cakephp-2.1

问题 How we can write Join query using two table with where condition in cake php? 回答1: You can!! check this if it would help you $table = 'table_name'; $query['conditions'] = array($table.'.entity_id' => $entity_id, $table.'.is_active' => 1); $query['fields'] = array('creator.first_name AS cf_name', 'creator.last_name AS cl_name', 'creator.email AS c_email', 'usr.first_name', 'usr.last_name', $table.'.id AS id', $table.'.guid', $table.'.updated_date', 'usr.email AS email'); // To do joining to

问题 I am currently using CakePHP 2.1.x and what I would like to do is use one core install of cakePHP for many apps on the same domain. This is what my current structure looks like. /var/www /app/ /webroot /lib /Cake /plugins /vendors What I would like to be able to accomplish is mydomain.com/myApp1 mydomain.com/myApp2 mydomain.com/myApp3 /var/www /apps/ /myApp1 /myApp2 /myApp3 /lib /Cake /plugins /vendors in the above scenario I'm not sure where /webroot would go or if there would be multiple

问题 Hi Im trying to make a posts sections for a university project on video streaming website. I have a posts controller and model. I thought that I would create an element for the posts input box so that I could echo this out in the localhost/evolvids/uploads/watch/ section, however when I submit a post I get thrown a MISSING CONTROLLER error. This is my code for the Posts Element <div class="postsform"> <table> <tr> <td> <?php echo $this->Form->create('Posts', array('action'=>'add')); echo

问题 I am currently building a web app which has two models, Donor and Donation Models respectively. It has multiple user roles. When the staff user first registers a donor, I want him to be redirected to another form which allows him to fill in the Donation details(the donor is registered once the first donation is successful). Firs of all, should I create a donation controller, from which I would redirect the user using: return $this->redirect(array('controller'=>'donations','action'=>'add'));

问题 i am working on a Cakephp 2.3 .. here is my code class MessagesController extends AppController{ public $helpers = array('Time'); public function messages(){ $datetime= $this->Message->getDatetime($userid); //$datetime = 2013-06-14 10:28:00 $time = $this->Userinfo->getTimeZone($userid); //$time = 5.00 $finaltime = $this->Time->format('Y-m-d H:i:s', $dateTime, NULL, $time); exit(); } it is giving me Fatal error: Call to a member function format() on a non-object anybody know what i am doing

问题 I want to switch DB based on domain, selecting credentials from another DB, but I can't switch.. AppController.php // Select username, password and database based on domain $this->Company->find('first', [...]); if ($company) { // Connect to second database, droping connection from first. $dataSource = ConnectionManager::getDataSource('default'); $dataSource->config['login'] = $company['Company']['dbuser']; $dataSource->config['password'] = $company['Company']['dbpass']; $dataSource->config[

问题 How to submit a select value on click of a jquery function and save the values in database and return to the same page? I'm having this select box, This select box is not inside any form. <select name="myRating" class="rating" id="serialStar"> <option value="1">Alright</option> <option value="2">Ok</option> <option value="3">Getting Better</option> <option value="4">Awesome</option> <option value="5">Loved!</option> </select> I'm using below jQuery function to send the select value into a

问题 I can add a class ("myclass") to the link (previous or next) which does not contain a link.(current page) I just add the class following the default ones: echo $this->Paginator->prev('< ', array(), null, array('class' => 'prev disabled myClass')); And it generates something like this: <span class="prev myClass disabled ">prev</span> But, when the previous or the next links have a link, then i the class is not added and it generates something like this: <span class="prev"> <a href="/tickets

问题 I'd to use the cakephp console on EC2 by this code : sudo cd /path/to/cakephp/app && Console/cake But weirdly I got this message: No such file or directory 回答1: Solved by installing dos2unix . First you have to install dos2unix sudo yum install dos2unix or sudo apt-get install dos2unix Then run: sudo dos2unix /path/to/app/Console/cake Maybe you need also to grant chmod -R 777 app/Console 来源： https://stackoverflow.com/questions/31888769/using-console-cake-on-ec2

问题 Controller public function index(){ $this->Store->unbindModel( array('belongsTo' => array('Employee')), true ); $options=array( 'joins' => array( array( 'table' => 'Employee', 'alias' => 'Employee', 'foreignKey' => true, 'conditions'=> array('Employee.employee_store = Store.store_name') ) )); $coupons = $this->Store->find('all', $options); } Model class Store extends AppModel { var $useTable = 'store'; } Sql : SELECT `Store`.`id`, `Store`.`store_name`, `Store`.`store_address`, `Store`.`store