[FIXED] How to handle Query builder with date formats

Issue

I would like to show only the events for which startdate begins as of tomorrow. How can I proceed ? For the moment it's taking h:i:s into account as well.

public function display()
{
    $now = date('Y-m-d H:i:s');

    $this->loadModel('Events');
    $events = $this->Events
        ->find()
        ->contain(['Categories', 'Venues'])
        ->limit(5)
        ->order(['events.startdate' => 'ASC'])
        ->where(['events.startdate >' => $now]);
    $this->set('upcomingEvents', $events);
}

I tried this but it's not working:

public function display()
{
    $now = date('Y-m-d');

    $this->loadModel('Events');
    $events = $this->Events
        ->find()
        ->contain(['Categories', 'Venues'])
        ->limit(5)
        ->order(['events.startdate' => 'ASC'])
        ->where(['date_format("%Y-%m-%d", events.startdate)' => $now]);
    $this->set('upcomingEvents', $events);
}

Solution

No need to format the date in the query and lose the indexing functionality, instead compare against a datetime string that uses 00:00:00 for the time part, eg the start of the day.

You can achieve this easily with date('Y-m-d 00:00:00'), however that would be today, not tomorrow, you'd need to feed date() with a timestamp like date('Y-m-d 00:00:00', strtotime('+1 day')).

You can also use the datetime functionality that ships with CakePHP. Today \Cake\Chronos\Date::tomorrow() would create a date object representing 2021-10-06 where the time part will always be 00:00:00.

->where(['Events.startdate >=' => \Cake\Chronos\Date::tomorrow()])

The ORM will automatically format the date object accordingly to the column type, so for your I suppose DATETIME column, it would end up as 2021-10-06 00:00:00 in the SQL.

See also

• Cookbook > Date & Time
• Chronos Cookbook

Answered By - ndm