Question

这是行动网址

http://localhost/carsdirectory/cars/home

cars_controller.php（控制器）

public function home(){

    $this->set('latest_cars', $this->Car->find('all', array(
        'order' => array(
            'Car.modified DESC',
            'Car.created Desc'              
        ),
        'limit' => '3'
     )));

    $this->set('galleries', $this->Gallery->find('all'));


 }

car.php（模型）

  public $hasMany = array(
    'Gallery' => array(
        'className' => 'Gallery',
        'foreignKey' => 'car_id',
        'dependent' => true
    )
);

gallery.php（模型）

    var $belongsTo = array(
        'Car' => array(
            'className' => 'Car',
            'foreignKey' => 'car_id',
)
   );

home.ctp（视图）

<?php foreach($latest_cars as $latest_car){ ?>

  <img src="img/car-listings.jpg" />     // now it's static

     <h4><?php echo $latest_car['Car']['car_name']; ?></h4>  // it's dynamic it's coming car table 

     <span>$<?php echo $latest_car['Car']['car_price']; ?></span>  // it's dynamic it's coming car table 


<?php } ?>

我已经替换了该行

 <img src="img/car-listings.jpg" />

用那一行

  <?php $this->Html->image('/media/filter/small/'.$latest_cars['Gallery']['dirname'].'/'.$latest_cars['Gallery']['basename']);?>

但我得到了那个错误

未定义索引：图库[APP \ views \ cars \ home.ctp，第226行]

 <img src="img/car-listings.jpg" />  this line i want to make dynamic , so my question how to use join in cars_controller or any other idea and i want to fetch data from galleries table

这是画廊表结构

id - 1

basename - chrysanthemum_10.jpg

car_id - 1

提前致谢

Answer 1

因此，您拥有Car模型和Gallery模型。由于Gallery具有car_id属性，因此可以用它来形成这些CakePHP Associations：

图库属于汽车

和

Car hasOne Gallery

您可以选择实际需要的关联，并在模型中定义它们。在您的情况下，您想要在查询汽车时显示汽车的画廊，所以：

// in car.php

var $hasOne = 'Gallery';

然后，您可以选择是否要使用Containable来控制哪些关联包含在查询中，或者只使用recursive来包含所有关联：

// in cars_controller.php

$this->set('latest_cars', $this->Car->find('all', array(
    'recursive' => 1,
    'order' => array(
        'Car.modified DESC',
        'Car.created Desc'              
    ),
    'limit' => '3'
 )));

然后在您的视图中，使用$latest_car['Car']访问汽车属性，使用$latest_car['Gallery']访问图库属性

修改

如果Car有多个Gallery，那么您应该期待这种结构：

[0] => Car => (first car) Gallery => [0] => (first gallery of first car) [1] => (second gallery of first car) [1] => Car => (second car) Gallery => [0] => (first gallery of second car) etc.

以便在您的视图中访问它：

<?php foreach($latest_cars as $latest_car){ foreach ($latest_car['Gallery'] as $gallery) echo $this->Html->image('/media/filter/small/'.$gallery['dirname'].'/'.$gallery['basename']); ?>

Answer 2

在Controller find中使用join，根据需要指定字段

$results= $this->Car->find('all',
array('fields'=> array('Car.*','galleries.*'), 
'joins'=> array( 
array('table'=>'galleries', 'type'=>'inner',
'conditions'=>array('Car.car_id=galleries.car_id'))
))
)

Answer 3

在控制器文件中添加此行

  $this->loadModel('Table');//table is your model name in singular
  $this->set('table', $this->Table->find('all'));

你可以直接在视图中使用$ table，如果你与table有关系，你可以使用CakePHP提供的默认关系

谢谢

如何从Cake php中的两个表中获取数据

3 个答案: