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  • HttpWebRequest possibly slowing website

    - by Steven Smith
    Using Visual studio 2012, C#.net 4.5 , SQL Server 2008, Feefo, Nopcommerce Hey guys I have Recently implemented a new review service into a current site we have. When the change went live the first day all worked fine. Since then though the sending of sales to Feefo hasnt been working, There are no logs either of anything going wrong. In the OrderProcessingService.cs in Nop Commerce's Service, i call a HttpWebrequest when an order has been confirmed as completed. Here is the code. var email = HttpUtility.UrlEncode(order.Customer.Email.ToString()); var name = HttpUtility.UrlEncode(order.Customer.GetFullName().ToString()); var description = HttpUtility.UrlEncode(productVariant.ProductVariant.Product.MetaDescription != null ? productVariant.ProductVariant.Product.MetaDescription.ToString() : "product"); var orderRef = HttpUtility.UrlEncode(order.Id.ToString()); var productLink = HttpUtility.UrlEncode(string.Format("myurl/p/{0}/{1}", productVariant.ProductVariant.ProductId, productVariant.ProductVariant.Name.Replace(" ", "-"))); string itemRef = ""; try { itemRef = HttpUtility.UrlEncode(productVariant.ProductVariant.ProductId.ToString()); } catch { itemRef = "0"; } var url = string.Format("feefo Url", login, password,email,name,description,orderRef,productLink,itemRef); var request = (HttpWebRequest)WebRequest.Create(url); request.KeepAlive = false; request.Timeout = 5000; request.Proxy = null; using (var response = (HttpWebResponse)request.GetResponse()) { if (response.StatusDescription == "OK") { var stream = response.GetResponseStream(); if(stream != null) { using (var reader = new StreamReader(stream)) { var content = reader.ReadToEnd(); } } } } So as you can see its a simple webrequest that is processed on an order, and all product variants are sent to feefo. Now: this hasnt been happening all week since the 15th (day of the implementation) the site has been grinding to a halt recently. The stream and reader in the the var content is there for debugging. Im wondering does the code redflag anything to you that could relate to the process of website? Also note i have run some SQL statements to see if there is any deadlocks or large escalations, so far seems fine, Logs have also been fine just the usual logging of Bots. Any help would be much appreciated! EDIT: also note that this code is in a method that is called and wrapped in A try catch UPDATE: well forget about the "not sending", thats because i was just told my code was rolled back last week

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  • First record does not show in pagination script

    - by whitstone86
    This is my pagination script which extracts info for my TV guide project that I am working on. Currently I've been experimenting with different PHP/MySQL before it becomes a production site. This is my current script: <?php /*********************************** * PhpMyCoder Paginator * * Created By PhpMyCoder * * 2010 PhpMyCoder * * ------------------------------- * * You may use this code as long * * as this notice stays intact and * * the proper credit is given to * * the author. * ***********************************/ // set the default timezone to use. Available since PHP 5.1 putenv("TZ=US/Eastern") ?> <head> <title> Pagination Test - Created By PhpMyCoder</title> <style type="text/css"> #nav { font: normal 13px/14px Arial, Helvetica, sans-serif; margin: 2px 0; } #nav a { background: #EEE; border: 1px solid #DDD; color: #000080; padding: 1px 7px; text-decoration: none; } #nav strong { background: #000080; border: 1px solid #DDD; color: #FFF; font-weight: normal; padding: 1px 7px; } #nav span { background: #FFF; border: 1px solid #DDD; color: #999; padding: 1px 7px; } </style> </head> <?php //Require the file that contains the required classes include("pmcPagination.php"); //PhpMyCoder Paginator $paginator = new pmcPagination(20, "page"); //Connect to the database mysql_connect("localhost","root","PASSWORD"); //Select DB mysql_select_db("mytvguide"); //Select only results for today and future $result = mysql_query("SELECT programme, channel, airdate, expiration, episode, setreminder FROM lagunabeach where expiration >= now() order by airdate, 3 ASC LIMIT 0, 100;"); //You can also add reuslts to paginate here mysql_data_seek($queryresult,0) ; while($row = mysql_fetch_array($result)) { $paginator->add(new paginationData($row['programme'], $row['channel'], $row['airdate'], $row['expiration'], $row['episode'], $row['setreminder'])); } ?> <?php //Show the paginated results $paginator->paginate (); ?><? include("pca-footer1.php"); ?> <?php //Show the navigation $paginator->navigation(); ?> Despite me having two records for the programmes airing today, it only shows records from the second one onwards - the programme that airs at 8:35pm UK time GMT does not show, but the later 11:25pm UK time GMT one does show. How should I fix this? Above is my code if that is of any use! Thanks

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  • assign a model's attribute through association

    - by justcode
    I'm new to rails and working on a rails app and I'm stuck pondering this issue. I have three models class product < ActiveRecord::Base attr_accessible :name, :issn, :category validates_presence_of :name, :issn, :category validates_numericality_of :issn, :message => "has to be a number" has_many :user_products has_many :users, :through => :user_products end class UserProduct < ActiveRecord::Base attr_accessible :price, :category validates_presence_of :price, :category validates_numericality_of :price, :message = "has to be a number" belongs_to :user belongs_to :product end class user < ActiveRecord::Base # devise authentication here # Setup accessible (or protected) attributes for your model attr_accessible :email, :password, :password_confirmation, :remember_me has_many :user_products has_many :products, :through = :user_products end here is my new.html.erb <div class="MainBodyWrapper"> <div class="span8"> <div id="listBoxWrapper"> <fieldset> <%= form_for(@product, :html => { :class => "form-inline" }, :style => "margin-bottom: 60px" ) do |f| %> <div class="control-group"> <label class="control-label" for="name">name</label> <div class="controls"> <%= f.text_field :price, :class => 'input-xlarge input-name', :id => "name" %> </div> </div> <div class="listingButtons"> <button class="btn btn-info"></i>Add</button> <a class="btn">Upload Pictures (anytime)</a> </div> </fieldset> </div> </div> There are reasons why I want to setup the models this way. So the question is this: I want the user to enter the info for the product in the form but it also involves putting in the price of the product which exists in a different model/table (user_product) that is associated with product. How can I do this? You can see that my form_for uses @product. Any help will be appreciated.

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  • Cannot extend a class located in another file, PHP

    - by NightMICU
    I am trying to set up a class with commonly used tasks, such as preparing strings for input into a database and creating a PDO object. I would like to include this file in other class files and extend those classes to use the common class' code. However, when I place the common class in its own file and include it in the class it will be used in, I receive an error that states the second class cannot be found. For example, if the class name is foo and it is extending bar (the common class, located elsewhere), the error says that foo cannot be found. But if I place the code for class bar in the same file as foo, it works. Here are the classes in question - Common Class abstract class coreFunctions { protected $contentDB; public function __construct() { $this->contentDB = new PDO('mysql:host=localhost;dbname=db', 'username', 'password'); } public function cleanStr($string) { $cleansed = trim($string); $cleansed = stripslashes($cleansed); $cleansed = strip_tags($cleansed); return $cleansed; } } Code from individual class include $_SERVER['DOCUMENT_ROOT'] . '/includes/class.core-functions.php'; $mode = $_POST['mode']; if (isset($mode)) { $gallery = new gallery; switch ($mode) { case 'addAlbum': $gallery->addAlbum($_POST['hash'], $_POST['title'], $_POST['description']); } } class gallery extends coreFunctions { private function directoryPath($string) { $path = trim($string); $path = strtolower($path); $path = preg_replace('/[^ \pL \pN]/', '', $path); $path = preg_replace('[\s+]', '', $path); $path = substr($path, 0, 18); return $path; } public function addAlbum($hash, $title, $description) { $title = $this->cleanStr($title); $description = $this->cleanStr($description); $path = $this->directoryPath($title); if ($title && $description && $hash) { $addAlbum = $this->contentDB->prepare("INSERT INTO gallery_albums (albumHash, albumTitle, albumDescription, albumPath) VALUES (:hash, :title, :description, :path)"); $addAlbum->execute(array('hash' => $hash, 'title' => $title, 'description' => $description, 'path' => $path)); } } } The error when I try it this way is Fatal error: Class 'gallery' not found in /home/opheliad/public_html/admin/photo-gallery/includes/class.admin_photo-gallery.php on line 10

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  • join 03 table in the database codeIgniter

    - by python
    with my table. person_id serial NOT NULL, firstname character varying(30) NOT NULL, lastname character varying(30), email character varying(50), username character varying(20) NOT NULL, "password" character varying(100) NOT NULL, gender character varying(10), dob date, accesslevel smallint NOT NULL, company_id integer NOT NULL,//Reference to table company position_id integer NOT NULL,//Reference to table position company_id serial NOT NULL, company_name character varying(80) NOT NULL, description character varying(255), address character varying(100) NOT NULL, In my controller ........................ // load data $persons = $this->person_model->get_paged_list(10,0); // generate table data $this->load->library('table'); $this->table->set_empty("&nbsp;"); $this->table->set_heading('No', 'FirstName', 'LastName','E-mail','Company''Gender', 'Date of Birth', 'Actions'); foreach ($persons as $person){ $this->table->add_row(++$i, $person->firstname, $person->lastname, $person->email, $person->company_name, //HOW CAN I GOT THE POSITION TITLE ?, strtoupper($person->gender)=='M'? 'Male':'Female', date('d-m-Y',strtotime($person->dob)), } My model <?php class Person_Model extends Model { private $person= 'person'; function Person(){ parent::Model(); } function list_all(){ $this->db->order_by('person_id','asc'); return $this->db->get($person); } function count_all(){ return $this->db->count_all($this->person); } function get_paged_list($limit = 0, $offset = 0) { $this->db->limit($limit, $offset); $this->db->select("person.*, company.company_name as company"); $this->db->from('person'); $this->db->join('company','person.company_id = company.company_id','left'); //MY QUESTION:? CAN I JOIN MORE WITH TABLE POSITION? $query = $this->db->get(); return $query->result(); } function get_by_id($id){ $this->db->where('person_id', $id); return $this->db->get($this->person); } function save($person){ $this->db->insert($this->person, $person); return $this->db->insert_id(); } function update($id, $person){ $this->db->where('person_id', $id); $this->db->update($this->person, $person); } function delete($id){ $this->db->where('person_id', $id); $this->db->delete($this->person); } } ?>

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  • Authlogic Help! Registering a new user when currently logged-in as a user not working.

    - by looloobs
    Hi Just as a disclaimer I am new to rails and programming in general so apologize for misunderstanding something obvious. I have Authlogic with activation up and running. So for my site I would like my users who are logged in to be able to register other users. The new user would pick their login and password through the activation email, but the existing user needs to put them in by email, position and a couple other attributes. I want that to be done by the existing user. The problem I am running into, if I am logged in and then try and create a new user it just tries to update the existing user and doesn't create a new one. I am not sure if there is some way to fix this by having another session start??? If that is even right/possible I wouldn't know how to go about implementing it. I realize without knowing fully about my application it may be difficult to answer this, but does this even sound like the right way to go about this? Am I missing something here? Users Controller: class UsersController < ApplicationController before_filter :require_no_user, :only => [:new, :create] before_filter :require_user, :only => [:show, :edit, :update] def new @user = User.new end def create @user = User.new if @user.signup!(params) @user.deliver_activation_instructions! flash[:notice] = "Your account has been created. Please check your e-mail for your account activation instructions!" redirect_to profile_url else render :action => :new end end def show @user = @current_user end def edit @user = @current_user end def update @user = @current_user # makes our views "cleaner" and more consistent if @user.update_attributes(params[:user]) flash[:notice] = "Account updated!" redirect_to profile_url else render :action => :edit end end end My User_Session Controller: class UserSessionsController < ApplicationController before_filter :require_no_user, :only => [:new, :create] before_filter :require_user, :only => :destroy def new @user_session = UserSession.new end def create @user_session = UserSession.new(params[:user_session]) if @user_session.save flash[:notice] = "Login successful!" if @user_session.user.position == 'Battalion Commander' : redirect_to battalion_path(@user_session.user.battalion_id) else end else render :action => :new end end def destroy current_user_session.destroy flash[:notice] = "Logout successful!" redirect_back_or_default new_user_session_url end end

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  • Creating a mySQL query using PHP form dropdowns - If user ignores dropdown, do not filter by that pa

    - by user303043
    Hello, I am creating a simple MySQL query that will be built from the user selecting options from 2 dropdowns. The issue I am having is that I would like each drop down to default that if they do not actually choose an option, do not filter by that dropdown parameter. So, if they come in, and simply hit submit without choosing from a dropdown they should see everything. If they come in and pick from only one of the dropdowns, the query will basically ignore filtering by the other dropdown. I tried making <OPTION VALUE='any'>Choose but my query didn't know what to do with the 'any' and just shows no results. Here is my code. Thank you very much for whatever help you can provide. FORM <form method="POST" action="<?php echo $_SERVER['REQUEST_URI']; ?>"> <select name="GameType"> <OPTION VALUE='any'>Choose Game Type <option value="Game1">Option 1</option> <option value="Game2">Option 2</option> <option value="Game3">Option 3</option> </select> <select name="Instructor"> <OPTION VALUE='any'>Choose Instructor <option VALUE="InstructorA">Instructor A</option> <option value="InstructorB">Instructor B</option> <option value="InstructorC">Instructor C</option> </select> <input type='submit' value='Search Videos'> </form> MYSQL <?PHP connection stuff $db_handle = mysql_connect($server, $user_name, $password); $db_found = mysql_select_db($database, $db_handle); if ($db_found) { $SQL = "SELECT * FROM Videos WHERE GameType=\"{$_POST['GameType']}\" AND Instructor=\"{$_POST['Instructor']}\""; $result = mysql_query($SQL); while ($db_field = mysql_fetch_assoc($result)) { echo $db_field['ShortDescription'] . ", "; echo $db_field['LongDescription'] . ", "; echo $db_field['GameType'] . ", "; echo $db_field['NumberOfPlayers'] . ", "; echo $db_field['Instructor'] . ", "; echo $db_field['Stakes'] . ", "; echo $db_field['UserPermissionLevel'] . ", "; echo $db_field['DateCreated'] . "<BR>"; } mysql_close($db_handle); } else { print "Database NOT Found "; mysql_close($db_handle); } ?>

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  • An error occurred creating the form. See Exception.InnerException for details. The error is: Object

    - by Ben
    I get this error when attempting to debug my form, I cannot see where at all the error could be (also does not highlight where), anyone have any suggestions? An error occurred creating the form. See Exception.InnerException for details. The error is: Object reference not set to an instance of an object. Public Class Form1 Dim dateCrap As String = "Date:" Dim IPcrap As String = "Ip:" Dim pcCrap As String = "Computer:" Dim programCrap As String = "Program:" Dim textz As String = TextBox1.Text Dim sep() As String = {vbNewLine & vbNewLine} Dim sections() As String = Text.Split(sep, StringSplitOptions.None) Dim NewArray() As String = TextBox1.Text.Split(vbNewLine) Private Sub Button2_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button2.Click For i = 0 To sections.Count - 1 TextBox2.Text = sections(i) Dim text2 As String = TextBox2.Text Dim sep2() As String = {vbNewLine & vbNewLine} Dim sections2() As String = Text.Split(sep, StringSplitOptions.None) Dim FTPinfo() As String = TextBox2.Text.Split(vbNewLine) Dim clsRequest As System.Net.FtpWebRequest = _ DirectCast(System.Net.WebRequest.Create(sections2(0).Replace("Url/Host:", "")), System.Net.FtpWebRequest) clsRequest.Credentials = New System.Net.NetworkCredential(sections2(1).Replace("Login:", ""), (sections2(2).Replace("Password:", ""))) clsRequest.Method = System.Net.WebRequestMethods.Ftp.UploadFile ' read in file... Dim bFile() As Byte = System.IO.File.ReadAllBytes(txtShellDir.Text) ' upload file... Dim clsStream As System.IO.Stream = clsRequest.GetRequestStream() clsStream.Write(bFile, 0, bFile.Length) clsStream.Close() clsStream.Dispose() Next End Sub Private Sub Button3_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button3.Click ListBox1.Items.Clear() Dim fdlg As OpenFileDialog = New OpenFileDialog() fdlg.Title = "Browse for the FTP List you wish to use." fdlg.InitialDirectory = Application.ExecutablePath fdlg.Filter = "All files (*.txt)|*.txt|txt files (*.txt)|*.txt" fdlg.FilterIndex = 2 fdlg.RestoreDirectory = True If fdlg.ShowDialog() = DialogResult.OK Then 'ListBox1.Items.AddRange(Split(My.Computer.FileSystem.ReadAllText(fdlg.FileName), vbNewLine)) TextBox1.Text = My.Computer.FileSystem.ReadAllText(fdlg.FileName) End If Dim tmp() As String = TextBox1.Text.Split(CChar(vbNewLine)) For Each line As String In tmp If line.Length > 1 Then TextBox1.AppendText(line & vbNewLine) End If Next TextBox1.Text = TextBox1.Text.Replace(" ", "") TextBox1.Text = TextBox1.Text.Replace("----------------------------------------------------------", vbNewLine) For Each item As String In NewArray ListBox1.Items.Add(item) Next Try For i = 0 To ListBox1.Items.Count - 1 If ListBox1.Items(i).Contains(dateCrap) Then ListBox1.Items.RemoveAt(i) End If Next Catch ex As Exception End Try Try For i = 0 To ListBox1.Items.Count - 1 If ListBox1.Items(i).Contains(IPcrap) Then ListBox1.Items.RemoveAt(i) End If Next Catch ex As Exception End Try Try For i = 0 To ListBox1.Items.Count - 1 If ListBox1.Items(i).Contains(pcCrap) Then ListBox1.Items.RemoveAt(i) End If Next Catch ex As Exception End Try Try For i = 0 To ListBox1.Items.Count - 1 If ListBox1.Items(i).Contains(programCrap) Then ListBox1.Items.RemoveAt(i) End If Next Catch ex As Exception End Try TextBox1.Text = "" For Each thing As Object In ListBox1.Items TextBox1.AppendText(thing.ToString & vbNewLine) Next End Sub Private Sub Button4_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) For i = 0 To ListBox1.SelectedItems.Count - 1 TextBox1.Text = TextBox1.Text & ListBox1.Items.Item(i).ToString() & vbNewLine Next End Sub End Class

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  • Rails nested form for belongs_to

    - by user1232533
    I'm new to rails and have some troubles with creating a nested form. My models: class User < ActiveRecord::Base belongs_to :company accepts_nested_attributes_for :company, :reject_if => :all_blank end class Company < ActiveRecord::Base has_many :users end Now i would like to create a new company from the user sign_up page (i use Devise btw) by given only a company name. And have a relation between the new User and new Company. In the console i can create a company for a existing User like this: @company = User.first.build_company(:name => "name of company") @company.save That works, but i can't make this happen for a new user, in my new user sign_up form i tried this (i know its wrong by creating a new User fist but im trying to get something working here..): <%= simple_form_for(resource, :as => resource_name, :html => { :class => 'form-horizontal' }, :url => registration_path(resource_name)) do |f| %> <%= f.error_notification %> <div class="inputs"> <% @user = User.new company = @user.build_company() %> <% f.fields_for company do |builder| %> <%= builder.input :name, :required => true, :autofocus => true %> <% end %> <%= f.input :email, :required => true, :autofocus => true %> <%= f.input :password, :required => true %> <%= f.input :password_confirmation, :required => true %> </div> <div class="form-actions"> <%= f.button :submit, :class => 'btn-primary', :value => 'Sign up' %> </div> I did my best to google for a solution/ example.. found some nested form examples but it's just not clear to me how to do this. Really hope somebody can help me with this. Any help on this would be appreciated. Thanks in advance! Greets, Daniel

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  • Criticise/Recommendations for my code

    - by aLk
    Before i go any further it would be nice to know if there is any major design flaws in my program so far. Is there anything worth changing before i continue? Model package model; import java.sql.*; import java.util.*; public class MovieDatabase { @SuppressWarnings({ "rawtypes", "unchecked" }) public List queryMovies() throws SQLException { Connection connection = null; java.sql.Statement statement = null; ResultSet rs = null; List results = new ArrayList(); try { DriverManager.registerDriver(new com.mysql.jdbc.Driver()); connection = DriverManager.getConnection("jdbc:mysql://localhost:3306/test", "root", "password"); statement = connection.createStatement(); String query = "SELECT * FROM movie"; rs = statement.executeQuery(query); while(rs.next()) { MovieBean bean = new MovieBean(); bean.setMovieId(rs.getInt(1)); bean.setTitle(rs.getString(2)); bean.setYear(rs.getInt(3)); bean.setRating(rs.getInt(4)); results.add(bean); } } catch(SQLException e) { } return results; } } Servlet public class Service extends HttpServlet { @SuppressWarnings("rawtypes") protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { response.setContentType("text/html"); PrintWriter out = response.getWriter(); out.println("Movies!"); MovieDatabase movies = new MovieDatabase(); try { List results = movies.queryMovies(); Iterator it = results.iterator(); while(it.hasNext()) { MovieBean movie = new MovieBean(); movie = (MovieBean)it.next(); out.println(movie.getYear()); } } catch(SQLException e) { } } } Bean package model; @SuppressWarnings("serial") public class MovieBean implements java.io.Serializable { protected int movieid; protected int rating; protected int year; protected String title; public MovieBean() { } public void setMovieId(int movieidVal) { movieid = movieidVal; } public void setRating(int ratingVal) { rating = ratingVal; } public void setYear(int yearVal) { year = yearVal; } public void setTitle(String titleVal) { title = titleVal; } public int getMovieId() { return movieid; } public int getRating() { return rating; } public int getYear() { return year; } public String getTitle() { return title; } }

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  • PHP problems when transfering code from Windows to OS X

    - by Makka95
    I have recently bought a new MacBook Pro. Before I had my MacBook Pro I was working on a website on my desktop computer. And now I want to transfer this code to my new MacBook Pro. The problem is that when I transfered the code (I put it on Dropbox and simply downloaded it on my MacBook Pro) I started to see lots of error messages in my PHP code. The error message I”m receiving is: Warning: Cannot modify header information - headers already sent by (output started at /some/file.php:1) in /some/file.php on line 23 I have done some research on this and it seems that this error is most frequently caused by a new line, simple whitespace or any output before the <?php sign. I have looked through all the places where I have cookies that are being sent in the HTTP request and also where I'm using the header() function. I haven’t detected any output or whitespace that possibly could interfere and cause this problem. Noteworthy is that the error always says that the output is started at line 1. Which got me thinking if there is some kind of coding differences in the way that the Mac OS X and Windows operating systems handle new lines or white spaces? Or could the Dropbox transfer messed something up? The code on one of the sites(login.php) which produces the error: <?php include "mysql_database.php"; login(); $id = $_SESSION['Loggedin']; setcookie("login", $id, (time()+60*60*24*30)); header('Location: ' . $_SERVER['HTTP_REFERER']); ?> login function: function login() { $connection = connecttodatabase(); $pass = ""; $user = ""; $query = ""; if (isset($_POST['user']) && $_POST['user'] != null) { $user = $_POST['user']; if (isset($_POST['pass']) && $_POST['pass'] != null) { $pass = md5($_POST['pass']); $query = "SELECT ID FROM Anvandare WHERE Nickname='$user' AND Password ='$pass'"; } } if ($query != "") { $id = $connection->query($query); $id = mysqli_fetch_assoc($id); $id = $id['ID']; $_SESSION['Loggedin'] = $id; } closeconnection($connection); } Complete error: Warning: Cannot modify header information - headers already sent by (output started at /Users/name/GitHub/website/login.php:1) in /Users/namn/GitHub/website/login.php on line 9

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  • pdo connection scope

    - by Scarface
    Hey guys I have a connection class I found for pdo. I am calling the connection method on the page that the file is included on. The problem is that within functions the $conn variable is not defined even though I stated the method was public (bare with me I am very new to OOP), and I was wondering if anyone had an elegant solution other then using global in every function. Any suggestions are greatly appreciated. CONNECTION class PDOConnectionFactory{ // receives the connection public $con = null; // swich database? public $dbType = "mysql"; // connection parameters // when it will not be necessary leaves blank only with the double quotations marks "" public $host = "localhost"; public $user = "user"; public $senha = "password"; public $db = "database"; // arrow the persistence of the connection public $persistent = false; // new PDOConnectionFactory( true ) <--- persistent connection // new PDOConnectionFactory() <--- no persistent connection public function PDOConnectionFactory( $persistent=false ){ // it verifies the persistence of the connection if( $persistent != false){ $this->persistent = true; } } public function getConnection(){ try{ // it carries through the connection $this->con = new PDO($this->dbType.":host=".$this->host.";dbname=".$this->db, $this->user, $this->senha, array( PDO::ATTR_PERSISTENT => $this->persistent ) ); // carried through successfully, it returns connected return $this->con; // in case that an error occurs, it returns the error; }catch ( PDOException $ex ){ echo "We are currently experiencing technical difficulties. We have a bunch of monkies working really hard to fix the problem. Check back soon: ".$ex->getMessage(); } } // close connection public function Close(){ if( $this->con != null ) $this->con = null; } } PAGE USED ON include("includes/connection.php"); $db = new PDOConnectionFactory(); $conn = $db->getConnection(); function test(){ try{ $sql = 'SELECT * FROM topic'; $stmt = $conn->prepare($sql); $result=$stmt->execute(); } catch(PDOException $e){ echo $e->getMessage(); } } test();

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  • Please Describe Your Struggles with Minimizing Use of Global Variables

    - by MetaHyperBolic
    Most of the programs I write are relatively flowchartable processes, with a defined start and hoped-for end. The problems themselves can be complex but do not readily lean towards central use of objects and event-driven programming. Often, I am simply churning through great varied batches of text data to produce different text data. Only occasionally do I need to create a class: As an example, to track warnings, errors, and debugging message, I created a class (Problems) with one instantiation (myErr), which I believe to be an example of the Singleton design pattern. As a further factor, my colleagues are more old school (procedural) than I and are unacquainted with object-oriented programming, so I am loath to create things they could not puzzle through. And yet I hear, again and again, how even the Singleton design pattern is really an anti-pattern and ought to be avoided because Global Variables Are Bad. Minor functions need few arguments passed to them and have no need to know of configuration (unchanging) or program state (changing) -- I agree. However, the functions in the middle of the chain, which primarily control program flow, have a need for a large number of configuration variables and some program state variables. I believe passing a dozen or more arguments along to a function is a "solution," but hardly an attractive one. I could, of course, cram variables into a single hash/dict/associative array, but that seems like cheating. For instance, connecting to the Active Directory to make a new account, I need such configuration variables as an administrative username, password, a target OU, some default groups, a domain, etc. I would have to pass those arguments down through a variety of functions which would not even use them, merely shuffle them off down through a chain which would eventually lead to the function that actually needs them. I would at least declare the configuration variables to be constant, to protect them, but my language of choice these days (Python) provides no simple manner to do this, though recipes do exist as workarounds. Numerous Stack Overflow questions have hit on the why? of the badness and the requisite shunning, but do not often mention tips on living with this quasi-religious restriction. How have you resolved, or at least made peace with, the issue of global variables and program state? Where have you made compromises? What have your tricks been, aside from shoving around flocks of arguments to functions?

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  • Converting my lightweight MySQL DB wrapper into MySQLi. Pesky Problems

    - by Chaplin
    Here is the original code: http://pastebin.com/DNxtmApY. I'm not that interested in prepared statements at the moment, I just want this wrapper updating to MySQLi so once MySQL becomes depreciated I haven't got to update a billion websites. Here is my attempt at converting to MySQLi. <? $database_host = "127.0.0.1"; $database_user = "user"; $database_pass = "pass"; $database_name = "name"; $db = new database($database_host, $database_user, $database_pass, $database_name); class database { var $link, $result; function database($host, $user, $pass, $db) { $this->link = mysqli_connect($host, $user, $pass, $db) or $this->error(); mysqli_select_db($db, $this->link) or $this->error(); } function query($query) { $this->result = mysqli_query($query, $this->link) or $this->error(); $this->_query_count++; return $this->result; } function countRows($result = "") { if ( empty( $result ) ) $result = $this->result; return mysqli_num_rows($result); } function fetch($result = "") { if ( empty( $result ) ) $result = $this->result; return mysqli_fetch_array($result); } function fetch_num($result = "") { if ( empty( $result ) ) $result = $this->result; return mysqli_fetch_array($result, mysqli_NUM); } function fetch_assoc($result = "") { if ( empty( $result ) ) $result = $this->result; return mysqli_fetch_array($result, mysqli_ASSOC); } function escape($str) { return mysqli_real_escape_string($str); } function error() { if ( $_GET["debug"] == 1 ){ die(mysqi_error()); } else { echo "Error in db code"; } } } function sanitize($data) { //apply stripslashes if magic_quotes_gpc is enabled if(get_magic_quotes_gpc()) $data = stripslashes($data); // a mysqli connection is required before using this function $data = trim(mysqli_real_escape_string($data)); return $data; } However it chucks all sorts of errors: Warning: mysql_query(): Access denied for user 'www-data'@'localhost' (using password: NO) in /home/count/Workspace/lib/classes/user.php on line 7 Warning: mysql_query(): A link to the server could not be established in /home/count/Workspace/lib/classes/user.php on line 7 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/count/Workspace/lib/classes/user.php on line 8 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given in /home/count/Workspace/lib/classes/database.php on line 31

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  • Get return values from a stored procedure in c# (login process)

    - by Jin
    Hi all, I am trying to use a Stored Procedure which takes two parameters (login, pw) and returns the user info. If I execute the SP manually, I get Session_UID User_Group_Name Sys_User_Name ------------------------------------ -------------------------------------------------- - NULL Administrators NTMSAdmin No rows affected. (1 row(s) returned) @RETURN_VALUE = 0 Finished running [dbo].[p_SYS_Login]. But with the code below, I only get the return value. do you know how to get the other values shown above like Session_UID, User_Group_Name, and Sys_User_Name ? if you see the commented part below code. I tried to add some output parameters but it doesn't work with incorrect number of parameters error. string strConnection = Settings.Default.ConnectionString; using (SqlConnection conn = new SqlConnection(strConnection)) { using (SqlCommand cmd = new SqlCommand()) { SqlDataReader rdr = null; cmd.Connection = conn; cmd.CommandText = "p_SYS_Login"; //cmd.CommandText = "p_sys_Select_User_Group"; cmd.CommandType = CommandType.StoredProcedure; SqlParameter paramReturnValue = new SqlParameter(); paramReturnValue.ParameterName = "@RETURN_VALUE"; paramReturnValue.SqlDbType = SqlDbType.Int; paramReturnValue.SourceColumn = null; paramReturnValue.Direction = ParameterDirection.ReturnValue; //SqlParameter paramGroupName = new SqlParameter("@User_Group_Name", SqlDbType.VarChar, 50); //paramGroupName.Direction = ParameterDirection.Output; //SqlParameter paramUserName = new SqlParameter("@Sys_User_Name", SqlDbType.VarChar, 50); //paramUserName.Direction = ParameterDirection.Output; cmd.Parameters.Add(paramReturnValue); //cmd.Parameters.Add(paramGroupName); //cmd.Parameters.Add(paramUserName); cmd.Parameters.AddWithValue("@Sys_Login", textUserID.Text); cmd.Parameters.AddWithValue("@Sys_Password", textPassword.Text); try { conn.Open(); object result = cmd.ExecuteNonQuery(); int returnValue = (int)cmd.Parameters["@RETURN_VALUE"].Value; if (returnValue == 0) { Hide(); Program.MapForm.Show(); } else if (returnValue == 1) { MessageBox.Show("The username or password you entered is incorrect", "NTMS Login", MessageBoxButtons.OK, MessageBoxIcon.Warning); } else if (returnValue == 2) { MessageBox.Show("This account is disabled", "NTMS Login", MessageBoxButtons.OK, MessageBoxIcon.Warning); } else { MessageBox.Show("Database error. Please contact administrator", "NTMS Login", MessageBoxButtons.OK, MessageBoxIcon.Warning); } } catch (Exception ex) { string message = ex.Message; string caption = "MAVIS Exception"; MessageBoxButtons buttons = MessageBoxButtons.OK; MessageBox.Show( message, caption, buttons, MessageBoxIcon.Warning, MessageBoxDefaultButton.Button1); } } } Thanks for your help.

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  • Capistrano Error

    - by Casey van den Bergh
    I'm Running CentOS 5 32 bit version. This is my deploy.rb file on my local computer: #======================== #CONFIG #======================== set :application, "aeripets" set :scm, :git set :git_enable_submodules, 1 set :repository, "[email protected]:aeripets.git" set :branch, "master" set :ssh_options, { :forward_agent => true } set :stage, :production set :user, "root" set :use_sudo, false set :runner, "root" set :deploy_to, "/var/www/#{application}" set :app_server, :passenger set :domain, "aeripets.co.za" #======================== #ROLES #======================== role :app, domain role :web, domain role :db, domain, :primary => true #======================== #CUSTOM #======================== namespace :deploy do task :start, :roles => :app do run "touch #{current_release}/tmp/restart.txt" end task :stop, :roles => :app do # Do nothing. end desc "Restart Application" task :restart, :roles => :app do run "touch #{current_release}/tmp/restart.txt" end end And this the error I get on my local computer when I try to cap deploy. executing deploy' * executingdeploy:update' ** transaction: start * executing deploy:update_code' executing locally: "git ls-remote [email protected]:aeripets.git master" command finished in 1297ms * executing "git clone -q [email protected]:aeripets.git /var/www/seripets/releases/20111126013705 && cd /var/www/seripets/releases/20111126013705 && git checkout -q -b deploy 32ac552f57511b3ae9be1d58aec54d81f78f8376 && git submodule -q init && git submodule -q sync && export GIT_RECURSIVE=$([ ! \"git --version\" \\< \"git version 1.6.5\" ] && echo --recursive) && git submodule -q update --init $GIT_RECURSIVE && (echo 32ac552f57511b3ae9be1d58aec54d81f78f8376 > /var/www/seripets/releases/20111126013705/REVISION)" servers: ["aeripets.co.za"] Password: [aeripets.co.za] executing command ** [aeripets.co.za :: err] sh: git: command not found command finished in 224ms *** [deploy:update_code] rolling back * executing "rm -rf /var/www/seripets/releases/20111126013705; true" servers: ["aeripets.co.za"] [aeripets.co.za] executing command command finished in 238ms failed: "sh -c 'git clone -q [email protected]:aeripets.git /var/www/seripets/releases/20111126013705 && cd /var/www/seripets/releases/20111126013705 && git checkout -q -b deploy 32ac552f57511b3ae9be1d58aec54d81f78f8376 && git submodule -q init && git submodule -q sync && export GIT_RECURSIVE=$([ ! \"git --version`\" \< \"git version 1.6.5\" ] && echo --recursive) && git submodule -q update --init $GIT_RECURSIVE && (echo 32ac552f57511b3ae9be1d58aec54d81f78f8376 /var/www/seripets/releases/20111126013705/REVISION)'" on aeripets.co.za

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  • Allow users to pull temporary data then delete table?

    - by JM4
    I don't know the best way to title this question but am trying to accomplish the following goal: When a client logs into their profile, they are presented with a link to download data from an existing database in CSV format. The process works, however, I would like for this data to be 'fresh' each time they click the link so my plan was - once a user has clicked the link and downloaded the CSV file, the database table would 'erase' all of its data and start fresh (be empty) until the next set of data populated it. My EXISTING CSV creation code: <?php $host = 'localhost'; $user = 'username'; $pass = 'password'; $db = 'database'; $table = 'tablename'; $file = 'export'; $link = mysql_connect($host, $user, $pass) or die("Can not connect." . mysql_error()); mysql_select_db($db) or die("Can not connect."); $result = mysql_query("SHOW COLUMNS FROM ".$table.""); $i = 0; if (mysql_num_rows($result) > 0) { while ($row = mysql_fetch_assoc($result)) { $csv_output .= $row['Field'].", "; $i++; } } $csv_output .= "\n"; $values = mysql_query("SELECT * FROM ".$table.""); while ($rowr = mysql_fetch_row($values)) { for ($j=0;$j<$i;$j++) { $csv_output .= '"'.$rowr[$j].'",'; } $csv_output .= "\n"; } $filename = $file."_".date("Y-m-d",time()); header("Content-type: application/vnd.ms-excel"); header("Content-disposition: csv" . date("Y-m-d") . ".csv"); header( "Content-disposition: filename=".$filename.".csv"); print $csv_output; exit; ?> any ideas?

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  • HTML format using Java mail in android

    - by TheDevMan
    I am trying to implement an HTML format mail using the Java mail in android. I would like to get results like this: When I look at the html format sent from lookout in my GMAIL. I don't see any link, but just has this format: [image: Lookout_logo] [image: Signal_flare_icon] Your battery level is really low, so we located your device with Signal Flare. I was trying the following: Properties props = System.getProperties(); props.put("mail.smtp.starttls.enable", "true"); // added this line props.put("mail.smtp.host", host); props.put("mail.smtp.user", from); props.put("mail.smtp.password", pass); props.put("mail.smtp.port", "587"); props.put("mail.smtp.auth", "true"); javax.mail.Session session = javax.mail.Session.getDefaultInstance(props, null); MimeMessage message = new MimeMessage(session); message.setFrom(new InternetAddress(from)); InternetAddress[] toAddress = new InternetAddress[to.length]; // To get the array of addresses for( int i=0; i < to.length; i++ ) { // changed from a while loop toAddress[i] = new InternetAddress(to[i]); } message.setRecipients(Message.RecipientType.BCC, toAddress); message.setSubject(sub); //message.setText(body); body = "<!DOCTYPE html><html><body><img src=\"http://en.wikipedia.org/wiki/Krka_National_Park#mediaviewer/File:Krk_waterfalls.jpg\">"; message.setContent(body, "text/html; charset=utf-8"); Transport transport = session.getTransport("smtp"); transport.connect(host, from, pass); transport.sendMessage(message, message.getAllRecipients()); transport.close(); When I look at the html format sent with the above code. I get the following: <!DOCTYPE html><html><body><img src="http://en.wikipedia.org/wiki/Krka_National_Park#mediaviewer/File:Krk_waterfalls.jpg> How to make sure the user will not be able to see any html code or URL link like the mail sent by LOOKOUT? Thanks!

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  • What should I learn & use to become a pro in PHP & Python Web development?

    - by pecker
    Hello, I'll just show some code to show how I do web development in PHP. <html> <head> <title>Example #3 TDavid's Very First PHP Script ever!</title> </head> <? print(Date("m/j/y")); require_once("somefile.php"); $mysql_db = "DATABASE NAME"; $mysql_user = "YOUR MYSQL USERNAME"; $mysql_pass = "YOUR MYSQL PASSWORD"; $mysql_link = mysql_connect("localhost", $mysql_user, $mysql_pass); mysql_select_db($mysql_db, $mysql_link); $result = mysql_query("SELECT impressions from tds_counter where COUNT_ID='$cid'", $mysql_link); if(mysql_num_rows($result)) { mysql_query("UPDATE tds_counter set impressions=impressions+1 where COUNT_ID='$cid'", $mysql_link); $row = mysql_fetch_row($result); if(!$inv) { print("$row[0]"); } } ?> <body> </body> </html> Thats it. I write every file like this. Recently, I learnt OOP and started using classes & objects in PHP. I hear that there are many frameworks there for PHP. They say that one must use these libraries. But I feel they are just making things complicated. Anyway, this is how I've been doing my web development. Now, I want to improve this. and make it professional. Also I want to move to Python. I searched SO archives and found everyone suggesting Django. But, can any one give me some idea about how web development in Python works? user (client) request for page --- webserver(-embedded PHP interpreter) ---- Server side(PHP) Script --- MySQL Server. Now, is it that instead of PHP interpreter there is python interpreter & instead of php script there is python script, which contains both HTML & python (embedded in some kind of python tags). Python script connects to database server and fetches some data which will be printed as HTML. or is it different in python world? Is this Django thing like frameworks for PHP? Can't one code in python without using Django. Because, I never encountered any post without django Please give me some kick start.

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  • Delete duplicate rows, do not preserve one row

    - by Radley
    I need a query that goes through each entry in a database, checks if a single value is duplicated elsewhere in the database, and if it is - deletes both entries (or all, if more than two). Problem is the entries are URLs, up to 255 characters, with no way of identifying the row. Some existing answers on Stackoverflow do not work for me due to performance limitations, or they use uniqueid which obviously won't work when dealing with a string. Long Version: I have two databases containing URLs (and only URLs). One database has around 3,000 urls and the other around 1,000. However, a large majority of the 1,000 urls were taken from the 3,000 url database. I need to merge the 1,000 into the 3,000 as new entries only. For this, I made a third database with combined URLs from both tables, about 4,000 entries. I need to find all duplicate entries in this database and delete them (Both of them, without leaving either). I have followed the query of a few examples on this site, but whenever I try to delete both entries it ends up deleting all the entries, or giving sql errors. Alternatively: I have two databases, each containing the separate database. I need to check each row from one database against the other to find any that aren't duplicates, and then add those to a third database. Edit: I've got my own PHP solution which is pretty hacky, but works. I cannot answer my own question for 8 hours because I'm new, so here it is for now: I went with a PHP script to accomplish this, as I'm more familiar with PHP than MySQL. This generates a simple list of urls that only exist in the target database, but not both. If you have more than 7,000 entries to parse this may take awhile, and you will need to copy/paste the results into a text file or expand the script to store them back into a database. I'm just doing it manually to save time. Note: Uses MeekroDB <pre> <?php require('meekrodb.2.1.class.php'); DB::$user = 'root'; DB::$password = ''; DB::$dbName = 'testdb'; $all = DB::query('SELECT * FROM old_urls LIMIT 7000'); foreach($all as $row) { $test = DB::query('SELECT url FROM new_urls WHERE url=%s', $row['url']); if (!is_array($test)) { echo $row['url'] . "\n"; }else{ if (count($test) == 0) { echo $row['url'] . "\n"; } } } ?> </pre>

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  • How do I defer execution of some Ruby code until later and run it on demand in this scenario?

    - by Kyle Kaitan
    I've got some code that looks like the following. First, there's a simple Parser class for parsing command-line arguments with options. class Parser def initialize(&b); ...; end # Create new parser. def parse(args = ARGV); ...; end # Consume command-line args. def opt(...); ...; end # Declare supported option. def die(...); ...; end # Validation handler. end Then I have my own Parsers module which holds some metadata about parsers that I want to track. module Parsers ParserMap = {} def self.make_parser(kind, desc, &b) b ||= lambda {} module_eval { ParserMap[kind] = {:desc => "", :validation => lambda {} } ParserMap[kind][:desc] = desc # Create new parser identified by `<Kind>Parser`. Making a Parser is very # expensive, so we defer its creation until it's actually needed later # by wrapping it in a lambda and calling it when we actually need it. const_set(name_for_parser(kind), lambda { Parser.new(&b) }) } end # ... end Now when you want to add a new parser, you can call make_parser like so: make_parser :db, "login to database" do # Options that this parser knows how to parse. opt :verbose, "be verbose with output messages" opt :uid, "user id" opt :pwd, "password" end Cool. But there's a problem. We want to optionally associate validation with each parser, so that we can write something like: validation = lambda { |parser, opts| parser.die unless opts[:uid] && opts[:pwd] # Must provide login. } The interface contract with Parser says that we can't do any validation until after Parser#parse has been called. So, we want to do the following: Associate an optional block with every Parser we make with make_parser. We also want to be able to run this block, ideally as a new method called Parser#validate. But any on-demand method is equally suitable. How do we do that?

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  • My php script only reads the first row from mysql and doesn't check the rest of the rows for matches

    - by RobertH
    I'm trying to write a script for users to register to a club, and it does all the validation stuff properly and works great until it gets to the part where its supposed to check for duplicates. I'm not sure what is going wrong. HELP PLEASE!!! Thank you in Advance, <?php mysql_connect ("sqlhost", "username", "password") or die(mysql_error()); mysql_select_db ("databasename") or die(mysql_error()); $errormsgdb = ""; $errordb = "Sorry but that "; $error1db = "Name"; $error2db = "email"; $error3db = "mobile number"; $errordbe = " is already registered"; $pass1db = "No Matching Name"; $pass2db = "No Matching Email"; $pass3db = "No Matching Mobile"; $errorcount = 0; $qResult = mysql_query ("SELECT * FROM table"); $nRows = mysql_num_rows($qResult); for ($i=1; $i< $nRows+1; $i++){ $result = mysql_query("SELECT id,fname,lname,dob,email,mobile,agree,code,joindate FROM table WHERE fname = '$ffname"); if ($result > 0) { $errorcount = $errorcount++; $passdb = 0; $errormsgdb = $error1db; echo "<div class=\"box red\">$errordb $errormsgdb } else { $pass = 1; $errormsgdb = $pass1db; echo "<div class=\"box green\">$errormsgdb</div><br />"; } //--------------- Check if DB checks returned errors ------------------------------------> if($errorcount <= 0){ $dobp = $_REQUEST['day'].'/'.$_REQUEST['month'].'/'.$_REQUEST['year']; $dob = $_REQUEST['year'].$_REQUEST['month'].$_REQUEST['day']; //header('Location: thankyou.php?ffname='.$ffname.'&flname='.$flname.'&dob='.$dob.'&femail='.$femail.'&fmobile='.$fmobile.'&agree='.$agree.'&code='.$code.'&dobp='.$dobp); echo "<div class='box green'>Form completed! Error Count = $errorcount</div>"; } else { echo "<div class='box red'>There was an Error! Error Count = $errorcount</div>"; } } ?>

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  • session fixation

    - by markiv
    Hi All, I am new to web development, and trying to get a hold on security issues. I went through this article on http://guides.rubyonrails.org/security.html these are some of the steps the author has mentioned how an attacker fixes session. 1. The attacker creates a valid session id: He loads the login page of the web application where he wants to fix the session, and takes the session id in the cookie from the response (see number 1 and 2 in the image). 2. He possibly maintains the session. Expiring sessions, for example every 20 minutes, greatly reduces the time-frame for attack. Therefore he accesses the web application from time to time in order to keep the session alive. 3. Now the attacker will force the user’s browser into using this session id (see number 3 in the image). As you may not change a cookie of another domain (because of the same origin policy), the attacker has to run a JavaScript from the domain of the target web application. Injecting the JavaScript code into the application by XSS accomplishes this attack. Here is an example: <script>?document.cookie="_session_id=16d5b78abb28e3d6206b60f22a03c8d9";?</script>. Read more about XSS and injection later on. 4. The attacker lures the victim to the infected page with the JavaScript code. By viewing the page, the victim’s browser will change the session id to the trap session id. 5. As the new trap session is unused, the web application will require the user to authenticate. 6. From now on, the victim and the attacker will co-use the web application with the same session: The session became valid and the victim didn’t notice the attack. I dont understand couple of points. i) why is user made to login in step5, since session is sent through. ii) I saw possible solutions on wiki, like user properties check and others why cant we just reset the session for the user whoever is login in when they enter username and password in step5? Thanks in advance Markiv

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  • c# .net MVC4 Model to represent table or form

    - by Matthew Chambers
    Hello I am a little confused with regards to models in mvc 4 and thought someone may be able to point me in the right direction. This would be most appreciated. For example if i have a table that has the following fields [StringLength(100, ErrorMessage = "The {0} must be at least {2} characters long.", MinimumLength = 5)] public string UserName { get; set; } [Required(ErrorMessage="Email Address is Required")] [StringLength(15, ErrorMessage = "Email Address must be between {0} and {1} in size",MinimumLength = 5 )] [DataType(DataType.EmailAddress)] [Display(Name="Email")] public string Email { get; set; } [MaxLength(25)] [Display(Name="Mobile Telephone Number")] public string Mobile {get;set;} [MaxLength(500)] [Display(Name="Headline")] public string Headline {get;set;} [Required] [StringLength(200)] [Display(Name = "First Name")] public string FirstName {get;set;} [Required] [StringLength(200)] [Display(Name="Surname")] public string Surname { get; set;} public virtual int? DayOfBirthId { get; set; } public virtual DayOfBirth DayOfBirth { get; set; } public virtual int? MonthOfBirthId { get; set; } public virtual MonthOfBirth MonthOfBirth { get; set; } public virtual int? YearOfBirthId { get; set; } public virtual YearOfBirth YearOfBirth{get;set;} This is my user profile table in the database. However I would like a form that the user registers to the site with. When they first register i do not need all the details such as telephone all i really need is there username, email address and password. Do i create another model for this. Or do i have one model and on the controller set the fields to null or empty string that are not required on registration. I have validation also so would this be set for data that has not been entered on the form. My question is ultimately should all forms represent models- should the database be redesigned to meet this required. Or should the controller set the values that are not required. Or should there be another model that represents the form be created which maps to this table. I am a little confused on this and clarification of anyone would be most appreciated.

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  • The big last_insert_id() problem, again.

    - by wretrOvian
    Note - this follows my question here: http://stackoverflow.com/questions/2983685/jdbc-does-the-connection-break-if-i-lose-reference-to-the-connection-object Now i have a created a class so i can deal with JDBC easily for the rest of my code - public class Functions { private String DB_SERVER = ""; private String DB_NAME = "test"; private String DB_USERNAME = "root"; private String DB_PASSWORD = "password"; public Connection con; public PreparedStatement ps; public ResultSet rs; public ResultSetMetaData rsmd; public void connect() throws java.io.FileNotFoundException, java.io.IOException, SQLException, Exception { String[] dbParms = Parameters.load(); DB_SERVER = dbParms[0]; DB_NAME = dbParms[1]; DB_USERNAME = dbParms[2]; DB_PASSWORD = dbParms[3]; // Connect. Class.forName("com.mysql.jdbc.Driver").newInstance(); con = DriverManager.getConnection("jdbc:mysql://" + DB_SERVER + "/" + DB_NAME, DB_USERNAME, DB_PASSWORD); } public void disconnect() throws SQLException { // Close. con.close(); } } As seen Parameters.load() refreshes the connection parameters from a file every-time, so that any changes to the same may be applied on the next immediate connection. An example of this class in action - public static void add(String NAME) throws java.io.FileNotFoundException, java.io.IOException, SQLException, Exception { Functions dbf = new Functions(); dbf.connect(); String query = "INSERT INTO " + TABLE_NAME + "(" + "NAME" + ") VALUES(?)"; PreparedStatement ps = dbf.con.prepareStatement(query); ps.setString(1, NAME); ps.executeUpdate(); dbf.disconnect(); } Now here is the problem - for adding a record to the table above, the add() method will open a connection, add the record - and then call disconnect() . What if i want to get the ID of the inserted record after i call add() -like this : Department.add("new dept"); int ID = getlastID(); Isn't it possible that another add() was called between those two statements?

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