforce password history policy setting so that several previous passwords are remembered. With this policy setting, users cannot use the same password when their password expires.  Define the Maximum password age policy setting so that passwords expire as often as necessary for your environment, typically, every 30 to 90 days. With this policy setting, if an attacker cracks a password, the attacker only has access to the network until the password expires.  Define the Minimum password age policy setting so that passwords cannot be changed until they are more than a certain number of days old. This policy setting works in combination with the Enforce password historypolicy setting. If a minimum password age is defined, users cannot repeatedly change their passwords to get around the Enforce password history policy setting and then use their original password. Users must wait the specified number of days to change their passwords.  Define a Minimum password length policy setting so that passwords must consist of at least a specified number of characters. Long passwords--seven or more characters--are usually stronger than short ones. With this policy setting, users cannot use blank passwords, and they have to create passwords that are a certain number of characters long.  Enable the Password must meet complexity requirements policy setting. This policy setting checks all new passwords to ensure that they meet basic strong password requirements.  Password must meet the following complexity requirement, when they are changed or created: Not contain the user's entire Account Name or entire Full Name. The Account Name and Full Name are parsed for delimiters: commas, periods, dashes or hyphens, underscores, spaces, pound signs, and tabs. If any of these delimiters are found, the Account Name or Full Name are split and all sections are verified not to be included in the password. There is no check for any character or any three characters in succession. Contain characters from three of the following five categories:  English uppercase characters (A through Z) English lowercase characters (a through z) Base 10 digits (0 through 9) Non-alphabetic characters (for example, !, $, #, %) A catch-all category of any Unicode character that does not fall under the previous four categories. This fifth category can be regionally specific.

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  • What is the origin of the name string? [closed]

    - by Andrej M.
    Possible Duplicate: Etymology of “String” Every programmer knows the meaning of the name string. In programming, it is traditionally a sequence of characters. But historically, who has decided that a sequence of characters will be called a string? Has there ever been an attempt to name a sequence of characters differently, but was ultimately abandoned due to the rising popularity of the name string?

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  • How to Handle frame rates and synchronizing screen repaints

    - by David Kroukamp
    I would first off say sorry if the title is worded incorrectly. Okay now let me give the scenario I'm creating a 2 player fighting game, An average battle will include a Map (moving/still) and 2 characters (which are rendered by redrawing a varying amount of sprites one after the other). Now at the moment I have a single game loop limiting me to a set number of frames per second (using Java): Timer timer = new Timer(0, new AbstractAction() { @Override public void actionPerformed(ActionEvent e) { long beginTime; //The time when the cycle begun long timeDiff; //The time it took for the cycle to execute int sleepTime; //ms to sleep (< 0 if we're behind) int fps = 1000 / 40; beginTime = System.nanoTime() / 1000000; //execute loop to update check collisions and draw gameLoop(); //Calculate how long did the cycle take timeDiff = System.nanoTime() / 1000000 - beginTime; //Calculate sleep time sleepTime = fps - (int) (timeDiff); if (sleepTime > 0) {//If sleepTime > 0 we're OK ((Timer)e.getSource()).setDelay(sleepTime); } } }); timer.start(); in gameLoop() characters are drawn to the screen ( a character holds an array of images which consists of their current sprites) every gameLoop() call will change the characters current sprite to the next and loop if the end is reached. But as you can imagine if a sprite is only 3 images in length than calling gameLoop() 40 times will cause the characters movement to be drawn 40/3=13 times. This causes a few minor anomilies in the sprited for some charcters So my question is how would I go about delivering a set amount of frames per second in when I have 2 characters on screen with varying amount of sprites?

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  • Ios Game with many animated Nodes,performance issues

    - by user31929
    I'm working in a large map upside-down game(not tiled map),the map i use is a city. I have to insert many node to create the "life of the city",something like people that cross the streets,cars,etc... Some of this characters are involved in physics and game logic but others are only graphic characters. For what i know the only way i can achive this result is to create each character node with or without physic body and animate each character with a texture atlas. In this way i think that i'll have many performance problems, (the characters will be something like 100/150) even if i'll apply all the performance tips that i know... My question is: with large numbers of characters there another programming pattern that i must follow ? What is the approch of game like simcity,simpsons tapped out for ios,etc... that have so many animation at the same time?

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  • Why doesn't my implementation of ElGamal work for long text strings?

    - by angstrom91
    I'm playing with the El Gamal cryptosystem, and my goal is to be able to encipher and decipher long sequences of text. I have come up with a method that works for short sequences, but does not work for long sequences, and I cannot figure out why. El Gamal requires the plaintext to be an integer. I have turned my string into a byte[] using the .getBytes() method for Strings, and then created a BigInteger out of the byte[]. After encryption/decryption, I turn the BigInteger into a byte[] using the .toByteArray() method for BigIntegers, and then create a new String object from the byte[]. This works perfectly when i call ElGamalEncipher with strings up to 129 characters. With 130 or more characters, the output produced from ElGamalDecipher is garbled. Can someone suggest how to solve this issue? Is this an issue with my method of turning the string into a BigInteger? If so, is there a better way to turn my string of text into a BigInteger and back? Below is my encipher/decipher code with a program to demonstrate the problem. import java.math.BigInteger; public class Main { static BigInteger P = new BigInteger("15893293927989454301918026303382412" + "2586402937727056707057089173871237566896685250125642378268385842" + "6917261652781627945428519810052550093673226849059197769795219973" + "9423619267147615314847625134014485225178547696778149706043781174" + "2873134844164791938367765407368476144402513720666965545242487520" + "288928241768306844169"); static BigInteger G = new BigInteger("33234037774370419907086775226926852" + "1714093595439329931523707339920987838600777935381196897157489391" + "8360683761941170467795379762509619438720072694104701372808513985" + "2267495266642743136795903226571831274837537691982486936010899433" + "1742996138863988537349011363534657200181054004755211807985189183" + "22832092343085067869"); static BigInteger R = new BigInteger("72294619754760174015019300613282868" + "7219874058383991405961870844510501809885568825032608592198728334" + "7842806755320938980653857292210955880919036195738252708294945320" + "3969657021169134916999794791553544054426668823852291733234236693" + "4178738081619274342922698767296233937873073756955509269717272907" + "8566607940937442517"); static BigInteger A = new BigInteger("32189274574111378750865973746687106" + "3695160924347574569923113893643975328118502246784387874381928804" + "6865920942258286938666201264395694101012858796521485171319748255" + "4630425677084511454641229993833255506759834486100188932905136959" + "7287419551379203001848457730376230681693887924162381650252270090" + "28296990388507680954"); public static void main(String[] args) { FewChars(); System.out.println(); ManyChars(); } public static void FewChars() { //ElGamalEncipher(String plaintext, BigInteger p, BigInteger g, BigInteger r) BigInteger[] cipherText = ElGamal.ElGamalEncipher("This is a string " + "of 129 characters which works just fine . This is a string " + "of 129 characters which works just fine . This is a s", P, G, R); System.out.println("This is a string of 129 characters which works " + "just fine . This is a string of 129 characters which works " + "just fine . This is a s"); //ElGamalDecipher(BigInteger c, BigInteger d, BigInteger a, BigInteger p) System.out.println("The decrypted text is: " + ElGamal.ElGamalDecipher(cipherText[0], cipherText[1], A, P)); } public static void ManyChars() { //ElGamalEncipher(String plaintext, BigInteger p, BigInteger g, BigInteger r) BigInteger[] cipherText = ElGamal.ElGamalEncipher("This is a string " + "of 130 characters which doesn’t work! This is a string of " + "130 characters which doesn’t work! This is a string of ", P, G, R); System.out.println("This is a string of 130 characters which doesn’t " + "work! This is a string of 130 characters which doesn’t work!" + " This is a string of "); //ElGamalDecipher(BigInteger c, BigInteger d, BigInteger a, BigInteger p) System.out.println("The decrypted text is: " + ElGamal.ElGamalDecipher(cipherText[0], cipherText[1], A, P)); } } import java.math.BigInteger; import java.security.SecureRandom; public class ElGamal { public static BigInteger[] ElGamalEncipher(String plaintext, BigInteger p, BigInteger g, BigInteger r) { // returns a BigInteger[] cipherText // cipherText[0] is c // cipherText[1] is d SecureRandom sr = new SecureRandom(); BigInteger[] cipherText = new BigInteger[2]; BigInteger pText = new BigInteger(plaintext.getBytes()); // 1: select a random integer k such that 1 <= k <= p-2 BigInteger k = new BigInteger(p.bitLength() - 2, sr); // 2: Compute c = g^k(mod p) BigInteger c = g.modPow(k, p); // 3: Compute d= P*r^k = P(g^a)^k(mod p) BigInteger d = pText.multiply(r.modPow(k, p)).mod(p); // C =(c,d) is the ciphertext cipherText[0] = c; cipherText[1] = d; return cipherText; } public static String ElGamalDecipher(BigInteger c, BigInteger d, BigInteger a, BigInteger p) { //returns the plaintext enciphered as (c,d) // 1: use the private key a to compute the least non-negative residue // of an inverse of (c^a)' (mod p) BigInteger z = c.modPow(a, p).modInverse(p); BigInteger P = z.multiply(d).mod(p); byte[] plainTextArray = P.toByteArray(); return new String(plainTextArray); } }

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  • EPM 11.1.2 - Issues during configuration when using Oracle DB if not using UTF8

    - by Ahmed A
    If you see issues during configuration when using Oracle DB if not using UTF8: Workaround: a. During configuration of EPM products, a warning message is displayed if the Oracle DB is not UTF8 enabled. If you continue with the configuration, certain products will not work as they will not be able to read the contents in the tables as the format is wrong.b. The Oracle DB must be setup to use AL32UTF8 or a superset that contains AL32UTF8. c. The only difference between AL32UTF8 and UTF8 character sets is that AL32UTF8 stores characters beyond U+FFFF as four bytes (exactly as Unicode defines UTF-8). Oracle’s “UTF8” stores these characters as a sequence of two UTF-16 surrogate characters encoded using UTF-8 (or six bytes per character). Besides this storage difference, another difference is better support for supplementary characters in AL32UTF8 character set.

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  • Trouble typing accented letters at the terminal prompt after launching Python

    - by Nicojo
    Edit: Using Mac OSX 10.6, whether I use Terminal.app or iTerm.app, when I launch Python, I can no longer type accented letters (e.g.é or ä). Any ideas? ORIGINAL POST: I am using iTerm 0.10. I would like to type in a string with accented characters (e.g. é) but when I do so at the iTerm prompt, no character appears. This does not occur in Terminal. Could someone help me find out what the problem is, and eventually fix it? EDIT: In Terminal.app, I can use accented characters. However, when I launch the Python 2.71 prompt, I can no longer type in accented characters. When I quit python and return to the terminal prompt, I can again type accented characters. In iTerm, although I quit Python and restarted iTerm, I cannot type in accented characters (I do not know if I could before).

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  • Chinese Characters in email sent via PHP not showing up.

    - by rye
    hi All, a funny problem. I send mail via PHP from my testing server with Chinese chars in it and it sends perfectly. Encoding is utf-8. When I upload the same PHP file to another server and try to send from there, the e-mail will look 90% fine in one mail client (web-based mail actually, gmail), but in my mail client (Apple Mail) it's all gibberish even when I try changing the encoding in the mail client. I'm stuck here because everything works fine on one server, but not on another so I'm not sure where to start looking for solutions. What's even more puzzling is that on the production server, the mail looks somewhat ok (strange case of some characters not showing) but in other mail apps it looks like garbage. any idea where I can start looking to solve this? thanks for any help here! Regards.. php script $books = json_decode ($_POST['books']); $body = ' ?? ' . $_POST['name'] . ',?????????,????????,???????? '; $iLen = count($books); for ($i = 0; $i ' . $book-title . '' . $book-author . ''; $body .= '??: ' . $book-synopsis . ''; $body .= '???: ' . $book-age . ''; $body .= '??: ' . $book-setting . ''; $body .= '??: ' . $book-purpose . ''; $body .= '???: ' . $book-call . ''; $body .= '???: ' . $book-publisher . ''; } $body .= ' ????,Name '; $headers = 'MIME-Version: 1.0' . "\r\n"; $headers .= 'Content-type: text/html; charset=utf-8' . "\r\n"; $headers .= 'From: Name ' . "\r\n"; $ok = mail ($_POST['email'], '???????:???????????', $body, $headers); result ä? å¥? ryan, 以丗æ?¯ä? ä»/å–œä’ ç?Œç«?,ç»?å–©å–?è®”æ??亗è¯=稗,昕蜙æ±?ç°=䒜籟å?ŸåŸ? ç‘?ç‘?ævŒæ?˜å¤°çv±ä? 麜å?—å¸8é?·å°p, å±±å§? Synopsis: ç?—å?¯çv±ç°=å°?å?‰å®?å®?æ•/ä’v牨å®8痬瘒ç°=戒åp?å‚‘å?‰åœvåœv说å®8æ?˜å¤°çv±å®8ã•? Age Group: 4 - 6 å”™ Setting: ç=¤ä?„ Purpose: ä»·å•pè§?å‚‘ä¿8è¿?五å–?ç°=æ=ƒæ8? Call no: JP MAC Publisher: 麜å?—å¸8é?·å°p, å±±å§?. ç‘?ç‘?ævŒæ?˜å¤°çv±ä? .丅海 : 尌咴å=¿ç«¥åOºç˜vç¤=, 2005.

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  • Code Golf: Code 39 Bar Code

    - by gwell
    The challenge The shortest code by character count to draw an ASCII representation of a Code 39 bar code. Wikipedia article about Code 39: http://en.wikipedia.org/wiki/Code_39 Input The input will be a string of legal characters for Code 39 bar codes. This means 43 characters are valid: 0-9 A-Z (space) and -.$/+%. The * character will not appear in the input as it is used as the start and stop characters. Output Each character encoded in Code 39 bar codes have nine elements, five bars and four spaces. Bars will be represented with # characters, and spaces will be represented with the space character. Three of the nine elements will be wide. The narrow elements will be one character wide, and the wide elements will be three characters wide. A inter-character space of a single space should be added between each character pattern. The pattern should be repeated so that the height of the bar code is eight characters high. The start/stop character * (bWbwBwBwb) would be represented like this: # # ### ### # # # ### ### # # # ### ### # # # ### ### # # # ### ### # # # ### ### # # # ### ### # # # ### ### # ^ ^ ^^ ^ ^ ^ ^^^ | | || | | | ||| narrow bar -+ | || | | | ||| wide space ---+ || | | | ||| narrow bar -----+| | | | ||| narrow space ------+ | | | ||| wide bar --------+ | | ||| narrow space ----------+ | ||| wide bar ------------+ ||| narrow space --------------+|| narrow bar ---------------+| inter-character space ----------------+ The start and stop character * will need to be output at the start and end of the bar code. No quiet space will need to be included before or after the bar code. No check digit will need to be calculated. Full ASCII Code39 encoding is not required, just the standard 43 characters. No text needs to be printed below the ASCII bar code representation to identify the output contents. The character # can be replaced with another character of higher density if wanted. Using the full block character U+2588, would allow the bar code to actually scan when printed. Test cases Input: ABC Output: # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # Input: 1/3 Output: # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # Input: - $ (minus space dollar) Output: # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # Code count includes input/output (full program).

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  • Can get members, but not count of NSMutableArray

    - by Curyous
    I'm filling an NSMutableArray from a CoreData call. I can get the first object, but when I try to get the count, the app crashes with Program received signal: “EXC_BAD_ACCESS”. How can I get the count? Here's the relevant code - I've put a comment on the line where it crashes. - (void)viewDidLoad { [super viewDidLoad]; managedObjectContext = [[MySingleton sharedInstance] managedObjectContext]; if (managedObjectContext != nil) { charactersRequest = [[NSFetchRequest alloc] init]; charactersEntity = [NSEntityDescription entityForName:@"Character" inManagedObjectContext:managedObjectContext]; [charactersEntity retain]; [charactersRequest setEntity:charactersEntity]; [charactersRequest retain]; NSError *error; characters = [[managedObjectContext executeFetchRequest:charactersRequest error:&error] mutableCopy]; if (characters == nil) { NSLog(@"Did not get results for characters: %@", error.localizedDescription); } else { [characters retain]; NSLog(@"Found some character(s)."); Character* character = (Character *)[characters objectAtIndex:0]; NSLog(@"Name of first one: %@", character.name); NSLog(@"Found %@ character(s).", characters.count); // Crashes on this line with - Program received signal: “EXC_BAD_ACCESS”. } } } And previous declarations from the header file: @interface CrowdViewController : UITableViewController { NSManagedObjectContext *managedObjectContext; NSFetchRequest *charactersRequest; NSEntityDescription *charactersEntity; NSMutableArray *characters; } I'm a bit perplexed and would really appreciate finding out what is going on.

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  • Escaping In Expressions

    The expressions language is a C style syntax, so you may need to escape certain characters, for example: "C:\FolderPath\" + @VariableName Should be "C:\\FolderPath\\" + @VariableName Another use of the escape sequence allows you to specify character codes, like this \xNNNN, where NNNN is the Unicode character code that you want. For example the following expression will produce the same result as the previous example as the Unicode character code 005C equals a back slash character: "C:\x005CFolderPath\x005C" + @VariableName For more information about Unicode characters see http://www.unicode.org/charts/ Literals are also supported within expressions, both string literals using the common escape sequence syntax as well as modifiers which influence the handling of numeric values. See the "Literals (SSIS)":http://msdn2.microsoft.com/en-US/library/ms141001(SQL.90).aspx topic. Using the Unicode escaped character sequence you can make up for the lack of a CHAR function or equivalent.

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  • How do I edit keyboard preferences from the command line?

    - by jumpnett
    I want to swap the Caps Lock and Escape key as specified in this answer: Use the keyboard preferences to swap Caps Lock and Escape - seriously, how often do you use Caps Lock? Using vim you will be using Escape all the time, and having it available on the home row makes a huge difference. With the standard Ubuntu desktop, go through the menus: System - Preferences - Keyboard - Layouts tab. Then hit the "Layout Options" button, click on the triangle next to "Caps Lock key behaviour" and select "Swap ESC and CapsLock". but, I'm using Ubuntu Server with no gui, so how would I do this from the command line?

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  • HOWTO: Disable complex password policy on Hyper-V Server 2008?

    - by Ian Boyd
    How do you disable the password complexity requirements on a Microsoft Hyper-V Server 2008 R2? Keep in mind that when you log into the server, the only UI you have is: And you cannot run gpedit.msc: C:\Users\Administrator>gpedit.msc 'gpedit.msc' is not recognized as an internal or external command, operable program or batch file. because there are no .msc snap-ins installed with Microsoft Hyper-V Server 2008 R2. The problem comes when you're trying to add an account to the server, so you can manage it, but it doesn't like most passwords: And, predictably, typing NET HELPMSG 2245 gives you The password does not meet the password policy requirements. Check the minimum p assword length, password complexity and password history requirements. i hoped it would have been a friendly user experience, and either: offered to disable the password policy tell me how to disable the password policy tell me how to check the minimum password length, password complexity and password history requirements. Password Complexity Requirements The Microsoft's default password complexity for Server Core is: Passwords cannot contain the user’s account name or parts of the user’s full name that exceed two consecutive characters. Passwords must be at least six characters in length. Passwords must contain characters from three of the following four categories: 1.English uppercase characters (A through Z). 2.English lowercase characters (a through z). 3.Base 10 digits (0 through 9). 4.Non-alphabetic characters (for example, !, $, #, %). External links Technet Forums: Hyper-V Server disable complex passwords Technet: Passwords must meet complexity requirements of the installed password filter Update: 2k views? So many people keep coming coming to it: up-vote it!

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  • SEO issue red characters in source code? &gt; Why? Syntax highlighting? browser source code?

    - by judi
    SEO issue red characters Hi all I'm building webstes using dreamweaver, but when I look at the source code it is red for &quot; characters. I'm told anything appearing in red puts off Google's seo. Does anyone know why this appears in red? For example when I view code source on the site i get the gt; in red <a href="miss-sold-mortgages.html" class="darkblue">Find out more&gt;&gt;</a></span> </div> Thanks for your help Regards Judi

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  • In SQL Server changing column varchar(255) nvarchar

    - by JD
    Hi, I am using SQL server 2008 express and some of our columns are defined as varchar(255). Should I convert these columns to NvarChar(255) or nvarchar(max)? The reason I ask is I read that nvarchar(255) for unicode characters would actually store 1/2 the number of characters (since unicode characters are 2 bytes) whereas 255 with varchar() would allow me to store 255 characters (or is it 255 - 2 for the offset). Would there be any performance hits using nvarchar(max)? JDs

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  • Decode base64 data as array in Python

    - by skerit
    I'm using this handy Javascript function to decode a base64 string and get an array in return. This is the string: base64_decode_array('6gAAAOsAAADsAAAACAEAAAkBAAAKAQAAJgEAACcBAAAoAQAA') This is what's returned: 234,0,0,0,235,0,0,0,236,0,0,0,8,1,0,0,9,1,0,0,10,1,0,0,38,1,0,0,39,1,0,0,40,1,0,0 The problem is I don't really understand the javascript function: var base64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'.split(""); var base64inv = {}; for (var i = 0; i < base64chars.length; i++) { base64inv[base64chars[i]] = i; } function base64_decode_array (s) { // remove/ignore any characters not in the base64 characters list // or the pad character -- particularly newlines s = s.replace(new RegExp('[^'+base64chars.join("")+'=]', 'g'), ""); // replace any incoming padding with a zero pad (the 'A' character is zero) var p = (s.charAt(s.length-1) == '=' ? (s.charAt(s.length-2) == '=' ? 'AA' : 'A') : ""); var r = []; s = s.substr(0, s.length - p.length) + p; // increment over the length of this encrypted string, four characters at a time for (var c = 0; c < s.length; c += 4) { // each of these four characters represents a 6-bit index in the base64 characters list // which, when concatenated, will give the 24-bit number for the original 3 characters var n = (base64inv[s.charAt(c)] << 18) + (base64inv[s.charAt(c+1)] << 12) + (base64inv[s.charAt(c+2)] << 6) + base64inv[s.charAt(c+3)]; // split the 24-bit number into the original three 8-bit (ASCII) characters r.push((n >>> 16) & 255); r.push((n >>> 8) & 255); r.push(n & 255); } // remove any zero pad that was added to make this a multiple of 24 bits return r; } What's the function of those "<<<" and "" characters. Or is there a function like this for Python?

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  • Regular Expression for username

    - by neobie
    I need help on regular expression on the condition (4) below: Begin with a-z End with a-z0-9 allow 3 special characters like ._- The characters in (3) must be followed by alphanumeric characters, and it cannot be followed by any characters in (3) themselves. Not sure how to do this. Any help is appreciated, with the sample and some explanations.

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  • Must go through Windows Boot Loader to get to Grub

    - by Zach
    I just installed a fresh copy of Precise alongside Windows 7. I have to separate 750GB hard drives; /dev/sda holds the Windows partitions and /dev/sdb holds the Ubuntu partitions. Other than that, these are fresh installs of both Windows 7 and Ubuntu 12.04. Whenever I boot, Grub doesn't load, instead it goes to a black screen with a single blinking (horizontal bar) cursor in the top right corner. However, if I boot, hit escape right as the BIOS/POST screen finishes up, see the Windows Boot Loader and hit escape to make it go back to the BIOS screen. After the BIOS screen, grub shows up and everything functions normally; I can boot into Ubuntu or Win7. I don't want to have to do the Escape, Escape, Wait, Boot trick every time. I have no idea what would be wrong or what information I could give you guys to help diagnose. I have run a sudo update-grub and it found everything normally. I tried adding nomodeset flag in the /etc/default/grub line GRUB_CMDLINE_LINUX_DEFAULT which searching around made me think might work. Thoughts on what I could do to fix this? EDIT: I've tried changing the boot order so that both drives in the BIOS (both are labeled as "Internal HDD") have had a try booting first. I think the problem may be that every time I boot, the BIOS boot order is different... and I have to reset it. It seems to not be stable... but I'm not sure how to go about fixing that either. The machine has both traditional BIOS and UEFI. It came standard in "Legacy" mode; so it is currently set to boot through Legacy mode. I've reinstalled Ubuntu now, and now if I hit escape at the end of the BIOS/POST startup screen, it takes me to GRUB menu. Otherwise it automatically loads Windows. It seems like GRUB is now the acting bootloader, it just doesn't automatically start that unless I ask it to open a bootloader. In my other machines, it has always automatically started at the end of BIOS/POST. EDIT2: Using gparted, I just looked at my partitions, it would seem that my linux-swap partition is currently flagged as the boot partition for my Ubuntu install. I currently only have 2 partitions: one of "ext4" with a mount point of "/" and flag " "; and the "linux-swap" with mount point " " and flag "boot." If I change the boot flag to be on "/," it does not reliably solve the problem. After 10 boots: 2 Booted successfully to GRUB 5 Booted directly to Windows 7 3 booted to the black screen with the cursor and hung there Further research makes me think this is an issue of the BIOS not reliably booting hard drives in the same order or not finding both hard drives. If I ask it to create a "boot menu" sometimes it has 2 entries for "Internal HDD," sometimes 1. Also the list it creates changes order every time I bring it up; so it is not following a consistent boot sequence. Will report back if this is not an issue with GRUB.

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  • C# Regex - Replace multiple characters at once without overwriting?

    - by Everaldo Aguiar
    Hello guys, I'm implementing a c# program that should automatize a Mono-alphabetic substitution cipher. The functionality i'm working on at the moment is the simplest one: The user will provide a plain text and a cipher alphabet, for example: Plain text(input): THIS IS A TEST Cipher alphabet: A - Y, H - Z, I - K, S - L, E - J, T - Q Cipher Text(output): QZKL KL QJLQ I thought of using regular expressions since I've been programming in perl for a while, but I'm encountering some problems on c#. First I would like to know if someone would have a suggestion for a regular expression that would replace all occurrence of each letter by its corresponding cipher letter (provided by user) at once and without overwriting anything. Example: In this case, user provides plaintext "TEST", and on his cipher alphabet, he wishes to have all his T's replaced with E's, E's replaced with Y and S replaced with J. My first thought was to substitute each occurrence of a letter with an individual character and then replace that character by the cipherletter corresponding to the plaintext letter provided. Using the same example word "TEST", the steps taken by the program to provide an answer would be: 1 - replace T's with (lets say) @ 2 - replace E's with # 3 - replace S's with & 4 - Replace @ with E, # with Y, & with j 5 - Output = EYJE This solution doesn't seem to work for large texts. I would like to know if anyone can think of a single regular expression that would allow me to replace each letter in a given text by its corresponding letter in a 26-letter cipher alphabet without the need of splitting the task in an intermediate step as I mentioned. If it helps visualize the process, this is a print screen of my GUI for the program: http://img43.imageshack.us/img43/2118/11618743.jpg

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  • how can i convert a .tpl file to a .php file? [closed]

    - by kim
    What do I do?? I am building a site and there is a categories.tpl that I want to go where sitemap.php is. sorry i am brand new to all this. let me try to be more clear.id show you a picture but it is marking it as spam. i have a menu at the top of my site like with any retail site. [About Cart Account and Products]. when you click products it takes to you the sitemap.php file. however i need the content from the categories.tpl to appear instead. (Categories in prestashop is another way of saying products) here is the categories.tpl code: {include file=$tpl_dir./breadcrumb.tpl} {include file=$tpl_dir./errors.tpl} {if $category-id AND $category-active} {$category-name|escape:'htmlall':'UTF-8'} {$nb_products|intval} {if $nb_products1}{l s='products'}{else}{l s='product'}{/if} {if $scenes} <!-- Scenes --> {include file=$tpl_dir./scenes.tpl scenes=$scenes} {else} <!-- Category image --> {if $category->id_image} <img src="{$link->getCatImageLink($category->link_rewrite, $category->id_image, 'category')}" alt="{$category->name|escape:'htmlall':'UTF-8'}" title="{$category->name|escape:'htmlall':'UTF-8'}" id="categoryImage" /> {/if} {/if} {if $category->description} <div class="cat_desc">{$category->description}</div> {/if} {if isset($subcategories)} <!-- Subcategories --> <div id="subcategories"> <h3>{l s='Subcategories'}</h3> <ul class="inline_list"> {foreach from=$subcategories item=subcategory} <li> <a href="{$link->getCategoryLink($subcategory.id_category, $subcategory.link_rewrite)|escape:'htmlall':'UTF-8'}" title="{$subcategory.name|escape:'htmlall':'UTF-8'}"> {if $subcategory.id_image} <img src="{$link->getCatImageLink($subcategory.link_rewrite, $subcategory.id_image, 'medium')}" alt="" /> {else} <img src="{$img_cat_dir}default-medium.jpg" alt="" /> {/if} </a> <br /> <a href="{$link->getCategoryLink($subcategory.id_category, $subcategory.link_rewrite)|escape:'htmlall':'UTF-8'}">{$subcategory.name|escape:'htmlall':'UTF-8'}</a> </li> {/foreach} </ul> <br class="clear"/> </div> {/if} {if $products} {include file=$tpl_dir./product-sort.tpl} {include file=$tpl_dir./product-list.tpl products=$products} {include file=$tpl_dir./pagination.tpl} {elseif !isset($subcategories)} <p class="warning">{l s='There is no product in this category.'}</p> {/if} {elseif $category-id} {l s='This category is currently unavailable.'} {/if} and here is the sitemap.php include(dirname(FILE).'/config/config.inc.php'); include(dirname(FILE).'/header.php'); include(dirname(FILE).'/product-sort.php'); $nbProducts = intval(Product::getNewProducts(intval($cookie-id_lang), isset($p) ? intval($p) - 1 : NULL, isset($n) ? intval($n) : NULL, true)); include(dirname(FILE).'/pagination.php'); $smarty-assign(array( 'products' = Product::getNewProducts(intval($cookie-id_lang), intval($p) - 1, intval($n), false, $orderBy, $orderWay), 'nbProducts' = intval($nbProducts))); $smarty-display(_PS_THEME_DIR_.'new-products.tpl'); include(dirname(FILE).'/footer.php'); ?

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  • Microsoft Chart control converts \n in file names to newline characters.

    - by xpda
    I am using a Microsoft Chart control (system.windows.forms.datavisualization.charting.chart) in a Windows forms application, vb.net 2008. I use folder paths for the x values in a pie chart. Chart control converts a name like c:\newfolder into c:[newline]ewfolder. I tried adding a slash, making it c:\\newfolder, but this only changes it to c:\[newline]ewfolder. Is there a workaround for this behavior?

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