Cakephp join：从两个表中检索数据

Question

i am working on a Cakephp 2.x.. i have two tables into my db name Images and Audios .. both have a userid field ... first what i want is i want to retrive the data and sort by date field which is in both the tables.. on my view page i want to show the images and videos by date...

for example image1,video2,video3,image2

right now i am doing this in Images model

   function getImagesAndAudio($userid){
       $this->bindModel(array(
        'belongsTo' => array(
            'Audio' => array(
                'className' => 'Audio',
                'foreignKey' => false,
                'conditions' => array(

                    'Image.user_id = Audio.user_id',




                ),
                'type' => 'LEFT',

                'order'=>'Image.date',//error


            )
        )
    ), false);

    return $this->find('all', array('conditions' => array('Image.User_id' => $userid),
        'contain' => array('Audio' ),
        ));

the problem is right now it gives me error if i do this

      'order'=>'date',

and the other thing i dont know how can i show the images and videos in mixup order... i mean when i do this

   foreach ($datas as $data){
     echo $data['Image']['filename'];
     echo $data['Audio']['filename'];

      }

the problem is i have to manually write the code that show image now and now show the audio... i want to show the file with respect to date .. so if there is an image after the two audios file show an image and then so on

Answer 1

If I understand you correct i think this approach should work:

Define the proper Relations between Image/Audio and User (User hasMany Audio/Image and Audio/Image belongsTo User)

You can define how the related Images/Audio Files should be ordered (see CakePHP Documentation: http://book.cakephp.org/2.0/en/models/associations-linking-models-together.html#hasmany)

Now in your controller just set the user for your view and you can retrieve your data with

<?php
//Same goes for Image
foreach($userVariable[‘Audio‘] as $audio) {
    echo $audio['filename‘]
}
?>