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• CodeIgniter's index.php

  - by 01010011

  Hi, Following CodeIngiter's installation instructions, I extracted CodeIgniter into XAMPP's htdocs folder and placed its index.php at the root. When I navigated to the following url h t t p://127.0.0.1/index.php I got the following error messages: Warning: require_once(C:\xampp\htdocs/system/codeigniter/CodeIgniter.php) [function.require-once]: failed to open stream: No such file or directory in C:\xampp\htdocs\index.php on line 115 Fatal error: require_once() [function.require]: Failed opening required 'C:\xampp\htdocs/system/codeigniter/CodeIgniter.php' (include_path='.;C:\xampp\php\PEAR') in C:\xampp\htdocs\index.php on line 115 When I checked line 115 in the index.php, it stated: require_once BASEPATH.'codeigniter/CodeIgniter'.EXT; Experimenting a little, I put CodeIniter's "system" folder at the root as well, and it worked, providing me access to CodeIgniter's Welcome page, but the instructions did not say to do this. Experimenting further, when I put back the "system" folder and navigated to the original place where the index.php h t t p://127.0.0.1/codeigniter_1.7.2/index.php it worked. But myquestion is, how do I get to the CodeIgniter's welcome page by placing only the index.php in XAMPP's htdocs as instructed by CodeIgniter? Any assistance will be appreciated.

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• PHPMyAdmin works with https Only (not http)

  - by 01010011

  Hi I've been having a problem getting phpmyadmin to work consistently on my XP desktop and laptop computers for months now. When I type into Chrome's browser on both machines, localhost/phpmyadmin, I kept getting Error #1045 Access Denied for user at root@localhost (using password yes). Eventually, I realized that I had two (2) versions of mysql installed (XAMPP and MySQL Server 5.1) on both machines. So I uninstalled the MySQL Server 5.1I from the desktop and phpmyadmin worked. But when I uninstalled MySQL Server 5.1 from my laptop, it did not work. But I realized I could still get into MySQL Commandline Client using my password and that my databases were still intact. So I uninstalled and reinstalled XAMPP on the laptop and phpmyadmin worked after that. Now I have a new problem. On phpMyAdmin's home page has a message at the bottom: Your configuration file contains settings (root with no password) that correspond to the default MySQL privileged account. Your MySQL server is running with this default, is open to intrusion, and you really should fix this security hole by setting a password for user 'root'. So I located the following lines in config.inc.php file: /* Authentication type and info */ $cfg['Servers'][$i]['auth_type'] = 'config'; $cfg['Servers'][$i]['user'] = 'root'; $cfg['Servers'][$i]['password'] = ''; $cfg['Servers'][$i]['AllowNoPassword'] = true; and I just changed the last 2 lines as follows: $cfg['Servers'][$i]['password'] = 'mypassword'; $cfg['Servers'][$i]['AllowNoPassword'] = false; As soon as I did that and I tried to access phpmyadmin again, I got the Error #1045 message again, but when I tried https://localhost/phpmyadmin/ I got a red page saying this sites certificate is not trusted would you like to proceed anyway. And now it only works using https. I would really like to settle all my phpmyadmin problems once and for all so here are my questions: 1. Why does my laptop only access phpmyadmin via https? 2. How do I change my password in my configuration file? Also, if you have any other tips regarding phpMyAdmin, they are very welcome. Thanks in advance

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• CodeIgniter's Scaffolding not working

  - by 01010011

  Hi, I keep getting a 404 Page Not Found whenever I try to access CodeIgniter's Scaffolding page in my browser, like so: localhost/codeignitor/index.php/blog/scaffolding/mysecretword I can access localhost/codeignitor/index.php/blog just fine. I followed CodeIgnitor's instructions in their "Create a blog in 20 minutes" by storing my database settings in the database.php file; and automatically connecting to the database by inserting "database" in the core array of the autoload.php; and I've added both parent::Controller(); and $this-load-scaffolding('myTableName') to blog's constructor. It still gives me this 404. Any suggestions?

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• CodeIgnitor Error 403 Access forbidden

  - by 01010011

  Hi, I extracted CodeIgnitor to XAMPP's htdocs and when I tried to access index.php like this: h t t p://127.0.0.1/ci/index.php I get the following error message: Access forbidden! You don't have permission to access the requested object. It is either read-protected or not readable by the server. Error 403...... How can I troubleshoot this problem?

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• UML Class Diagram for User Login

  - by 01010011

  Hi, The diagram below is my very first attempt at creating a UML class diagram describing a user login into a website. I'm sure its a poor design and full of flaws, but I'm hoping to learn from you guys how you would design a simple login like this. I'm particularly interested in your use of design patterns and which patterns you would use, how you would implement it in the design, and why. Any advise, criticisms, comments and suggestions will be really appreciated. Thanks in advance.

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• Entity Relationship Diagram - Relationship Strength?

  - by 01010011

  Hi, I am trying to figure out under what circumstances I should use weak (non-identifying) relationships (where the primary key of the related entity does not contain a primary key component of the parent entity), verses when I should use strong (identifying) relationships (primary key of the related entity contains a primary key component of the parent entity). For example, when designing an Entity Relationship Diagram , if I have two entities, (e.g. book and purchaser), how do I know when to choose the solid Crows Foot or the dashed Crows Foot to connect the two entities? Any assistance will be appreciated. Thanks in advance.

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• Doctrine CodeIgniter MySQL CRUD errors

  - by 01010011

  Hi, I am using CI + Doctrine + MySQL and I am getting the following CRUD errors: (1) When trying to create a new record in the database with this code: $book_title = 'The Peloponnesian War'; $b = new Book(); $b-title = $book_title; $b-price = 10.50; $b-save(); I get this error: Fatal error: Uncaught exeption 'Doctrine_Connection_Mysql_Exception' with message 'SQLSTATE[42S22]: Column not found: 1054 Unknown column 'title' in 'field list' in ... (2) When trying to fetch a record from the database and display on my view page with this code: $book_title = 'The Peloponnesian War'; $title = $book_title; $search_results = Doctrine::getTable('Book')-findOneByTitle($title); echo $search_results-title; (in view file) I get this error: Fatal error: Uncaught exception 'Doctrine_Connection_Mysql_Exception' with message 'SQLSTATE[45S22]: Column not found: 1054 Unknown column 'b.id' in 'field list" in ... And finally, when I try to update a record as follows: $book_title = 'The Peloponnesian War'; $title = $book_title; $u = Doctrine::getTable('Book')-find($title); $u-title = $title; $u-save(); I get a similar error: Fatal error: Uncaught exception 'Doctrine_Connection_Mysql_Exception' with message 'SQLSTATE[42S22]: Column not found: 1054 Unknown column 'b.id' in 'field list''in ... Here is my Doctrine_Record model: class Book extends Doctrine_Record{ public function setTableDefinition() { $this->hasColumn('book_id'); $this->hasColumn('isbn10','varchar',20); $this->hasColumn('isbn13','varchar',20); $this->hasColumn('title','varchar',100); $this->hasColumn('edition','varchar',20); $this->hasColumn('author_f_name','varchar',20); $this->hasColumn('author_m_name','varchar',20); $this->hasColumn('author_l_name','varchar',20); $this->hasColumn('cond','enum',null, array('values' => array('as new','very good','good','fair','poor'))); $this->hasColumn('price','decimal',8, array('scale' =>2)); $this->hasColumn('genre','varchar',20); } public function setUp() { $this->setTableName('Book'); //$this->actAs('Timestampable'); } Any assistance will be really appreciated. Thanks in advance.

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• USE case to Class Diagram - How do I?

  - by 01010011

  Hi, I would like your guidance on how to create classes and their relationships (generalization, association, aggregation and composition) accurately from my USE case diagram (please see below). I am trying to create this class diagram so I can use it to create a simple online PHP application that allows the user to register an account, login and logout, and store, search and retrieve data from a MySQL database. Are my classes correct? Or should I create more classes? And if so, what classes are missing? What relationships should I use when connecting the register, login, logout, search_database and add_to_database to the users? I'm new to design patterns and UML class diagrams but from my understanding, the association relationship relates one object with another object; the aggregation relationship is a special kind of association that allows "a part" to belong to more than one "whole" (e.g. a credit card and its PIN - the PIN class can also be used in a debit card class); and a composition relationship is a special form of aggregation that allows each part to belong to only one whole at a time. I feel like I have left out some classes or something because I just can't seem to find the relationships from my understanding of relationships. Any assistance will be really appreciated. Thanks in advance. USE CASE DIAGRAM CLASS DIAGRAM

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• CodeIgniter's Scaffolding and Helper Functions Not Working

  - by 01010011

  Hi, I'm following CodeIgniter's tutorial "Create a blog in 20 minutes" and I am having trouble getting the helper, anchor and Scaffolding functions to work. I can't seem to create links on my HTML page using the helper and anchor functions. I put $this->load->helper('url'); $this->load->helper('form'); in the constructor under parent::Controller(); and <p>&lt;?php anchor('blog/comments','Comments'); ?&gt;</p> within the foreach loop as specified in the tutorial. But Im not getting the links to appear. Secondly, I keep getting a 404 Page Not Found error whenever I try to access CodeIgniter's Scaffolding page in my browser, like so: localhost/codeignitor/index.php/blog/scaffolding/mysecretword I can access localhost/codeignitor/index.php/blog just fine. I followed CodeIgnitor's instructions in their "Create a blog in 20 minutes" by storing my database settings in the database.php file; and automatically connecting to the database by inserting "database" in the core array of the autoload.php; and I've added both parent::Controller(); and $this->load->scaffolding('myTableName') to blog's constructor. It still gives me this 404. Any assistance will be appreciated. Thanks in advance?

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• CodeIgniter Unable to Access Error Message Error

  - by 01010011

  Hi, I've created this registration form for registering new users to a website using CodeIgniter. My problem is, whenever I enter a username that already exists in my database, instead of giving me my error message which explains this to the user, it instead gives me this error message: Unable to access an error message corresponding to your field name Here are snippets of the code from my controller. Any assistance will be appreciated: function register() $this->load->library('form_validation'); $this->form_validation->set_rules('username', 'Username','trim|required|alpha_numeric|min_length[6]|xss_clean|strtolower|callback_username_not_exists); ... } function username_not_exists($username) { $this->form_validation->set_message('username','That %s already exists.'); if($this->User_model->check_exists_username($username)) { return false; } else { return true; }

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• MySQl Error #1064

  - by 01010011

  Hi, I keep getting this error: MySQL said: #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 'INSERT INTO books.book(isbn10,isbn13,title,edition,author_f_name,author_m_na' at line 15 with this query: USE books; DROP TABLE IF EXISTS book; CREATE TABLE `books`.`book`( `book_id` INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY, `isbn10` VARCHAR(15) NOT NULL, `isbn13` VARCHAR(15) NOT NULL, `title` VARCHAR(50) NOT NULL, `edition` VARCHAR(50) NOT NULL, `author_f_name` VARCHAR(50) NOT NULL, `author_m_name` VARCHAR(50) NOT NULL, `author_l_name` VARCHAR(50) NOT NULL, `cond` ENUM('as new','very good','good','fair','poor') NOT NULL, `price` DECIMAL(8,2) NOT NULL, `genre` VARCHAR(50) NOT NULL, `quantity` INT NOT NULL) INSERT INTO books.book(isbn10,isbn13,title,edition,author_f_name,author_m_name,author_l_name,cond,price,genre,quantity)** VALUES ('0136061699','978-0136061694','Software Engineering: Theory and Practice','4','Shari','Lawrence','Pfleeger','very good','50','Computing','2'); Any idea what the problem is?

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• CodeIgniter's form_hidden

  - by 01010011

  Hi, How do I achieve this: //view.php <input type="hidden" name="Name" value="<?php echo $variable->id; ?> using CodeIgniter’s form_hidden helper: //view.php <?php echo form_hidden('Name','<?php echo $variable->id;') ?> The first one works fine when I display $variable-id but CI's form_hidden doesn't work.

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• CodeIgniter Authentication Library for Beginners?

  - by 01010011

  Hi, Can anyone recommend a great CodeIgniter Registration/Login/Logout library for beginners to web development? I read on stackoverflow that Redux was highly recommended at one time. What is the current recommendation?

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• UML Class Relationships

  - by 01010011

  Hi, I would like to confirm whether I am on the right track when identifying common UML class relationships. For example, is the relationship between: 1 a stackoverflow member and his/her stackoverflow user account categorized as a composition relationship or an aggregation relationship? At first I thought it was an association because this member "has a" account. However on second thought, I am thinking its composition because each "part" (user account) belongs to only one whole (user) at a time, meaning for as long as I am logged into stackoverflow, I have to use this one and only account until I log off. If I log back onto stackoverflow with a different account then its composition again. Do you agree? 2 a database and a person's user account an aggregation relationship? I think so because 1 database (the whole) can store 0...* number of user accounts (the parts) but another database can store the same user accounts. Finally, can anyone recommend a website that specializes in designing code using UML? Thanks in advance

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• Count Records Returned MySQL Doctrine

  - by 01010011

  Hi, How do I check the number of records returned from a search of my MySQL database with a statement like this: $searchKey = 'Something to search for'; $searchResults = Doctrine::getTable('TableName')->createQuery('t')- >where('columnName LIKE ?','%'.$searchKey.'%')->execute();

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• HTML5 svg not working

  - by 01010011

  Hi, I'm using Chrome version 5.0.375.55 and Firefox version 3.5.9 but I can't get the HTML5 code below to display a box. <!DOCTYPE html> <!-- this tells browser, this is HTML5 --> <html> <body> <svg width="200" height="200"> <rect x="0" y="0" width="100" height="100" fill="blue" stroke="red" stroke-width="5px" rx="8" ry="8" id="myRect" class="chart" /> </svg> </body> </html> The following sites stated that my browsers support HTML5 and svg so what gives? http://caniuse.com/ http://www.html5test.com/

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• CodeIgniter's Scaffolding not working

  - by 01010011

  Hi, I keep getting a 404 Page Not Found whenever I try to access CodeIgniter's Scaffolding page in my browser, like so: localhost/codeignitor/index.php/blog/scaffolding/mysecretword I can access localhost/codeignitor/index.php/blog just fine. I followed CodeIgnitor's instructions in their "Create a blog in 20 minutes" by storing my database settings in the database.php file; and automatically connecting to the database by inserting "database" in the core array of the autoload.php; and I've added both parent::Controller(); and $this-load-scaffolding('myTableName') to blog's constructor. It still gives me this 404. Any suggestions?

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• Doctrine Default Primary Key Problem (Again)

  - by 01010011

  Hi, Should I change all of my uniquely-named MySQL database primary keys to 'id' to avoid getting errors related to Doctrine's default primary key set in the plugin 'doctrine_pi.php'? To further elaborate, I am getting the following reoccurring error, this time after trying to login to my login page: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'u.book_id' in 'field list'' in... I suspect the problem resides at a MySQL table used for my login, of which has a primary key called id Marc B originally solved an identical problem for me in this post http://stackoverflow.com/questions/2702229/doctrine-codeigniter-mysql-crud-errors when I had the same problem with a different table within the same database. Following his suggestion, I changed the default primary key located at system/application/plugins/doctrine_pi.php from 'id' to 'book_id': <?php // system/application/plugins/doctrine_pi.php ... // set the default primary key to be named 'id', integer, 4 bytes Doctrine_Manager::getInstance()->setAttribute( Doctrine::ATTR_DEFAULT_IDENTIFIER_OPTIONS, array('name' => 'book_id', 'type' => 'integer', 'length' => 4)); and that solved my previous problem. However, my login page stopped working. So what is the safe thing to do? Change all of my primary keys to 'id' (will that solve the problem without causing some other problem I am not aware of). Or should I add some lines of code in doctrine_pi.php?

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• Passing Info From One View to Another

  - by 01010011

  EDIT 1. The results of a query from my database are displayed on my view page. Next to each unique result is a button: // first_view.php <?php echo form_open('controller/method'); ?> <?php foreach($results_found as $item): ?> <p><?php echo $item->name ?></p> <p><?php form_submit('buy','Buy This'); ?></p> When a user clicks on one of these buttons (lets say button 4), I would like to display the user's choice from the array onto another view. I've tried using this: // first_view.php <?php echo $this->session->set_userdata('choice',$item); ?> immediately before // first_view.php <?php echo form_close(); ?> thinking that the user's final choice would be stored there so I can display it on another view like this: // second_controller.php $c = $this->session->userdata('choice'); // second_view.php echo 'Your Choose: '. $c; However, what actually displays is the last result displayed on first_view.php, not what the user choose. My question is, if the user clicked on button 4, how do I get that particular choice displayed onto another view? END OF EDIT1 ORIGINAL QUESTION My view page (CodeIgniter) displays the contents of an array in the form of several links. If a user clicks on a link, I want the user to be taken to another view page (via a controller) that gives more information concerning that particular link (which is stored in that particular position in the array) My question is, how do I access and display the contents of that particular location within the array on the second view file without getting a "Undefined variable error" on the second view? Here is what I am trying to do in code form // my_controller.php $array_name['variable_name']; $this->load->view('first_view.php'); // first_view.php <?php foreach($variable_name as $vn): ?> <?php echo anchor('controller_name' $vn->info ?> // if user clicks on 3rd link <?php endforeach ?> // I want to beable to access // index 2 of the array in the // second view file so I can // display more info // second_view.php <?php $vn[2]->info ?>

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• Error in SQL Syntax ERROR1064

  - by 01010011

  Hi, Everytime I try to create the following table in MySQL command line: CREATE TABLE book(book_id int NOT NULL AUTO_INCREMENT PRIMARY KEY, isbn char(20), title char(20), author_f_name char(20), author_l_name char(20), condition ENUM("as new","very good","good","fair","poor"), price decimal(8,2), genre char(20)); I keep getting this error message: ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL version for the right syntax to use near 'condition ENUM("as new","very good","good","fair","poor"), price decimal(8,2), g' at line 6 I've tried using single quotes and double quotes for the ENUM options. Where did I go wrong?

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• PHPSecurityScanner & SpikePHPSecAudit - How do I?

  - by 01010011

  Hi, How do I run PHP Security Scanner and SpikePHPSecAudit? I've already extracted them at the root of my website and thought it could be run like phpSecInfo where you just navigate to www.mySite.com/phpsecinfo/index.php Any assistance will be appreciated. ps I am using Windows XP and XAMPP

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• MySQL Password Not Working

  - by 01010011

  Hi, When I enter my password into MySQL Command Line Client, it rejects my password, gives 1 beep and closes the window. Can anyone provide me with a basic troubleshooting list of steps (from simple to progressively more technical) to regain entry into MySQL once and for all. Also, can anyone tell me what can cause my password to suddenly be rejected? I am also running XAMPP and the MySQL service is both installed and running. Thanks in advance. I remember when I first got this problem some months ago, if I recall correctly, I solved it by starting the MySQL service from Windows XP's Administrative Tool (something I did stopped the MySQL service and to this day I don't know what it was). Now the problem has started back again but this time, when I checked Services, the MySQL service was already 'started'. Any assistance will be appreciated. Thanks

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• Uploading CI Website

  - by 01010011

  Hi, This is the first time I’m building a website and using CodeIgniter and I was wondering whether you have any tips on uploading CI to a free web host. Can I just upload the entire CI folder? Or do I have to upload individual files (God no!)? What are my options? What about my MySQL database - do I just upload my mysqldump to the webhost? Also, can you recommend a good free webhost. I was thinking about 000webhost. Any other suggestions will be more than welcome. Thanks!

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• SQL Syntax Error 1064

  - by 01010011

  Hi, I keep getting the following error message ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right systax to use near ''isbn10','isbn13','title','edition','author_f_name','author_m_name','author_l_na' at line 1 when trying to populate my MySQL database from the command line with the following command: source C:\myFilePath\myFileName.sql Here is an excerpt from my mysqldump (showing the table structure for book). Where did I go wrong? Any assistance will be appreciated: -- -- Table structure for table book DROP TABLE IF EXISTS book; /*!40101 SET @saved_cs_client = @@character_set_client /; /!40101 SET character_set_client = utf8 */; CREATE TABLE book ( book_id int(11) NOT NULL AUTO_INCREMENT, isbn10 char(20) DEFAULT NULL, isbn13 char(20) DEFAULT NULL, title char(20) DEFAULT NULL, edition char(20) DEFAULT NULL, author_f_name char(20) DEFAULT NULL, author_m_name char(20) DEFAULT NULL, author_l_name char(20) DEFAULT NULL, cond enum('as new','very good','good','fair','poor') DEFAULT NULL, price decimal(8,2) DEFAULT NULL, genre char(20) DEFAULT NULL, PRIMARY KEY (book_id) ) ENGINE=MyISAM DEFAULT CHARSET=latin1; /*!40101 SET character_set_client = @saved_cs_client */; -- -- Dumping data for table book

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• Blackberry/iPhone Development Sim on Windows

  - by 01010011

  Hi, I'm looking for a free solution just to learn how to develop applications for Blackberry and iPhone on Windows that includes a Blackberry and iPhone simulator. The closest program I came across to this was DragonFireSDK but its not free. Any suggestions?

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