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• XML-RPC over SSL with Ruby: end of file reached (EOFError)

  - by Michael Conigliaro

  Hello, I have some very simple Ruby code that is attempting to do XML-RPC over SSL: require 'xmlrpc/client' require 'pp' server = XMLRPC::Client.new2("https://%s:%d/" % [ 'api.ultradns.net', 8755 ]) pp server.call2('UDNS_OpenConnection', 'sponsor', 'username', 'password') The problem is that it always results in the following EOFError exception: /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/1.8/net/protocol.rb:135:in `sysread': end of file reached (EOFError) So it appears that after doing the POST, I don't get anything back. Interestingly, this is the behavior I would expect if I tried to make an HTTP connection on the HTTPS port (or visa versa), and I actually do get the same exact exception if I change the protocol. Everything I've looked at indicates that using "https://" in the URL is enough to enable SSL, but I'm starting wonder if I've missed something. Note that Even though the credentials I'm using in the RPC are made up, I'm expecting to at least get back an XML error page (similar to if you access https://api.ultradns.net:8755/ with a web browser). I've tried running this code on OSX and Linux with the exact same result, so I have to conclude that I'm just doing something wrong here. Does anyone have any examples of doing XML-RPC over SSL with Ruby?

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• Deterministic and non uniform long string generation from seed

  - by Limonup

  I had this weird idea for an encryption that I wanted to try out, it may be bad, and it may have done before, but I'm just doing it for fun. The short version of the question is: Is it possible to generate a long, deterministic and non-uniformly distributed string/sequence of numbers from a small seed? Long(er) version: I was thinking to encrypt a text by changing encoding. The new encoding would be generated via Huffman algorithm. To work well, the Huffman algorithm would need a fairly long text with non uniform distribution. Then characters can have different bit-lengths which would be the primary strength of this encryption. The problem is that its impractical to enter in/remember a long text each time you want to decrypt the text. So I was wondering if it was possible to generate a text from password seed? It doesn't matter what the text is, as long as it has non uniform distribution of characters and that the exact same sequence can be recreated each time you give it the same seed. Preferably, are there any functions/extensions in Python that can do this? EDIT: To expand on the "strength" of varying bit length: if I have a string "test", ASCII values 116, 101, 115, 116, which gives bit values of 1110100 1100101 1110011 1110100 Then, say my Huffman algorithm generates encoding like t = 101 e = 1100111 s = 10001 The final string is 101 1100111 10001 101, if we encode this back to ASCII, we get 1011100 1111000 1101000, which is 3 entirely different characters. Obviously its impossible to perform any kind of frequency analysis or something like that on this.

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• Rails nested models and data separation by scope

  - by jobrahms

  I have Teacher, Student, and Parent models that all belong to User. This is so that a Teacher can create Students and Parents that can or cannot log into the app depending on the teacher's preference. Student and Parent both accept nested attributes for User so a Student and User object can be created in the same form. All four models also belong to Studio so I can do data separation by scope. The current studio is set in application_controller.rb by looking up the current subdomain. In my students controller (all of my controllers, actually) I'm using @studio.students.new instead of Student.new, etc, to scope the new student to the correct studio, and therefore the correct subdomain. However, the nested User does not pick up the studio from its parent - it gets set to nil. I was thinking that I could do something like params[:student][:user_attributes][:studio_id] = @student.studio.id in the controller, but that would require doing attr_accessible :studio_id in User, which would be bad. How can I make sure that the nested User picks up the same scope that the Student model gets when it's created? student.rb class Student < ActiveRecord::Base belongs_to :studio belongs_to :user, :dependent => :destroy attr_accessible :user_attributes accepts_nested_attributes_for :user, :reject_if => :all_blank end students_controller.rb def create @student = @studio.students.new @student.attributes = params[:student] if @student.save redirect_to @student, :notice => "Successfully created student." else render :action => 'new' end end user.rb class User < ActiveRecord::Base belongs_to :studio accepts_nested_attributes_for :studio attr_accessible :email, :password, :password_confirmation, :remember_me, :studio_attributes devise :invitable, :database_authenticatable, :recoverable, :rememberable, :trackable end

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• How to connect to Oracle using JRuby & JDBC

  - by Rob

  First approach: bare metal require 'java' require 'rubygems' require "c:/ruby/jruby-1.2.0/lib/ojdbc14.jar" # should be redundant, but tried it anyway odriver = Java::JavaClass.for_name("oracle.jdbc.driver.OracleDriver") puts odriver.java_class url = "jdbc:oracle:thin:@myhost:1521:mydb" puts "About to connect..." con = java.sql.DriverManager.getConnection(url, "myuser", "mypassword"); if con puts " connection good" else puts " connection failed" end The result of the above is: sqltest.rb:4: cannot load Java class oracle.jdbc.driver.OracleDriver (NameError) Second approach: Active Record require 'rubygems' gem 'ActiveRecord-JDBC' require 'jdbc_adapter' require 'active_record' require 'active_record/version' require "c:/ruby/jruby-1.2.0/lib/ojdbc14.jar" # should be redundant... ActiveRecord::Base.establish_connection( :adapter => 'jdbc', :driver => 'oracle.jdbc.driver.OracleDriver', :url => 'jdbc:oracle:thin:@myhost:1521:mydb', :username=>'myuser', :password=>'mypassword' ) ActiveRecord::Base.connection.execute("SELECT * FROM mytable") The result of this is: C:/ruby/jruby-1.2.0/lib/ruby/gems/1.8/gems/activerecord-jdbc-adapter-0.9.1/lib/active_recordconnection_adapters/jdbc_adapter.rb:330:in `initialize': The driver encountered an error: cannot load Java class oracle.jdbc.driver.OracleDriver (RuntimeError) Essentially the same error no matter how I go about it. I'm using JRuby 1.2.0 and I have ojdbc14.jar in my JRuby lib directory Gems: ActiveRecord-JDBC (0.5) activerecord-jdbc-adapter (0.9.1) activerecord (2.2.2) What am I missing? Thanks,

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• SQL to CodeIgniter Array Missing Data Issue

  - by SamD

  $query = $this->db->query("SELECT t1.numberofbets, t1.profit, t2.seven_profit, t3.28profit, user.user_id, username, password, email, balance, user.date_added, activation_code, activated FROM user LEFT JOIN (SELECT user_id, SUM(amount_won) AS profit, count(tip_id) AS numberofbets FROM tip GROUP BY user_id) as t1 ON user.user_id = t1.user_id LEFT JOIN (SELECT user_id, SUM(amount_won) AS seven_profit FROM tip WHERE date_settled > '$seven_daystime' GROUP BY user_id) as t2 ON user.user_id = t2.user_id LEFT JOIN (SELECT user_id, SUM(amount_won) AS 28profit FROM tip WHERE date_settled > '$twoeight_daystime' GROUP BY user_id) as t3 ON user.user_id = t3.user_id where activated = 1 GROUP BY user.user_id ORDER BY user.date_added DESC"); return $query->result_array(); The query works fine running it in phpMyAdmin and returns complete results (in image attached). However, printing the array in CodeIgniter, it has no value for one field ,seven_profit, where it is there in the SQL query ran in phpMyAdmin, just the discrepancy in this one field, from sql to php array... I just can’t see why, when printing the array, that one field, which should have value of 26, contains nothing? Any ideas? I changed the field name from starting with a number in attempt to fix it, but no difference. I know this is complex and looks horrible, any help or just people coming across something similar would be great to know about, thanks. Sam

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• MYSQL not running on Ubuntu OS - Error 2002.

  - by mgj

  Hi, I am a novice to mysql DB. I am trying to run the MYSQL Server on Ubuntu 10.04. Through Synaptic Package Manager I am have installed the mysql version: mysql-client-5.1 I wonder that how was the database password set for the mysql-client software that I installed through the above way.It would be nice if you could enlighten me on this. When I tried running this database, I encountered the error given below: mohnish@mohnish-laptop:/var/lib$ mysql ERROR 2002 (HY000): Can't connect to local MySQL server through socket '/var/run/mysqld/mysqld.sock' (2) mohnish@mohnish-laptop:/var/lib$ I referred to a similar question posted by another user. I didn't find a solution through the proposed answers. For instance when I tried the solutions posted for the similar question I got the following: mohnish@mohnish-laptop:/var/lib$ service start mysqld start: unrecognized service mohnish@mohnish-laptop:/var/lib$ ps -u mysql ERROR: User name does not exist. ********* simple selection ********* ********* selection by list ********* -A all processes -C by command name -N negate selection -G by real group ID (supports names) -a all w/ tty except session leaders -U by real user ID (supports names) -d all except session leaders -g by session OR by effective group name -e all processes -p by process ID T all processes on this terminal -s processes in the sessions given a all w/ tty, including other users -t by tty g OBSOLETE -- DO NOT USE -u by effective user ID (supports names) r only running processes U processes for specified users x processes w/o controlling ttys t by tty *********** output format ********** *********** long options *********** -o,o user-defined -f full --Group --User --pid --cols --ppid -j,j job control s signal --group --user --sid --rows --info -O,O preloaded -o v virtual memory --cumulative --format --deselect -l,l long u user-oriented --sort --tty --forest --version -F extra full X registers --heading --no-heading --context ********* misc options ********* -V,V show version L list format codes f ASCII art forest -m,m,-L,-T,H threads S children in sum -y change -l format -M,Z security data c true command name -c scheduling class -w,w wide output n numeric WCHAN,UID -H process hierarchy mohnish@mohnish-laptop:/var/lib$ which mysql /usr/bin/mysql mohnish@mohnish-laptop:/var/lib$ mysql ERROR 2002 (HY000): Can't connect to local MySQL server through socket '/var/run/mysqld/mysqld.sock' (2) I even tried referring to http://forums.mysql.com/read.php?11,27769,84713#msg-84713 but couldn't find anything useful. Please let me know how I could tackle this error. Thank you very much..

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• Powershell 2.0 Hang When Run From MsDeploy pre- post- ops using c/

  - by SonOfNun

  I am trying to invoke powershell during the preSync call in a MSDeploy command, but powershell does not exit the process after it has been called. The command (from command line): "tools/MSDeploy/msdeploy.exe" -verb:sync -preSync:runCommand="powershell.exe -NoLogo -NoProfile -NonInteractive -ExecutionPolicy Unrestricted -Command C:/MyInstallPath/deploy.ps1 Set-WebAppOffline Uninstall-Service ",waitInterval=60000 -usechecksum -source:dirPath="build/for-deployment" -dest:wmsvc=BLUEPRINT-X86,username=deployer,password=deployer,dirPath=C:/MyInstallPath I used a hack here (http://therightstuff.de/2010/02/06/How-We-Practice-Continuous-Integration-And-Deployment-With-MSDeploy.aspx) that gets the powershell process and kills it but that didn't work. I also tried taskkill and the sysinternals equivalent, but nothing will kill the process so that MSDeploy errors out. The command is executed, but then just sits there. Any ideas what might be causing powershell to hang like this? I have found a few other similar issues around the web but no answers. Environment is Win 2K3, using Powershell 2.0. UPDATE: Here is a .vbs script I use to invoke my powershell command now. Invoke using 'cscript.exe path/to/script.vbs': Option Explicit Dim oShell, appCmd,oShellExec Set oShell = CreateObject("WScript.Shell") appCmd = "powershell.exe -NoLogo -NoProfile -NonInteractive -ExecutionPolicy Unrestricted -Command ""&{ . c:/development/materialstesting/deploy/web/deploy.ps1; Set-WebAppOffline }"" " Set oShellExec = oShell.Exec(appCmd) oShellExec.StdIn.Close()

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• How to make a request from an android app that can enter a Spring Security secured webservice method

  - by johnrock

  I have a Spring Security (form based authentication) web app running CXF JAX-RS webservices and I am trying to connect to this webservice from an Android app that can be authenticated on a per user basis. Currently, when I add an @Secured annotation to my webservice method all requests to this method are denied. I have tried to pass in credentials of a valid user/password (that currently exists in the Spring Security based web app and can log in to the web app successfully) from the android call but the request still fails to enter this method when the @Secured annotation is present. The SecurityContext parameter returns null when calling getUserPrincipal(). How can I make a request from an android app that can enter a Spring Security secured webservice method? Here is the code I am working with at the moment: Android call: httpclient.getCredentialsProvider().setCredentials( //new AuthScope("192.168.1.101", 80), new AuthScope(null, -1), new UsernamePasswordCredentials("joeuser", "mypassword")); String userAgent = "Android/" + getVersion(); HttpGet httpget = new HttpGet(MY_URI); httpget.setHeader("User-Agent", userAgent); httpget.setHeader("Content-Type", "application/xml"); HttpResponse response; try { response = httpclient.execute(httpget); HttpEntity entity = response.getEntity(); ... parse xml Webservice Method: @GET @Path("/payload") @Produces("application/XML") @Secured({"ROLE_USER","ROLE_ADMIN","ROLE_GUEST"}) public Response makePayload(@Context Request request, @Context SecurityContext securityContext){ Payload payload = new Payload(); payload.setUsersOnline(new Long(200)); if (payload == null) { return Response.noContent().build(); } else{ return Response.ok().entity(payload).build(); } }

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• Cannot send email in ASP.NET through Godaddy servers.

  - by Jared

  I have an ASP.NET application hosted on Godaddy that I want to send email from. When it runs, I get: Mailbox name not allowed. The server response was: sorry, relaying denied from your location. The important parts of the code and Web.config are below: msg = new MailMessage("[email protected]", email); msg.Subject = "GreekTools Registration"; msg.Body = "You have been invited by your organization to register for the GreekTools recruitment application.<br/><br/>" + url + "<br/><br/>" + "Sincerely,<br/>" + "The GreekTools Team"; msg.IsBodyHtml = true; client = new SmtpClient(); client.Host = "relay-hosting.secureserver.net"; client.Send(msg); <system.net> <mailSettings> <smtp from="[email protected]"> <network host="relay-hosting.secureserver.net" port="25" userName="********" password="*********" /> </smtp> </mailSettings>

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• Simple continuously running XMPP client in python

  - by tom

  I'm using python-xmpp to send jabber messages. Everything works fine except that every time I want to send messages (every 15 minutes) I need to reconnect to the jabber server, and in the meantime the sending client is offline and cannot receive messages. So I want to write a really simple, indefinitely running xmpp client, that is online the whole time and can send (and receive) messages when required. My trivial (non-working) approach: import time import xmpp class Jabber(object): def __init__(self): server = 'example.com' username = 'bot' passwd = 'password' self.client = xmpp.Client(server) self.client.connect(server=(server, 5222)) self.client.auth(username, passwd, 'bot') self.client.sendInitPresence() self.sleep() def sleep(self): self.awake = False delay = 1 while not self.awake: time.sleep(delay) def wake(self): self.awake = True def auth(self, jid): self.client.getRoster().Authorize(jid) self.sleep() def send(self, jid, msg): message = xmpp.Message(jid, msg) message.setAttr('type', 'chat') self.client.send(message) self.sleep() if __name__ == '__main__': j = Jabber() time.sleep(3) j.wake() j.send('[email protected]', 'hello world') time.sleep(30) The problem here seems to be that I cannot wake it up. My best guess is that I need some kind of concurrency. Is that true, and if so how would I best go about that? EDIT: After looking into all the options concerning concurrency, I decided to go with twisted and wokkel. If I could, I would delete this post.

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• svnsync looses revision properties although hook installed

  - by roesslerj

  Hello all! I have a pretty weird problem. We have setup an SVN-Mirror via cronjob (because it needs to go from inside to outside of a firewall, so no post-commit-hook possible) and svnsync. We installed a pre-revprop-hook just as told. Everything seems to work fine, except that it doesn't. E.g. when manually executing the script. # svnsync --non-interactive sync file://<path-to-mirror> --source-username <usr> --source-password <pwd> Committed revision 19817. Copied properties for revision 19817. No error, no complaints. But if checking for the revision properties it says: # svnlook info <path-to-mirror> 0 # svn info -r HEAD file://<path-to-mirror> 2>&1 Path: <root-of-mirror> URL: file://<path-to-mirror> Repository Root: file://<path-to-mirror> Repository UUID: <uid> Revision: 19817 Node Kind: directory Last Changed Rev: 19817 So somehow the author and timestamp information gets lost. But we need that information for our internal processes. Since no error or warning is produced I have absolutely no idea even where to start to look. Everything is local (except for the remote master), so there are no server-logs to look at. Any ideas how I could approach that problem, or even better -- how to solve it? Any ideas appreciated.

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• cakephp read and display from more than one related table

  - by z22

  I have 2 tables Users and Artists. User: id name password city_id state_id country_id Artist: id user_id description I wish to create add and edit form(single form acts as add and edit). But on click of edit link, I wish to fill in the form fields from user as well as artist table. How do i do that? Below is my controller code: public function artist_register() { $this->country_list(); if ($this->request->is('post')) { $this->request->data['User']['group_id'] = 2; if ($this->{$this->modelClass}->saveAll($this->request->data)) { $this->redirect(array('controller' => 'Users', 'action' => 'login')); } } $this->render('artist_update'); } public function artist_update($id) { $artist = $this->{$this->modelClass}->findByuser_id($id); $artist_id = $artist[$this->modelClass]['id']; if($this->request->is('get')) { $this->request->data = $this->{$this->modelClass}->read(null, $artist_id); $this->request->data = $this->{$this->modelClass}->User->read(null, $id); } else { if($this->{$this->modelClass}->save($this->request->data)) { $this->Session->setFlash('Your Profile has been updated'); $this->redirect(array('action' => 'index')); } } } The code doesnt work. how do i solve it?

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• How to exit current activity to homescreen (without using "Home" button)?

  - by steff

  Hi everyone, I am sure this will have been answered but I proved unable to find it. So please excuse my redundancy. What I am trying to do is emulating the "Home" button which takes one back to Android's homescreen. So here is what causes me problems: I have 3 launcher activities. The first one (which is connected to the homescreen icon) is just a (password protected) configuration activity. It will not be used by the user (just admin) One of the other 2 (both accessed via an app widget) is a questionnaire app. I'm allowing to jump back between questions via the Back button or a GUI back button as well. When the questionnaire is finished I sum up the answers given and provide a "Finish" button which should take the user back to the home screen. For the questionnaire app I use a single activity (called ItemActivity) which calls itself (is that recursion as well when using intents?) to jump from one question to another: Questionnaire.serializeToXML(); Intent i = new Intent().setClass(c, ItemActivity.class); if(Questionnaire.instance.getCurrentItemNo() == Questionnaire.instance.getAmountOfItems()) { Questionnaire.instance.setCompleted(true); } else Questionnaire.instance.nextItem(); startActivity(i); The final screen shows something like "Thank you for participating" as well as the formerly described button which should take one back to the homescreen. But I don't really get how to exit the Activity properly. I've e.g. used this.finish(); but this strangely brings up the "Thank you" screen again. So how can I just exit by jumping back to the homescreen?? Sorry for the inconvinience. Regards, Steff

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• General JDBC Setup

  - by AeroDroid

  So I have a MySQL database set up on a Debian server and it works fine from a phpMyAdmin client. I'm currently working on a project to write a Java server that would be able to use the MySQL database that is already on this server through a JDBC connection. I've looked at many tutorials and documentations but all of them seem to just explain how to do client-side code, but I have yet to figure out how to even successfully open a JDBC connection to the server. As far as I am concerned, I believe that program has the drivers properly set up because it's not crashing anymore (I simply direct the Java Build Path of my program to the Connector/J provided by MySQL). As far as my program goes, this is what it looks like... import java.sql.*; public class JDBCTest { public static void main(String[] args) { System.out.println("Started!"); try { DriverManager.registerDriver(new com.mysql.jdbc.Driver()); System.out.println("Driver registered. Connecting..."); Connection conn = DriverManager.getConnection("jdbc:mysql://localhost/", "root", "password"); System.out.println("Connected!"); conn.close(); } catch (SQLException e) { System.out.println("Error!"); e.printStackTrace(); } } } This is what's printed... Started! Driver registered. Connecting... It's as if the DriverManager.getConnection(String) just freezes there. I'm sure this is a problem with the server because when I intentionally misspell localhost, or an IP address, the program crashes within 20 seconds. This just hangs there forever. Sorry about this wall of text, but my final question is if anyone has any information what I should do or install on the server to get this to work? Thank you so much!

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• Attachment empty/blank when sending mail using phpmailer

  - by chupinette

  Hello! I am using phpmailer class to send email and i am attaching a file to the mail. The email is sent successfully but the attachment which is an sql file is empty which should not be the case. I have tried with an image file, but it seems that every file that i attach is empty. Can anyone help me solve this problem please? $mail = new PHPMailer(); $body = "Reminder"; $mail->IsSMTP(); $mail->Host = "mail.yourdomain.com"; $mail->SMTPDebug = 1; $mail->SMTPAuth = true; $mail->SMTPSecure = "ssl"; $mail->Host = "smtp.gmail.com"; $mail->Port = 465; $mail->Username = "[email protected]"; $mail->Password = "abc"; $mail->SetFrom('[email protected]', 'blbla'); $mail->AddReplyTo("[email protected]","First Last"); $mail->Subject = "Your order has been successfully placed" $mail->MsgHTML($body); $mail->AddAddress("[email protected]","xyz"); $mail->AddAttachment("D:\b2\shop3.sql","shop3.sql"); I have tried to display the size of the file before attaching it, and it actually displays the file size. Can anyone help me please?

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• import the data in xls file and open them without Microsoft Excel

  - by user3669577

  I need to perform an application that cath values from SQL database after the esecution of a query. I must import the data in xls file and open them without Microsoft Excel. I'm a beginner and have too many problem. Can anyone help me. This is my code, at the moment: Option Infer On Imports System.Linq Imports System.Data.SqlClient Imports System Imports System.IO Imports System.Drawing Imports System.Drawing.Printing Imports System.Windows.Forms Imports ExcelLibrary.SpreadSheet Public Class frmLottiCaricati Dim CnSql As SqlConnection Private Sub frmLottiCaricati_Load(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Load Me.MdiParent = Inizio 'TB_MinusValenza.Text = VariazionePrezzi.MinusValenza 'TB_Periodo.Text = VariazionePrezzi.Periodo 'DG_Prodotti.AutoGenerateColumns = False Try Dim StringaSql = "Integrated Security=SSPI;Persist Security Info=False;Initial Catalog=" + Inizio.DatabaseSql + ";Data Source=" + Inizio.ServerSql + ";User ID=" + Inizio.UtenteSql + ";Password=" + Inizio.PwdSql CnSql = New SqlConnection(StringaSql) CnSql.Open() Dim command As SqlCommand Dim dadapter As New SqlDataAdapter Dim DS_Prodotti As New Data.DataSet Dim qry_Prodotti = "SELECT sistemaf.prodscadenze.Ministeriale, sistemaf.prodscadenze.Lotto, sistemaf.prodscadenze.Scadenza " & _ "FROM sistemaf.Prodscadenze " 'INNER JOIN sistemaf.Prodscadenze ON sistemaf.prodbase.Cod39 = sistemaf.prodscadenze.Ministeriale ;" command = New SqlCommand(qry_Prodotti, CnSql) dadapter.SelectCommand = command dadapter.Fill(DS_Prodotti) DG_Prodotti.DataSource = DS_Prodotti.Tables(0) 'DG_Prodotti.Columns("Descrizione").Width = 220 'DG_Prodotti.Columns("Ministeriale").Width = 60 DG_Prodotti.Columns("Lotto").Width = 60 'DG_Prodotti.Columns("Descrizione").AutoSizeMode = DataGridViewAutoSizeColumnMode.AllCells 'DG_Prodotti.Columns("Totale").DefaultCellStyle.Alignment = DataGridViewContentAlignment.MiddleRight Catch ex As Exception MessageBox.Show(ex.Message) End Try End Sub End Class I can open the data only with Microsoft Excel now. Have any suggestions?

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• authentication question (security code generation logic)

  - by Stick it to THE MAN

  I have a security number generator device, small enough to go on a key-ring, which has a six digit LCD display and a button. After I have entered my account name and password on an online form, I press the button on the security device and enter the security code number which is displayed. I get a different number every time I press the button and the number generator has a serial number on the back which I had to input during the account set-up procedure. I would like to incorporate similar functionality in my website. As far as I understand, these are the main components: Generate a unique N digit aplha-numeric sequence during registration and assign to user (permanently) Allow user to generate an N (or M?) digit aplha-numeric sequence remotely For now, I dont care about the hardware side, I am only interested in knowing how I may choose a suitable algorithm that will allow the user to generate an N (or M?) long aplha-numeric sequence - presumably, using his unique ID as a seed Identify the user from the number generated in step 2 (which decryption method is the most robust to do this?) I have the following questions: Have I identified all the steps required in such an authentication system?, if not please point out what I have missed and why it is important What are the most robust encryption/decryption algorithms I can use for steps 1 through 3 (preferably using 64bits)?

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• UITableView only pushes to one controller

  - by Mitochondria

  So I have my UITableView, with 2 sections and 1 cell in each, and if I click the first one, it works, then the second one, it goes to the first controller. RootViewController is a navigationController, trying to push to ViewControllers. Here's the code for the tableView: // Customize the number of sections in the table view. - (NSInteger)numberOfSectionsInTableView:(UITableView *)tableView { return 2; } // Customize the number of rows in the table view. - (NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section { if(section == 0) return 1; else return 1; } - (NSString *)tableView:(UITableView *)tableView titleForHeaderInSection:(NSInteger)section{ if(section == 0){ return @"Terminal/SSH Guides"; }else{ return @"Cydia Tutorials"; } } // Customize the appearance of table view cells. - (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath { static NSString *CellIdentifier = @"Cell"; UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:CellIdentifier]; if (cell == nil) { cell = [[[UITableViewCell alloc] initWithFrame:CGRectZero reuseIdentifier:CellIdentifier] autorelease]; } // Set up the cell... if(indexPath.section == 0){ cell.text = @"Changing Password for root"; } else { cell.text = @"Hiding Sections"; } return cell; } - (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath { id newController; switch (indexPath.row) { case 0: newController = [[rootpassword alloc] initWithNibName:@"rootpassword" bundle:nil]; break; case 1: newController = [[hidingsections alloc] initWithNibName:@"hidingsections" bundle:nil]; break; default: break; } [self.navigationController pushViewController:newController animated:TRUE]; [tableView deselectRowAtIndexPath:indexPath animated:TRUE]; } I'm also having trouble adding more sections and rows/cells to sections. Thanks.

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• How to handle dynamic role or username changes in JSF?

  - by roadrunner

  I have a JSF application running on glassfish 2.1 with a EJB 3 backend. For authentication I use a custom realm. The user authenticates using the e-mail-address and password he specified on registration. Everything is working quite well. Now I have two related problems: 1) The user can edit his profile and -- naturally -- he can also change his e-mail-address. Unfortunately when I perform operations based on the current user's identity using ExternalContext.getUserPrincipal().getName(), I will receive the previous e-mail-address the user used on login. At the moment I handle this by forcing the user to reauthenticate after he changed his e-mail-address, but is there another more graceful possibility? 2) Same for user roles. E.g. I have the user roles MEMBER and PREMIUM_MEMBER. A MEMBER may become a PREMIUM_MEMBER during his current session. Unfortunately the role seems to be only determined at login. Is there any possibility, that JSF and EJB recognize the new user role without the need for the user to re-authenticated?

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• Doctrine/symfony: getSqlQuery() output in phpMyAdmin/SQL tab

  - by user248959

  Hi, i have created this query that works OK: $q1 = Doctrine_Query::create() ->from('Usuario u') ->leftJoin('u.AmigoUsuario a ON u.id = a.user2_id OR u.id = a.user1_id') ->where("a.user2_id = ? OR a.user1_id = ?", array($id,$id)) ->andWhere("u.id <> ?", $id) ->andWhere("a.estado LIKE ?", 1); echo $q1->getSqlQuery(); The calling to getSqlQuery outputs this clause: SELECT s.id AS s_id, s.username AS s_username, s.algorithm AS s_algorithm, s.salt AS s_salt, s.password AS s__password, s.is_active AS s__is_active, s.is_super_admin AS s__is_super_admin, s.last_login AS s__last_login, s.email_address AS s__email_address, s.nombre_apellidos AS s__nombre_apellidos, s.sexo AS s__sexo, s.fecha_nac AS s__fecha_nac, s.provincia AS s_provincia, s.localidad AS s_localidad, s.fotografia AS s_fotografia, s.avatar AS s_avatar, s.avatar_mensajes AS s__avatar_mensajes, s.created_at AS s__created_at, s.updated_at AS s__updated_at, a.id AS a__id, a.user1_id AS a__user1_id, a.user2_id AS a__user2_id, a.estado AS a__estado FROM sf_guard_user s LEFT JOIN amigo_usuario a ON ((s.id = a.user2_id OR s.id = a.user1_id)) WHERE ((a.user2_id = ? OR a.user1_id = ?) AND s.id < ? AND a.estado LIKE ?) If i take that clause to phpmyadmin SQL tab i get this error 1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '? OR a.user1_id = ?) AND s.id < ? AND a.estado LIKE ?) LIMIT 0, 30' at line 1 Why i'm getting this error? Regards Javi

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• Lazarus Pascal - DB Connection - clarification

  - by itsols

  The following code is from the docs here: Program ConnectDB var AConnection : TSQLConnection; Procedure CreateConnection; begin AConnection := TIBConnection.Create(nil); AConnection.Hostname := 'localhost'; AConnection.DatabaseName := '/opt/firebird/examples/employee.fdb'; AConnection.UserName := 'sysdba'; AConnection.Password := 'masterkey'; end; begin CreateConnection; AConnection.Open; if Aconnection.Connected then writeln('Succesful connect!') else writeln('This is not possible, because if the connection failed, ' + 'an exception should be raised, so this code would not ' + 'be executed'); AConnection.Close; AConnection.Free; end. The main body of the code makes sense to me BUT I don't get where TSQLConnection came from. I cannot use CTRL + Space to autocomplete it either, which means my program has no reference to it. I'm trying to connect to Postgres by the way. Can someone please state what TSQLConnection is? Thanks!

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• Yii user rights extension

  - by jailed abroad

  i have installed the yii rights extension and here is my code after installation, database tables are created after installation. 'modules'=>array( // uncomment the following to enable the Gii tool 'rights'=>array( 'superuserName'=>'Admin', // Name of the role with super user privileges. 'authenticatedName'=>'Authenticated', //// Name of the authenticated user role. 'userIdColumn'=>'id',// Name of the user id column in the database. 'userNameColumn'=>'username', // Name of the user name column in the database. 'enableBizRule'=>true, // Whether to enable authorization item business rules. 'enableBizRuleData'=>false, //Whether to enable data for business rules. 'displayDescription'=>true, // Whether to use item description instead of name. ' // Key to use for setting success flash messages. 'flashErrorKey'=>'RightsError', / Key to use for setting error flash messages. // 'install'=>true, // Whether to install rights. 'baseUrl'=>'/rights', // Base URL for Rights. Change if module is nested. 'layout'=>'rights.views.layouts.main', // Layout to use for displaying Rights. 'appLayout'=>'application.views.layouts.main', //Application layout. 'cssFile'=>'rights.css', // Style sheet file to use for Rights. ' 'install'=>false, // Whether to enable installer. 'debug'=>false, ), 'gii'=>array( 'class'=>'system.gii.GiiModule', 'password'=>'1234', // If removed, Gii defaults to localhost only. Edit carefully to taste. 'ipFilters'=>array('127.0.0.1','::1'), ), ), But when i type url http://localhost/rightsTest/index.php/rights then it says There must be at least one superuser! I have tried many things but unable to find answer. Thanks for your help.

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• Java MySQL Query Problem MySQLSyntaxErrorException When Creating a Table

  - by Aqib Mushtaq

  I fairly new to MySQL with Java, but I have executed a few successful INSERT queries however cannot seem to get the CREATE TABLE query to execute without getting the 'MySQLSyntaxErrorException' exception. My code is as follows: import java.sql.*; Statement stmt; Class.forName("com.mysql.jdbc.Driver"); String url = "jdbc:mysql://localhost:3306/mysql"; Connection con = DriverManager.getConnection(url, "root", "password"); stmt = con.createStatement(); String tblSQL = "CREATE TABLE IF NOT EXISTS \'dev\'.\'testTable\' (\n" + " \'id\' int(11) NOT NULL AUTO_INCREMENT,\n" + " \'date\' smallint(6) NOT NULL\n" + ") ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;"; stmt.executeUpdate(tblSQL); stmt.close(); con.close(); And the error is as follows: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''dev'.'testTable' ( 'id' int(11) NOT NULL AUTO_INCREMENT, 'date' smallint(6) N' at line 1 I would appreciate it if anyone could spot the mistake in this query, as I've tried executing this within phpMyAdmin and it works as it should. Thanks in advance.

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• Fast, easy, and secure method to perform DB actions with GET

  - by rob - not a robber

  Hey All, Sort of a methods/best practices question here that I am sure has been addressed, yet I can't find a solution based on the vague search terms I enter. I know starting off the question with "Fast and easy" will probably draw out a few sighs, so my apologies. Here is the deal. I have a logged in area where an ADMIN can do a whole host of POST operations to input data relating to their profile. The way I have data structured is pretty distinct and well segmented in most tables as it relates to the ID of the admin. Now, I have a table where I dump one type of data into and differentiate this data by assigning the ADMIN's unique ID to each record. In other words, all ADMINs have this one type of data writing to this table. I just differentiate by the ADMIN ID with each record. I was planning on letting the ADMIN remove these records by clicking on a link with a query string - obviously using GET. Obviously, the query structure is in the link so any logged in admin could then exploit the URL and delete a competitor's records. Is the only way to safely do this through POST or should I pass through the session info that includes password and validate it against the ADMIN ID that is requesting the delete? This is obviously much more work for me. As they said in the auto repair biz I used to work in... there are 3 ways to do a job: Fast, Good, and Cheap. You can only have two at a time. Fast and cheap will not be good. Good and cheap will not have fast turnaround. Fast and good will NOT be cheap. haha I guess that applies here... can never have Fast, Easy and Secure all at once ;) Thanks in advance...

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• Spring 2.0.0/2.0.6 to 3.0.5 migration stories

  - by Pangea

  We are in the process of migrating to 3.0.5 of spring from 2.0.x. We mainly use spring in below scenarios custom scope: thread local scope persistence: jdbc+hibernate 3.6 (but moving to mix of ejb 3.0+jpa 2.0+hibernate, not sure if all 3 can co-exist in 1 app) transactions: local (but planning to use jta due to the necessity of using multiple persistence inits, and has to use ejb+jpa+hibernate in 1 single trans), declarative trans mgmt parent-child contexts cxf annotations+xml OracleLobHandler Resource/ResourceBundleMessageResource JSF/Facelets with FacesSpringVariableResolver ActiveMQ integration Quartz integration TaskExecutor JMX exporter HttpExporter/Invoker Appreciate if someone can share their experiences like what to watch out for head aches/pain points which ones to drop for better alternate choices in new 3.0.5 release Is it better to switch from commons/iscreen validator to Hibernate Validator (Spec impl) or Spring Validator Is there a bean mapping framework in spring that i can use instead of Dozer XSLT transformation helper: currently we have small homegrown framework to cache xslts during load. if spring can do that for me then I would like to drop this Encryption/Decryption support. Password generation support. Authentication with SALT any SAML (or claims based secur New ideas Suggestions Switch to latest version of aspectj Upgrade guide from 2.5 to 3.0.5

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