Search Results

Search found 12055 results on 483 pages for 'password complexity'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 57/483 | < Previous Page | 53 54 55 56 57 58 59 60 61 62 63 64  | Next Page >

  • How to pass in password to pg_dump?

    - by Mark
    I'm trying to create a cronjob to back up my database every night before something catastrophic happens. It looks like this command should meet my needs: pg_dump dbname | gzip > $(date +%Y-%m-%d).psql.gz Except after running that, it expects me to type in a password. I can't do that if I run it from cron. How can I pass one in automatically?

    Read the article

  • Android Password GUI

    - by ranjanarr
    I am looking for Android GUI which is wheel type number lock, with numbers on the wheel. The password GUI looks like a wheel with numbers on it and user can roll the wheel to select a number when wheel stops, only the numbers are visible that are on the surface of wheel.

    Read the article

  • Password-protected sharing allows access to users who have no account?

    - by romkyns
    Running Win7 on two computers in my LAN. Computer A has password-protected sharing enabled, and shares a folder. It has a single user account "Bob", and the Guest account is turned off. The network is workgroup-based. According to the descriptions of the "password-protected sharing" I could find, the only people who can access the shared folder via the LAN are those who know the username+password for the "Bob" account. However a second computer on the LAN is able to view this shared folder by simply browsing to Computer A. They don't need to enter any passwords or anything. The only user account registered on that PC is called "Jim", and has a different password from "Bob". How on earth is computer B able to view this shared folder? Is the popular description of the "password-protected sharing" feature inaccurate / did I misunderstand it big time? P.S. There is a possibility that the password for "Bob" has been entered on that PC once, and possibly the "remember password" box was checked. I've looked in the "Credential Manager" on both computers and there is nothing saved anywhere.

    Read the article

  • Google 2-step verification: Should my phone know my password? [closed]

    - by Sir Code-A-Lot
    Hi, Just enabled 2-step verification for my Google account. I have installed Google Authenticator on my Android phone, and I set up an application specific password for the Google account associated on my phone. This works great when just using installed apps like Gmail, Calendar and Google Reader. But if I want to access Google Docs, Google Tasks or any other website that requires a Google login, I don't seem to be able to use a application specific password. I have to use my real password and then use Google Authenticator to make a code for the next step. This means if my phone is stolen, revoking the password to my phone is pointless. The phone have already been verified, and all that is needed is my password, which the phones browser will have remembered. I realize that I can take measures to ensure the phones browser doesn't remember my password, but that's just not convenient at all. Am I missing something, or is there no elegant solution to this? Should I just let my phone know my real password? As I see it, being able to login with application specific passwords on websites (which apparently isn't possible) is the only way I can revoke my phones access in a meaningful way.

    Read the article

  • Trying to use Digest Authentication for Folder Protection

    - by Jon Hazlett
    StackOverflow users suggested I try my question here. I'm using Server 2008 EE and IIS 7. I've got a site that I've migrated over from XP Pro using IIS 5. On the old system, I was using IIS Password to use simple .htaccess files to control a couple of folders that I didn't want to be publicly viewable. Now that I'm running a full-blown DC with a more powerful version of IIS, I decided it'd be a good idea to start using something slightly more sophisticated. After doing my research and trying to keep things as cheap as possible with a touch of extra security, I decided that Digest Authentication would be the best way to go. My issue is this: With Anon access disabled and Digest enabled, I am never prompted for credentials. when on the server, viewing domain[dot]com/example will simply show my 401.htm page without prompting me for credentials. when on a different network/computer, viewing domain[dot]com/example again shows my 401.htm without prompting for credentials. At the site level I only have Anon enabled. Every subfolder, unless I want it protected, has just Anon enabled. Only the folders I want protected have Anon disabled and Digest enabled. I have tried editing the bindings to see if that would spark any kind of change... www.domain.com, domain.com, and localhost have all been tried. There was never a change in behavior at any permutation (aside from the page not being found when I un-bound localhost to the site). I might have screwed up when I deleted the default site from IIS. I didn't think I'd actually need it for anything, but some of what I have read online is telling me otherwise now. As for Digest settings, I have it pointed to local.domain.com, which is the name assigned to my AD Domain. I'm guessing that's right, but honestly have no clue about what a realm actually is. Would it matter that I have an A record for local.domain.com pointing to my IP address? I had problems initially with an absolute link for 401.htm pages, but have since resolved that. Instead of D:\HTTP\401.htm I've used /401.htm and all is well. I used to get error 500's because it couldn't find the custom 401.htm file, but now it loads just fine. As for some data, I was getting entries like this from access logs: 2009-07-10 17:34:12 10.0.0.10 GET /example/ - 80 - [workip] Mozilla/4.0+(compatible;+MSIE+7.0;+Windows+NT+5.1;+.NET+CLR+1.1.4322;+.NET+CLR+2.0.50727;+InfoPath.2) 401 2 5 132 But after correcting my 401.htm links now get logs like this: 2009-07-10 18:56:25 10.0.0.10 GET /example - 80 - [workip] Mozilla/5.0+(Windows;+U;+Windows+NT+5.1;+en-US;+rv:1.9.0.11)+Gecko/2009060215+Firefox/3.0.11 200 0 0 146 I don't know if that means anything or not. I still don't get any credential challenges, regardless of where I try to sign in from ( my workstation, my server, my cellphone even ). The only thing that's seemed to work is viewing localhost and I donno what could be preventing authentication from finding it's way out of the server. Thanks for any help! Jon

    Read the article

  • Hibernate - strange order of native SQL parameters

    - by Xorty
    Hello, I am trying to use native MySQL's MD5 crypto func, so I defined custom insert in my mapping file. <hibernate-mapping package="tutorial"> <class name="com.xorty.mailclient.client.domain.User" table="user"> <id name="login" type="string" column="login"></id> <property name="password"> <column name="password" /> </property> <sql-insert>INSERT INTO user (login,password) VALUES ( ?, MD5(?) )</sql-insert> </class> </hibernate-mapping> Then I create User (pretty simple POJO with just 2 Strings - login and password) and try to persist it. session.beginTransaction(); // we have no such user in here yet User junitUser = (User) session.load(User.class, "junit_user"); assert (null == junitUser); // insert new user junitUser = new User(); junitUser.setLogin("junit_user"); junitUser.setPassword("junitpass"); session.save(junitUser); session.getTransaction().commit(); What actually happens? User is created, but with reversed parameters order. He has login "junitpass" and "junit_user" is MD5 encrypted and stored as password. What did I wrong? Thanks EDIT: ADDING POJO class package com.xorty.mailclient.client.domain; import java.io.Serializable; /** * POJO class representing user. * @author MisoV * @version 0.1 */ public class User implements Serializable { /** * Generated UID */ private static final long serialVersionUID = -969127095912324468L; private String login; private String password; /** * @return login */ public String getLogin() { return login; } /** * @return password */ public String getPassword() { return password; } /** * @param login the login to set */ public void setLogin(String login) { this.login = login; } /** * @param password the password to set */ public void setPassword(String password) { this.password = password; } /** * @see java.lang.Object#toString() * @return login */ @Override public String toString() { return login; } /** * Creates new User. * @param login User's login. * @param password User's password. */ public User(String login, String password) { setLogin(login); setPassword(password); } /** * Default constructor */ public User() { } /** * @return hashCode * @see java.lang.Object#hashCode() */ @Override public int hashCode() { final int prime = 31; int result = 1; result = prime * result + ((null == login) ? 0 : login.hashCode()); result = prime * result + ((null == password) ? 0 : password.hashCode()); return result; } /** * @param obj Compared object * @return True, if objects are same. Else false. * @see java.lang.Object#equals(java.lang.Object) */ @Override public boolean equals(Object obj) { if (this == obj) { return true; } if (obj == null) { return false; } if (!(obj instanceof User)) { return false; } User other = (User) obj; if (login == null) { if (other.login != null) { return false; } } else if (!login.equals(other.login)) { return false; } if (password == null) { if (other.password != null) { return false; } } else if (!password.equals(other.password)) { return false; } return true; } }

    Read the article

  • C++ file input/output search

    - by Brian J
    Hi I took the following code from a program I'm writing to check a user generated string against a dictionary as well as other validation. My problem is that although my dictionary file is referenced correctly,the program gives the default "no dictionary found".I can't see clearly what I'm doing in error here,if anyone has any tips or pointers it would be appreciated, Thanks. //variables for checkWordInFile #define gC_FOUND 99 #define gC_NOT_FOUND -99 // static bool certifyThat(bool condition, const char* error) { if(!condition) printf("%s", error); return !condition; } //method to validate a user generated password following password guidelines. void validatePass() { FILE *fptr; char password[MAX+1]; int iChar,iUpper,iLower,iSymbol,iNumber,iTotal,iResult,iCount; //shows user password guidelines printf("\n\n\t\tPassword rules: "); printf("\n\n\t\t 1. Passwords must be at least 9 characters long and less than 15 characters. "); printf("\n\n\t\t 2. Passwords must have at least 2 numbers in them."); printf("\n\n\t\t 3. Passwords must have at least 2 uppercase letters and 2 lowercase letters in them."); printf("\n\n\t\t 4. Passwords must have at least 1 symbol in them (eg ?, $, £, %)."); printf("\n\n\t\t 5. Passwords may not have small, common words in them eg hat, pow or ate."); //gets user password input get_user_password: printf("\n\n\t\tEnter your password following password rules: "); scanf("%s", &password); iChar = countLetters(password,&iUpper,&iLower,&iSymbol,&iNumber,&iTotal); iUpper = countLetters(password,&iUpper,&iLower,&iSymbol,&iNumber,&iTotal); iLower =countLetters(password,&iUpper,&iLower,&iSymbol,&iNumber,&iTotal); iSymbol =countLetters(password,&iUpper,&iLower,&iSymbol,&iNumber,&iTotal); iNumber = countLetters(password,&iUpper,&iLower,&iSymbol,&iNumber,&iTotal); iTotal = countLetters(password,&iUpper,&iLower,&iSymbol,&iNumber,&iTotal); if(certifyThat(iUpper >= 2, "Not enough uppercase letters!!!\n") || certifyThat(iLower >= 2, "Not enough lowercase letters!!!\n") || certifyThat(iSymbol >= 1, "Not enough symbols!!!\n") || certifyThat(iNumber >= 2, "Not enough numbers!!!\n") || certifyThat(iTotal >= 9, "Not enough characters!!!\n") || certifyThat(iTotal <= 15, "Too many characters!!!\n")) goto get_user_password; iResult = checkWordInFile("dictionary.txt", password); if(certifyThat(iResult != gC_FOUND, "Password contains small common 3 letter word/s.")) goto get_user_password; iResult = checkWordInFile("passHistory.txt",password); if(certifyThat(iResult != gC_FOUND, "Password contains previously used password.")) goto get_user_password; printf("\n\n\n Your new password is verified "); printf(password); //writing password to passHistroy file. fptr = fopen("passHistory.txt", "w"); // create or open the file for( iCount = 0; iCount < 8; iCount++) { fprintf(fptr, "%s\n", password[iCount]); } fclose(fptr); printf("\n\n\n"); system("pause"); }//end validatePass method int checkWordInFile(char * fileName,char * theWord){ FILE * fptr; char fileString[MAX + 1]; int iFound = -99; //open the file fptr = fopen(fileName, "r"); if (fptr == NULL) { printf("\nNo dictionary file\n"); printf("\n\n\n"); system("pause"); return (0); // just exit the program } /* read the contents of the file */ while( fgets(fileString, MAX, fptr) ) { if( 0 == strcmp(theWord, fileString) ) { iFound = -99; } } fclose(fptr); return(0); }//end of checkwORDiNFile

    Read the article

  • Proper way to let user enter password for a bash script using only the GUI (with the terminal hidden)

    - by MountainX
    I have made a bash script that uses kdialog exclusively for interacting with the user. It is launched from a ".desktop" file so the user never sees the terminal. It looks 100% like a GUI app (even though it is just a bash script). It runs in KDE only (Kubuntu 12.04). My only problem is handling password input securely and conveniently. I can't find a satisfactory solution. The script was designed to be run as a normal user and to prompt for the password when a sudo command is first needed. In this way, most commands, those not requiring sudo rights, are run as the normal user. What happens (when the script is run from the terminal) is that the user is prompted for their password once and the default sudo timeout allows the script to finish, including any additional sudo commands, without prompting the user again. This is how I want it to work when run behind the GUI too. The main problem is that using kdesudo to launch my script, which is the standard GUI way, means that the entire script is executed by the root user. So file ownerships get assigned to the root user, I can't rely upon ~/ in paths, and many other things are less than ideal. Running the entire script as the root user is just a very unsatisfactory solution and I think it is a bad practice. I appreciate any ideas for letting a user enter the sudo password just once via GUI while not running the whole script as root. Thanks.

    Read the article

  • Facebook - Isn't this a big vulnerability risk for users? (After Password Change)

    - by Trufa
    I would like to know you opinions as programmers / developers. When I changed my Facebook password yesterday, by mistake I entered the old one and got this: Am I missing something here or this is a big potencial risk for users. In my opinion this is a problem BECAUSE it is FaceBook and is used by, well, everyone and the latest statistics show that 76.3% of the users are idiots [source:me], that is more that 3/4!! All kidding aside: Isn't this useful information for an attacker? It reveals private information about the user! It could help the attacker gain access to another site in which the user used the same password Granted, you should't use use the same password twice (but remember: 76.3%!!!) Doesn't this simply increase the surface area for attackers? It increases the chances of getting useful information at least. In a site like Facebook 1st choice for hackers and (bad) people interested in valued personal information shouldn't anything increasing the chance of a vulnerability be removed? Am I missing something? Am I being paranoid? Will 76.3% of the accounts will be hacked after this post? Thanks in advance!! BTW if you want to try it out, a dummy account: user: [email protected] (old) password: hunter2

    Read the article

  • When decomposing a large function, how can I avoid the complexity from the extra subfunctions?

    - by missingno
    Say I have a large function like the following: function do_lots_of_stuff(){ { //subpart 1 ... } ... { //subpart N ... } } a common pattern is to decompose it into subfunctions function do_lots_of_stuff(){ subpart_1(...) subpart_2(...) ... subpart_N(...) } I usually find that decomposition has two main advantages: The decomposed function becomes much smaller. This can help people read it without getting lost in the details. Parameters have to be explicitly passed to the underlying subfunctions, instead of being implicitly available by just being in scope. This can help readability and modularity in some situations. However, I also find that decomposition has some disadvantages: There are no guarantees that the subfunctions "belong" to do_lots_of_stuff so there is nothing stopping someone from accidentally calling them from a wrong place. A module's complexity grows quadratically with the number of functions we add to it. (There are more possible ways for things to call each other) Therefore: Are there useful convention or coding styles that help me balance the pros and cons of function decomposition or should I just use an editor with code folding and call it a day? EDIT: This problem also applies to functional code (although in a less pressing manner). For example, in a functional setting we would have the subparts be returning values that are combined in the end and the decomposition problem of having lots of subfunctions being able to use each other is still present. We can't always assume that the problem domain will be able to be modeled on just some small simple types with just a few highly orthogonal functions. There will always be complicated algorithms or long lists of business rules that we still want to correctly be able to deal with. function do_lots_of_stuff(){ p1 = subpart_1() p2 = subpart_2() pN = subpart_N() return assembleStuff(p1, p2, ..., pN) }

    Read the article

  • Disable browser 'Save Password' functionality

    - by mattsmith321
    One of the joys of working for a government healthcare agency is having to deal with all of the paranoia around dealing with PHI (Protected Health Information). Don't get me wrong, I'm all for doing everything possible to protect people's personal information (health, financial, surfing habits, etc.), but sometimes people get a little too jumpy. Case in point: One of our state customers recently found out that the browser provides the handy feature to save your password. We all know that it has been there for a while and is completely optional and is up to the end user to decide whether or not it is a smart decision to use or not. However, there is a bit of an uproar at the moment and we are being demanded to find a way to disable that functionality for our site. Question: Is there a way for a site to tell the browser not to offer to remember passwords? I've been around web development a long time but don't know that I have come across that before. Any help is appreciated. Thanks, Matt

    Read the article

  • What is the difference between these two nloglog(n) sorting algorithms? (Andersson et al., 1995 vs.

    - by Yktula
    Swanepoel's comment here lead me to this paper. Then, searching for an implementation in C, I came across this, which referenced another paper on an algorithm described here. Both papers describe integer sorting algorithms that run in O(nloglog(n)) time. What is the difference between the two? Have there been any more recent findings about this topic? Andersson et al., 1995 Han, 2004

    Read the article

  • Big 0 theta notation

    - by niggersak
    Can some pls help with the solution Use big-O notation to classify the traditional grade school algorithms for addition and multiplication. That is, if asked to add two numbers each having N digits, how many individual additions must be performed? If asked to multiply two N-digit numbers, how many individual multiplications are required? Suppose f is a function that returns the result of reversing the string of symbols given as its input, and g is a function that returns the concatenation of the two strings given as its input. If x is the string hrwa, what is returned by g(f(x),x)? Explain your answer - don't just provide the result!

    Read the article

  • Proving that a function f(n) belongs to a Big-Theta(g(n))

    - by PLS
    Its a exercise that ask to indicate the class Big-Theta(g(n)) the functions belongs to and to prove the assertion. In this case f(n) = (n^2+1)^10 By definition f(n) E Big-Theta(g(n)) <= c1*g(n) < f(n) < c2*g(n), where c1 and c2 are two constants. I know that for this specific f(n) the Big-Theta is g(n^20) but I don't know who to prove it properly. I guess I need to manipulate this inequality but I don't know how

    Read the article

  • Stress Test tool for Password Protected Website

    - by Jason
    We need to run a stress test on a password protection section of a website we host. What tool (paid or free) would be best for us to use for this? We'd like to be able to create several 'scripts' and then have the stress test simulate X number of users. Each script will have us login as a specific user and then click on some links and submit forms to simulate an actual user. Ideally the software would also create some nice data exports/charts. Server is a linux web server, but we could run this on linux or Windows so software that will run on either is fine.

    Read the article

  • Amazing families of algorithms over implicit graphs

    - by Diego de Estrada
    Dynamic programming is, almost by definition, to find a shortest/longest path on an implicit dag. Every DP algorithm just does this. An Holographic algorithm can be loosely described as something that counts perfect matchings in implicit planar graphs. So, my question is: are there any other families of algorithms that use well-known algorithms over implicit graphs to achieve a considerable speedup?

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 53 54 55 56 57 58 59 60 61 62 63 64  | Next Page >