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  • Did anyone have this issue with a simple Facebook app or know how to solve it?

    - by Jian Lin
    I have a really simple few lines of Facebook app, using the new Facebook API: <pre> <?php require 'facebook.php'; // Create our Application instance. $facebook = new Facebook(array( 'appId' => '117676584930569', 'secret' => '**********', // hidden here on the post... 'cookie' => true, )); var_dump($facebook); ?> but it is giving me the following output: http://apps.facebook.com/woolaladev/i2.php would give out object(Facebook)#1 (6) { ["appId:protected"]=> string(15) "117676584930569" ["apiSecret:protected"]=> string(32) "**********" <--- just hidden on this post ["session:protected"]=> NULL ["sessionLoaded:protected"]=> bool(false) ["cookieSupport:protected"]=> bool(true) ["baseDomain:protected"]=> string(0) "" } Session is NULL for some reason, but I am logged in and can access my home and profile and run other apps on Facebook (to see that I am logged on). I am following the sample on: http://github.com/facebook/php-sdk/blob/master/examples/example.php http://github.com/facebook/php-sdk/blob/master/src/facebook.php (download using raw URL: wget http://github.com/facebook/php-sdk/raw/master/src/facebook.php ) Trying on both hosting companies at dreamhost.com and netfirms.com, and the results are the same.

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  • Using PHP and cURL to login to indyarocks.com

    - by Divya
    I am new to cURL and don't know much about it. I basically want to login to my account on www.indyarocks.com through libcurl for PHP. I don't know what type of authentication it uses (I don't know how to find that out.). When I go to http://www.indyarocks.com, I get a login form which asks for my username and password. I put in my username and password and click login and everything is good. I tried to automate this using cURL. This is a snippet of my code. curl_setopt($curl_connection, CURLOPT_URL, "http://www.indyarocks.com/loginchk.php"); curl_setopt($curl_connection, CURLOPT_POST, 1); curl_setopt($curl_connection, CURLOPT_HTTPAUTH, CURLAUTH_ANY); curl_setopt($curl_connection, CURLOPT_USERPWD, $username.':'.$password); I looked at the source of the login page and found out the address of the page to which the username and password are sent (the action attribute of the form) which is "http://www.indyarocks.com/loginchk.php" and set it as the target url. When I run this, I get username or password is wrong error and the login fails. My username and password is correct. I don't know what the problem is. Can the password be encrypted? Can that be responsible for this failure? Please help me get around this problem. I'll be really thankful. Thanks in advance.

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  • SSL + Jquery + Ajax

    - by chobo2
    Hi I starting too look at a bit of security into my site. My site I would consider a very low security risk as it has really no personal information from the user other than email. However the security risk will go up a bit as I am partnering with a company and the initial password for this companies users will be the same password they use essentially to get onto the network and every piece of software. So I have up my security( what is fine by me...I wanted to get around to this anyways). So one of my security concerns is this. A user logs in. form submit(non ajax is done). Password is hashed & Salted and compared to one in the database. Reject or let them proceed. So this uses no jquery or ajax but is just asp.net mvc and C#. Still if my understanding is right the password is sent in clear text. So if a use SSL and I would not need to worry about that is this correct? If that is true is that all I need? Second the user can change their password at anytime. This is done through ajax. So when the password is sent it is sent in clear text( and I can verify this by looking at firebug). So if I have SSL enabled on this page is that all I need or do I need to do more? So I am just kinda confused of what I need to make the password being sent to the server(both ajax and full post ways secure). I am not sure if I need to do more then SSL or if that is enough and if it is not enough what is the next layer of security?

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  • Basic security, PHP mySQl

    - by yuudachi
    So I am making a basic log-in page. I have a good idea of what to do, but I'm still unsure of some things. I have a database full of students and a password column of course. I know I'm going to use md5 encryption in that column. The student enters their e-mail and student ID, and they get e-mailed a password if correct. But, where do I create the password? Do I have to manually add the password (which is just a randomly generated string) in mySQL to all the students? And I am suppose to send the password to the student; how will I know what to send the student if the password is encrypted? I was thinking about generating the password when the student first enters their e-mail and student ID. They get an e-mail of the random string, and at the same time, I add the same random string to the database, encrypted. Is that how it's suppose to work though? And it feels unsafe doing that all on the same page. Sorry for the long-winded, newbish question. I find this all facisnating at the same time as well (AES and RSA encryption :O)

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  • Error occurs while validating form input using jQuery in Firebug

    - by Param-Ganak
    I have written a custom validation code in jQuery, which is working fine. I have a login form which has two fields, i.e. userid and password. I have written a custom code for client side validation for these fields. This code is working fine and gives me proper error messages as per the situation. But the problem with this code is that when I enter the invalid data in any or both field and press submit button of form then it displays the proper error message but at the same time when I checked it in Firebug it displays following error message when submit button of the form is clicked validate is not defined function onclick(event) { javascript: return validate(); } (click clientX=473, clientY=273) Here is the JQUERY validation code $(document).ready(function (){ $("#id_login_form").validate({ rules: { userid: { required: true, minlength: 6, maxlength: 20, // basic: true }, password: { required: true, minlength: 6, maxlength: 15, // basic: true } }, messages: { userid: { required: " Please enter the username.", minlength: "User Name should be minimum 6 characters long.", maxlength: "User Name should be maximum 15 characters long.", // basic: "working here" }, password: { required: " Please enter the password.", minlength: "Password should be minimum 6 characters long.", maxlength: "Password should be maximum 15 characters long.", // basic: "working here too.", } }, errorClass: "errortext", errorLabelContainer: "#messagebox" } }); }); /* $.validator.addMethod('username_alphanum', function (value) { return /^(?![0-9]+$)[a-zA-Z 0-9_.]+$/.test(value); }, 'User name should be alphabetic or Alphanumeric and may contain . and _.'); $.validator.addMethod('alphanum', function (value) { return /^(?![a-zA-Z]+$)(?![0-9]+$)[a-zA-Z 0-9]+$/.test(value); }, 'Password should be Alphanumeric.'); $.validator.addMethod('basic', function (value) { return /^[a-zA-Z 0-9_.]+$/.test(value); }, 'working working working'); */ So please tell me where is I am wrong in my jQuery code. Thank You!

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  • How to add Remember me function at custom login box ?

    - by morningglory
    In my theme, there's custom page for the login. Login function at functions.php is like this function log_in($username, $password) { $user = parse_user($username); $username = $username; $password = $password; if(isEmptyString($username)) return new WP_Error('username', 'required'); if(isEmptyString($password)) return new WP_Error('password', "required"); if(!wp_check_password( $password, $user->user_pass ) ) return new WP_Error('wrong_password', "wrong"); wp_set_auth_cookie($user->ID, $remember); wp_login($username, $password); redirect_profile(); } function parse_user($info = null, $return = 'object') { if ( is_null( $info ) ) { global $current_user; if ( empty( $current_user->ID ) ) return null; $info = get_userdata( $current_user->ID ); } elseif ( empty( $info ) ) { return null; } if( $return == 'ID' ) { if ( is_object( $info ) ) return $info->ID; if ( is_numeric( $info ) ) return $info; } elseif( $return == 'object' ) { if ( is_object( $info ) && $info->ID) return $info; if ( is_object( $info )) return get_userdata( $info->ID ); if ( is_numeric( $info ) ) return get_userdata( $info ); if ( is_string( $info ) ) return get_userdatabylogin( $info ); } else { return null; } } I want to add remember me checkbox for user to logged in all the time until they logout. How can i add this ? Please kindly help me out. Thank you.

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  • Link Button on asp.net user control not firing

    - by andyriome
    Hi I have a user control, which is added to another user control. The nested user control is built up of a gridview, an image button and a link button. The nested user control is added to the outer control as a collection object based upon the results bound to the gridview. The problem that I have is that my link button doesn't work. I click on it and the event doesn't fire. Even adding a break point was not reached. As the nested user control is added a number of times, I have set image button to have unique ids and also the link button. Whilst image button works correctly with its java script. The link button needs to fire an event in the code behind, but despite all my efforts, I can't make it work. I am adding the link button to the control dynamically. Below is the relevant code that I am using: public partial class ucCustomerDetails : System.Web.UI.UserControl { protected override void CreateChildControls( ) { base.CreateChildControls( ); string strUniqueID = lnkShowAllCust.UniqueID; strUniqueID = strUniqueID.Replace('$','_'); this.lnkShowAllCust.ID = strUniqueID; this.lnkShowAllCust.Click += new EventHandler(this.lnkShowAllCust_Click); this.Controls.Add(lnkShowAllCust); } protected override void OnInit (EventArgs e) { CreateChildControls( ); base.OnInit(e); } protected override void OnLoad(EventArgs e) { base.EnsureChildControls( ); } protected void Page_Load(object sender, EventArgs e) { if (IsPostBack) { CreateChildControls( ); } } protected void lnkShowAllCust_Click(object sender, EventArgs e) { this.OnCustShowAllClicked(new EventArgs ( )); } protected virtual void OnCustShowAllClicked(EventArgs args) { if (this.ViewAllClicked != null) { this.ViewAllClicked(this, args); } } public event EventHandler ViewAllClicked; } I have been stuggling with this problem for the last 3 days and have had no success with it, and I really do need some help. Can anyone please help me?

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  • CustomButton Field not aligning in the center

    - by rupesh
    Hi all I have created a custom button and i am placing the bunch of Custombuttons in a verticalfieldManager , I have aligned the verticalField Manager in the center. when i am creating a default buttonField then verticalfield Manager is able to align the buttonfield in the center. but when i am assigning custombuttonfield in the verticalField Manager it's not aligning in the center. here is my custombuttoncode enter code herepublic CustomButtonField(String label,long style) { super(style); this.label = label; onPicture = Bitmap.getBitmapResource(onPicturePath); font = getFont(); this.setPadding(5,5, 5, 5); } public String getLabel() { return label; } public int getPreferredHeight() { return onPicture.getHeight(); } public int getPreferredWidth() { return onPicture.getWidth(); } protected void layout(int width , int height) { setExtent(Math.min(width, Display.getWidth()), Math.min(height,getPreferredHeight())); } protected void paint(Graphics graphics) { int texty =(getHeight()-getFont().getHeight())/2; if (isFocus()) { graphics.setColor(Color.BLACK); graphics.drawBitmap(0, 0, getWidth(), getHeight(),onPicture , 0, 0); graphics.setColor(Color.WHITE); graphics.setFont(font); graphics.drawText(label,0,texty,DrawStyle.ELLIPSIS,getWidth()); } else { graphics.drawBitmap(0, 0, getWidth(), getHeight(),onPicture , 0, 0); graphics.setColor(Color.WHITE); graphics.setFont(font); graphics.drawText(label,0,texty,DrawStyle.ELLIPSIS,getWidth()); } } public boolean isFocusable() { return true; } protected void onFocus(int direction) { super.onFocus(direction); invalidate(); } protected void onUnfocus() { super.onUnfocus(); invalidate(); } protected boolean navigationClick(int status, int time) { fieldChangeNotify(0); return true; } protected boolean keyChar(char character, int status, int time) { if (character == Keypad.KEY_ENTER) { fieldChangeNotify(0); return true; } return super.keyChar(character, status, time); } }

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  • Using h:outputFormat to message-format the f:selectItems of a h:selectOneRadio

    - by msharma
    I am having some trouble with using h:selectOneRadio. I have a list of objects which is being returned which needs to be displayed. I am trying something like this: <h:selectOneRadio id="selectPlan" layout="pageDirection"> <f:selectItems value="#{detailsHandler.planList}" /> </h:selectOneRadio> and planList is a List of Plans. Plan is defined as: public class Plan { protected String id; protected String partNumber; protected String shortName; protected String price; protected boolean isService; protected boolean isOption; //With all getters/setters } The text that must appear for each radio button is actually in a properties file, and I need to insert params in the text to fill out some value in the bean. For example the text in my properties file is: plan_price=The price of this plan is {0}. I was hoping to do something like this: <f:selectItems value="<h:outputFormat value="#{i18n.plan_price}"> <f:param value="#{planHandler.price}"> </h:outputFormat>" /> Usually if it's not a h:selectOneRadio component, if it's just text I use the h:outputFormat along with f:param tags to display the messages in my .property file called i18n above, and insert a param which is in the backing bean. here this does not work. Does anyone have any ideas how I can deal with this? I am being returned a list of Plans each with their own prices and the text to be displayed is held in property file. Any help much appreciated. Thanks!

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  • How to insert form data into MySQL database table

    - by Richard
    So I have this registration script: The HTML: <form action="register.php" method="POST"> <label>Username:</label> <input type="text" name="username" /><br /> <label>Password:</label> <input type="text" name="password" /><br /> <label>Gender:</label> <select name="gender"> <optgroup label="genderset"> <option value="Male">Male</option> <option value="Female">Female</option> <option value="Hermaphrodite">Hermaphrodite</option> <option value="Not Sure!!!">Not Sure!!!</option> </optgroup> </select><br /> <input type="submit" value="Register" /> </form> The PHP/SQL: <?php $username = $_POST['username']; $password = $_POST['password']; $gender = $_POST['gender']; mysql_query("INSERT INTO registration_info (username, password, gender) VALUES ('$username', '$password', '$gender') ") ?> The problem is, the username and password gets inserted into the "registration_info" table just fine. But the Gender input from the select drop down menu doesn't. Can some one tell me how to fix this, thanks.

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  • Stored procedure or function expects parameter which is not supplied

    - by user2920046
    I am trying to insert data into a SQL Server database by calling a stored procedure, but I am getting the error Procedure or function 'SHOWuser' expects parameter '@userID', which was not supplied. My stored procedure is called "SHOWuser". I have checked it thoroughly and no parameters is missing. My code is: public void SHOWuser(string userName, string password, string emailAddress, List preferences) { SqlConnection dbcon = new SqlConnection(conn); try { SqlCommand cmd = new SqlCommand(); cmd.Connection = dbcon; cmd.CommandType = System.Data.CommandType.StoredProcedure; cmd.CommandText = "SHOWuser"; cmd.Parameters.AddWithValue("@userName", userName); cmd.Parameters.AddWithValue("@password", password); cmd.Parameters.AddWithValue("@emailAddress", emailAddress); dbcon.Open(); int i = Convert.ToInt32(cmd.ExecuteScalar()); cmd.Parameters.Clear(); cmd.CommandText = "tbl_pref"; foreach (int preference in preferences) { cmd.Parameters.Clear(); cmd.Parameters.AddWithValue("@userID", Convert.ToInt32(i)); cmd.Parameters.AddWithValue("@preferenceID", Convert.ToInt32(preference)); cmd.ExecuteNonQuery(); } } catch (Exception) { throw; } finally { dbcon.Close(); } and the stored procedure is: ALTER PROCEDURE [dbo].[SHOWuser] -- Add the parameters for the stored procedure here ( @userName varchar(50), @password nvarchar(50), @emailAddress nvarchar(50) ) AS BEGIN INSERT INTO tbl_user(userName,password,emailAddress) values(@userName,@password,@emailAddress) select tbl_user.userID,tbl_user.userName,tbl_user.password,tbl_user.emailAddress, stuff((select ',' + preferenceName from tbl_pref_master inner join tbl_preferences on tbl_pref_master.preferenceID = tbl_preferences.preferenceID where tbl_preferences.userID=tbl_user.userID FOR XML PATH ('')),1,1,' ' ) AS Preferences from tbl_user SELECT SCOPE_IDENTITY(); END Pls help, Thankx in advance...

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  • Variable won't store in session

    - by Mittens
    So I'm trying to store the "rank" of a user when they log in to a control panel which displays different options depending on the given rank. I used the same method as I did for storing and displaying the username, which is displayed on the top of each page and works just fine. I can't for the life of me figure out why it won't work for the rank value, but I do know that it is not saving it in the session. Here is the bit that's not working; $username = ($_POST['username']); $password = hash('sha512', $_POST['password']); $dbhost = 'mysql:host=¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦;dbname=¦¦¦¦¦¦¦¦¦¦¦'; $dbuser = '¦¦¦¦¦¦¦¦¦¦¦'; $dbpassword = '¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦'; try { $db = new PDO($dbhost, $dbuser, $dbpassword); $statement = $db->prepare("select password from users where email = :name"); $statement->execute(array(':name' => $username)); $result = $statement->fetch(); $pass = $result[password]; $rank = $result[rank];} catch(PDOException $e) {echo $e->getMessage();} if ($password == $pass) { session_start(); $_SESSION['username'] = $username; $_SESSION['rank'] = $rank; header('Location: http://¦¦¦¦¦¦¦¦¦.ca/manage.php'); } else{ include'../../includes/head.inc'; echo '<h1>Incorrect username or password.</h1>'; include'../../includes/footer.inc'; } I'm also new to the whole PDO thing, hence why my method of authenticating the password is pretty sketchy.

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  • Pipelining String in Powershell

    - by ChvyVele
    I'm trying to make a simple PowerShell function to have a Linux-style ssh command. Such as: ssh username@url I'm using plink to do this, and this is the function I have written: function ssh { param($usernameAndServer) $myArray = $usernameAndServer.Split("@") $myArray[0] | C:\plink.exe -ssh $myArray[1] } If entered correctly by the user, $myArray[0] is the username and $myArray[1] is the URL. Thus, it connects to the URL and when you're prompted for a username, the username is streamed in using the pipeline. Everything works perfectly, except the pipeline keeps feeding the username ($myArray[0]) and it is entered as the password over and over. Example: PS C:\Users\Mike> ssh xxxxx@yyyyy login as: xxxxx@yyyyy's password: Access denied xxxxx@yyyyy's password: Access denied xxxxx@yyyyy's password: Access denied xxxxx@yyyyy's password: Access denied xxxxx@yyyy's password: Access denied xxxxx@yyyyy's password: FATAL ERROR: Server sent disconnect message type 2 (protocol error): "Too many authentication failures for xxxxx" Where the username has been substituted with xxxxx and the URL has been substituted with yyyyy. Basically, I need to find out how to stop the script from piping in the username ($myArray[0]) after it has been entered once. Any ideas? I've looked all over the internet for a solution and haven't found anything.

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  • Why can't we just use a hash of passphrase as the encryption key (and IV) with symmetric encryption algorithms?

    - by TX_
    Inspired by my previous question, now I have a very interesting idea: Do you really ever need to use Rfc2898DeriveBytes or similar classes to "securely derive" the encryption key and initialization vector from the passphrase string, or will just a simple hash of that string work equally well as a key/IV, when encrypting the data with symmetric algorithm (e.g. AES, DES, etc.)? I see tons of AES encryption code snippets, where Rfc2898DeriveBytes class is used to derive the encryption key and initialization vector (IV) from the password string. It is assumed that one should use a random salt and a shitload of iterations to derive secure enough key/IV for the encryption. While deriving bytes from password string using this method is quite useful in some scenarios, I think that's not applicable when encrypting data with symmetric algorithms! Here is why: using salt makes sense when there is a possibility to build precalculated rainbow tables, and when attacker gets his hands on hash he looks up the original password as a result. But... with symmetric data encryption, I think this is not required, as the hash of password string, or the encryption key, is never stored anywhere. So, if we just get the SHA1 hash of password, and use it as the encryption key/IV, isn't that going to be equally secure? What is the purpose of using Rfc2898DeriveBytes class to generate key/IV from password string (which is a very very performance-intensive operation), when we could just use a SHA1 (or any other) hash of that password? Hash would result in random bit distribution in a key (as opposed to using string bytes directly). And attacker would have to brute-force the whole range of key (e.g. if key length is 256bit he would have to try 2^256 combinations) anyway. So either I'm wrong in a dangerous way, or all those samples of AES encryption (including many upvoted answers here at SO), etc. that use Rfc2898DeriveBytes method to generate encryption key and IV are just wrong.

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  • PHP login, getting wrong count value from query / fetch array

    - by Chris
    Hello, *EDIT*Thanks to the comments below it has been figured out that the problem lies with the md5, without everything works as it should. But how do i implent the md5 then? I am having some troubles with the following code below to login. The database and register system are already working. The problem lies that it does not find any result at all in the query. IF the count is 0 it should redirect the user to a secured page. But this only works if i write count = 0, but this should be 0 , only if the user name and password is found he should be directed to the secure (startpage) of the site after login. For example root (username) root (password) already exists but i cannot seem to properly login with it. <?php session_start(); if (!empty($_POST["send"])) { $username = ($_POST["username"]); $password = (md5($_POST["password"])); $count = 0; $con = mysql_connect("localhost" , "root", ""); mysql_select_db("testdb", $con); $result = mysql_query("SELECT name, password FROM user WHERE name = '".$username."' AND password = '".$password."' ") or die("Error select statement"); $count = mysql_num_rows($result); if($count > 0) // always goes the to else, only works with >=0 but then the data is not found in the database, hence incorrect { $row = mysql_fetch_array($result); $_SESSION["username"] = $row["name"]; header("Location: StartPage.php"); } else { echo "Wrong login data, please try again"; } mysql_close($con); } ?>

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  • Returning Json object from controller action to jQuery

    - by PsychoCoder
    I'm attempting to get this working properly (2 days now). I'm working on a log in where I'm calling the controller action from jQuery, passing it a JSON object (utilizing json2.js) and returning a Json object from the controller. I'm able to call the action fine, but instead of being able to put the response where I want it it just opens a new window with this printed on the screen: {"Message":"Invalid username/password combination"} And the URL looks like http://localhost:13719/Account/LogOn so instead of calling the action and not reloading the page it's taking the user to the controller, which isn't good. So now for some code, first the controller code [HttpPost] public ActionResult LogOn(LogOnModel model, string returnUrl = "") { if (ModelState.IsValid) { var login = ObjectFactory.GetInstance<IRepository<PhotographerLogin>>(); var user = login.FindOne(x => x.Login == model.Username && x.Pwd == model.Password); if (user == null) return Json(new FailedLoginViewModel { Message = "Invalid username/password combination" }); else { if (!string.IsNullOrEmpty(returnUrl)) return Redirect(returnUrl); else return RedirectToAction("Index", "Home"); } } return RedirectToAction("Index", "Home"); } And the jQuery code $("#signin_submit").click(function () { var login = getLogin(); $.ajax({ type: "POST", url: "../Account/LogOn", data: JSON.stringify(login), dataType: 'json', contentType: 'application/json; charset=utf-8', error: function (xhr) { $("#message").text(xhr.statusText); }, success: function (result) { } }); }); function getLogin() { var un = $("#username").val(); var pwd = $("#password").val(); var rememberMe = $("#rememberme").val(); return (un == "") ? null : { Username: un, Password: pwd, RememberMe: rememberMe }; } In case you need to see the actual login form here that is as well <fieldset id="signin_menu"> <div> <span id="message"></span> </div> <% Html.EnableClientValidation(); %> <% using (Html.BeginForm("LogOn", "Account", FormMethod.Post, new { @id = "signin" })) {%> <% ViewContext.FormContext.ValidationSummaryId = "valLogOnContainer"; %> <%= Html.LabelFor(m => m.Username) %> <%= Html.TextBoxFor(m => m.Username, new { @class = "inputbox", @tabindex = "4", @id = "username" })%><%= Html.ValidationMessageFor(m => m.Username, "*")%> <p> <%= Html.LabelFor(m=>m.Password) %> <%= Html.PasswordFor(m => m.Password, new { @class = "inputbox", @tabindex = "5", @id = "password" })%><%= Html.ValidationMessageFor(m => m.Password, "*")%> </p> <p class="remember"> <input id="signin_submit" value="Sign in" tabindex="6" type="submit"/> <%= Html.CheckBoxFor(m => m.RememberMe, new { @class = "inputbox", @tabindex = "7", @id = "rememberme" })%> <%= Html.LabelFor(m => m.RememberMe) %> <p class="forgot"> <a href="#" id="forgot_password_link" title="Click here to reset your password.">Forgot your password?</a> </p> <p class="forgot-username"> <a href="#" id="forgot_username_link" title="Fogot your login name? We can help with that">Forgot your username?</a> </p> </p> <%= Html.ValidationSummaryJQuery("Please fix the following errors.", new Dictionary<string, object> { { "id", "valLogOnContainer" } })%> <% } %> </fieldset> The login form is loaded on the main page with <% Html.RenderPartial("LogonControl");%> Not sure if that has any bearing on this or not but thought I'd mention it. EDIT: The login form is loaded similar to the Twitter login, click a link and the form loads with the help of jQuery & CSS

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  • Solaris ldap Authentication

    - by Tman
    Iv been having a trouble trying to get my Solaris 10 server to authenticate against an eDir server.im managed to Set up my linux(RHeL,SLES) servers to authenticate against the ldap Server.which works fine. Here is my configuration Files. ldapclient list: NS_LDAP_FILE_VERSION= 2.0 NS_LDAP_BINDDN= cn=proxyuser,o=AEDev NS_LDAP_BINDPASSWD= {NS1}ecfa88f3a945c22222233 NS_LDAP_SERVERS= 192.168.0.19 NS_LDAP_SEARCH_BASEDN= ou=auth,o=AEDev NS_LDAP_AUTH= simple NS_LDAP_SEARCH_SCOPE= sub NS_LDAP_CACHETTL= 0 NS_LDAP_CREDENTIAL_LEVEL= anonymous NS_LDAP_SERVICE_SEARCH_DESC= group:ou=Groups,ou=auth,o=AEDev NS_LDAP_SERVICE_SEARCH_DESC= shadow:ou=users,ou=auth,o=AEDev?sub?objectClass=shadowAccount NS_LDAP_SERVICE_SEARCH_DESC= passwd:ou=auth,o=AEDev?sub?objectClass=posixAccount NS_LDAP_BIND_TIME= 10 NS_LDAP_SERVICE_AUTH_METHOD= pam_ldap:simple getent passwd works fine: root:x:0:0:Super-User:/:/sbin/sh daemon:x:1:1::/: bin:x:2:2::/usr/bin: sys:x:3:3::/: adm:x:4:4:Admin:/var/adm: lp:x:71:8:Line Printer Admin:/usr/spool/lp: uucp:x:5:5:uucp Admin:/usr/lib/uucp: nuucp:x:9:9:uucp Admin:/var/spool/uucppublic:/usr/lib/uucp/uucico smmsp:x:25:25:SendMail Message Submission Program:/: listen:x:37:4:Network Admin:/usr/net/nls: gdm:x:50:50:GDM Reserved UID:/: webservd:x:80:80:WebServer Reserved UID:/: postgres:x:90:90:PostgreSQL Reserved UID:/:/usr/bin/pfksh svctag:x:95:12:Service Tag UID:/: nobody:x:60001:60001:NFS Anonymous Access User:/: noaccess:x:60002:60002:No Access User:/: nobody4:x:65534:65534:SunOS 4.x NFS Anonymous Access User:/: tlla:x:2012:100::/home/tlla: test:x:2011:100::/home/test: thato:x:2010:100::/home/thato: pam.conf login auth sufficient pam_unix_auth.so.1 #server_policy login auth sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass login auth required pam_dial_auth.so.1 rlogin auth sufficient pam_rhosts_auth.so.1 rlogin auth requisite pam_authtok_get.so.1 rlogin auth required pam_dhkeys.so.1 rlogin auth required pam_unix_cred.so.1 rlogin auth sufficient pam_unix_auth.so.1 rlogin auth sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass rsh auth sufficient pam_rhosts_auth.so.1 rsh auth required pam_unix_cred.so.1 rsh auth sufficient pam_unix_auth.so.1 #server_policy rsh auth sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass other auth requisite pam_authtok_get.so.1 other auth required pam_dhkeys.so.1 other auth required pam_unix_cred.so.1 other auth sufficient pam_unix_auth.so.1 other auth sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass passwd auth required pam_passwd_auth.so.1 passwd auth sufficient pam_unix_auth.so.1 ssh account sufficient pam_unix.so.1 ssh account sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass other account requisite pam_roles.so.1 other account sufficient pam_unix_account.so.1 other account sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass other password required pam_dhkeys.so.1 other password requisite pam_authtok_get.so.1 other password requisite pam_authtok_check.so.1 other password required pam_authtok_store.so.1 other password sufficient pam_unix.so.1 other password sufficient /usr/lib/security/pam_ldap.so.1 try_first_pass Local Authentication Works But LDAP Authentication Doesn't Work.

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  • Bugzilla email issue

    - by xian
    My bugzilla system keep hit the following error: There was an error sending mail from '[email protected]' to '[email protected]':Can't send data I think that is some problem with my setting and configuration. First is the urlbase I have tried setting it to bugzilla.example.com, and http://127.0.0.1:81/, and http://10.0.0.236/ (My laptop IP address, I use this laptop to set up bugzilla) but the error still persists. Actually what should I put in the urlbase field? Parameter = Email Under mail_delivery_method, i choose SMTP. Under mailfrom, I put bugzilla-daemon. smtpserver, I tried leaving it blank, or setting it to 220.181.12.12 before, but could not solve my problem For my sql, the following is the data and command I used: C:\mysql\bin>mysql --user=root -p mysql Enter password: 1234 (When I install mysql into my laptop, it ask me to key an username and password, i have key in username as 'cvuser' and password as '1234', but here never ask me to key in any username) Welcome to the MySQL monitor. Commands end with ; or \g. Your MySQL connection id is 1 Server version: 5.5.15 MySQL Community Server (GPL) mysql> GRANT ALL PRIVILEGES ON bugs.* TO 'bugs'@'localhost' IDENTIFIED BY '123456'; Query OK, 0 rows affected (0.03 sec) In C:\Bugzilla\localconfig, I put the following info: # # How to access the SQL database: # $db_host = "localhost"; # where is the database? $db_port = 3306; # which port to use $db_name = "bugs"; # name of the MySQL database $db_user = "bugs"; # user to attach to the MySQL database # # Enter your database password here. It's normally advisable to specify # a password for your bugzilla database user. # If you use apostrophe (') or a backslash (\) in your password, you'll # need to escape it by preceding it with a \ character. (\') or (\\) # $db_pass = '123456'; Can someone tell me where my mistake is? I have googled for this issue for few days but still cannot find the solution.

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  • Can't login via ssh after upgrading to Ubuntu 12.10

    - by user42899
    I have an Ubuntu 12.04LTS instance on AWS EC2 and I upgraded it to 12.10 following the instructions at https://help.ubuntu.com/community/QuantalUpgrades. After upgrading I can no longer ssh into my VM. It isn't accepting my ssh key and my password is also rejected. The VM is running, reachable, and SSH is started. The problem seems to be about the authentication part. SSH has been the only way for me to access that VM. What are my options? ubuntu@alice:~$ ssh -v -i .ssh/sos.pem [email protected] OpenSSH_5.9p1 Debian-5ubuntu1, OpenSSL 1.0.1 14 Mar 2012 debug1: Reading configuration data /home/ubuntu/.ssh/config debug1: Reading configuration data /etc/ssh/ssh_config debug1: /etc/ssh/ssh_config line 19: Applying options for * debug1: Connecting to www.hostname.com [37.37.37.37] port 22. debug1: Connection established. debug1: identity file .ssh/sos.pem type -1 debug1: identity file .ssh/sos.pem-cert type -1 debug1: Remote protocol version 2.0, remote software version OpenSSH_5.9p1 Debian-5ubuntu1 debug1: match: OpenSSH_5.9p1 Debian-5ubuntu1 pat OpenSSH* debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_5.9p1 Debian-5ubuntu1 debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug1: kex: server->client aes128-ctr hmac-md5 none debug1: kex: client->server aes128-ctr hmac-md5 none debug1: sending SSH2_MSG_KEX_ECDH_INIT debug1: expecting SSH2_MSG_KEX_ECDH_REPLY debug1: Server host key: RSA 33:33:33:33:33:33:33:33:33:33:33:33:33:33 debug1: Host '[www.hostname.com]:22' is known and matches the RSA host key. debug1: Found key in /home/ubuntu/.ssh/known_hosts:12 debug1: ssh_rsa_verify: signature correct debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: SSH2_MSG_NEWKEYS received debug1: Roaming not allowed by server debug1: SSH2_MSG_SERVICE_REQUEST sent debug1: SSH2_MSG_SERVICE_ACCEPT received debug1: Authentications that can continue: publickey,password debug1: Next authentication method: publickey debug1: Trying private key: .ssh/sos.pem debug1: read PEM private key done: type RSA debug1: Authentications that can continue: publickey,password debug1: Next authentication method: password [email protected]'s password: debug1: Authentications that can continue: publickey,password Permission denied, please try again.

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  • Duplicity on a ReadyNAS

    - by Jason Swett
    Has anyone here run Duplicity on a ReadyNAS? I'm trying but here's what I get: duplicity full --encrypt-key="ABC123" /home/jason/ scp://[email protected]//gob Invalid SSH password Running 'sftp -oServerAliveInterval=15 -oServerAliveCountMax=2 [email protected]' failed (attempt #1) I've also found this post that says the "Invalid SSH password" message doesn't actually mean invalid SSH password. This would make sense because I'm not using an SSH password; I'm using a public key. I can ssh, ftp, sftp and rsync into my ReadyNAS just fine. (Actually, to be more accurate, I can get past authentication with ssh, ftp and sftp but I can't actually do anything past that. Regardless, that's enough to tell me that "Invalid SSH password" is bogus. Rsync works with no problems.) The post I found says the command will work as soon as the directory at the end of your scp command exists, but I don't know how to check for that. I know the share gob exists on my ReadyNAS and I know it's writable because I'm writing to it with rsync. Also, here is the verbose output: Using archive dir: /home/jason/.cache/duplicity/3bdd353b29468311ffa8485160da6873 Using backup name: 3bdd353b29468311ffa8485160da6873 Import of duplicity.backends.rsyncbackend Succeeded Import of duplicity.backends.sshbackend Succeeded Import of duplicity.backends.localbackend Succeeded Import of duplicity.backends.botobackend Succeeded Import of duplicity.backends.cloudfilesbackend Succeeded Import of duplicity.backends.giobackend Succeeded Import of duplicity.backends.hsibackend Succeeded Import of duplicity.backends.imapbackend Succeeded Import of duplicity.backends.ftpbackend Succeeded Import of duplicity.backends.webdavbackend Succeeded Import of duplicity.backends.tahoebackend Succeeded Main action: full ================================================================================ duplicity 0.6.10 (September 19, 2010) Args: /usr/bin/duplicity full --encrypt-key=ABC123 -v9 /home/jason/ scp://[email protected]//gob Linux gob 2.6.35-22-generic #33-Ubuntu SMP Sun Sep 19 20:34:50 UTC 2010 i686 /usr/bin/python 2.6.6 (r266:84292, Sep 15 2010, 15:52:39) [GCC 4.4.5] ================================================================================ Using temporary directory /tmp/duplicity-cridGi-tempdir Registering (mkstemp) temporary file /tmp/duplicity-cridGi-tempdir/mkstemp-ztuF5P-1 Temp has 86334349312 available, backup will use approx 34078720. Running 'sftp -oServerAliveInterval=15 -oServerAliveCountMax=2 [email protected]' (attempt #1) State = sftp, Before = '[email protected]'s' State = sftp, Before = '' Invalid SSH password Running 'sftp -oServerAliveInterval=15 -oServerAliveCountMax=2 [email protected]' failed (attempt #1) Any ideas as to what's going wrong?

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  • How to link to a subfolder of a share?

    - by Nicolas Raoul
    On my Windows XP server, a folder called Share2 is shared. It contains a subfolder called folder3. The guest account is protected by a password, which means network users have to type the guest password to access it. When a user types \\server\Share2 in his file explorer, he is prompted for a password. When a user types \\server\Share2\folder3 in his file explorer, an error appears. He is not even prompted for a password. This is problematic because I want to link to this particular folder. How can I link to folder3? Notes: - Both Desktop shortcuts and HTML links in IE7/8 give an error if I link to folder3, but work if I just link to Share2. - Using the file:// syntax instead of the \\ syntax leads to the same results. - Password setting per http://www.lancelhoff.com/how-to-password-protect-a-shared-folder - Not using "Simple File Sharing" - The error message is ???????????????????????? which means "could not find it. check the path and try again". No English Windows around to try, sorry! It is easy to reproduce the problem though, so can anyone post the English error message for the sake of searchability? Thanks!

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  • How do I remove encryption from a VMware Workstation 7 image?

    - by Chad
    I successfully encrypted a VM image and confirmed it still runs. I then closed the VM and reopened it and confirmed the encryption password was valid and worked. However, now I want to un-encrypt the VM. When I choose that option, it asks for "your password". I assume this means the password I created when I encrypted it. It doesn't work. I can still open the VM with the password and run it. But, it refuses to remove the encryption using that password. Am I missing something? Is there a password that I don't know about? Some details: I created this image (using standalone converter; physical machine source) I converted it to ACE Converted back to a normal VM (un-ACE'd it) Encrypted it Cannot remove the encryption but can open it and run it As you can see... I am exploring the VMware features. Thanks for any guidance you can give.

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  • Apache Server not working in MAMP

    - by jasonaburton
    Here's what I did before the problem started: I was creating a database for a site that I am working on in phpMyAdmin. I wrote some code to try to connect to the database I just created and I couldn't connect. I assumed it might be because I needed a password to connect to the database, so I created a password for it. Immediately after I created the password phpMyAdmin kicked me out saying: "#1045 - Access denied for user 'root'@'localhost' (using password: YES)" "phpMyAdmin tried to connect to the MySQL server, and the server rejected the connection. You should check the host, username and password in your configuration and make sure that they correspond to the information given by the administrator of the MySQL server." I found the php.ini file and searched for where I could change the password to match the one I just made, but couldn't find where I needed to change it. So I decided to scrap the database and uninstall MAMP from my computer and reinstall it hoping it would just reset all the defaults and I could go on my merry way. But now after reinstalling MAMP and trying to run the servers Apache won't start up and I have no idea why. One problem after another... Any advice or helpful ideas?

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  • How do I remove encryption from a VMware Workstation 7 image?

    - by Chad
    I successfully encrypted a VM image and confirmed it still runs. I then closed the VM and reopened it and confirmed the encryption password was valid and worked. However, now I want to un-encrypt the VM. When I choose that option, it asks for "your password". I assume this means the password I created when I encrypted it. It doesn't work. I can still open the VM with the password and run it. But, it refuses to remove the encryption using that password. Am I missing something? Is there a password that I don't know about? Some details: I created this image (using standalone converter; physical machine source) I converted it to ACE Converted back to a normal VM (un-ACE'd it) Encrypted it Cannot remove the encryption but can open it and run it As you can see... I am exploring the VMware features. Thanks for any guidance you can give.

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  • All application passwords lost on Windows 7

    - by Rynardt
    A couple of days ago I changed my Windows 7 login password. My laptop is on my company's domain, so password changes are done over the internal network. Since changing the password I noticed that all my saved Chrome passwords are missing. Also Skype, Windows Live, Internet Explorer and Outlook lost their saved passwords. I guess there could be more applications with lost passwords, but I have not opened them yet. This makes me think that most applications saves their passwords to a general password vault on the Windows system and this vault got somehow corrupted when I changed my domain login password for windows. Do anyone have any idea of how to fix this and prevent it from happening again? EDIT : More Info I do development work at the office, so most of the time I bypass the firewall and connect directly to the internet gateway. Now and then I would connect to the company wifi network to do printing and access files on a NAS. So by default my laptop does not connect to the wifi hotspot. On this occasion to update the password, I had to connect to the wifi. So referring to the comment by OmnipotentEntity below, could this have happened when the system rebooted without a connection to the network as the laptop does not auto connect to the wifi hotspot?

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