Question

当获得单个记录和相关记录（两个级别下降）时，我们是否可以将条件传递给CakePHP 3中包含的记录？在这种情况下，例如：

$user = $this->Users->get($this->Auth->user('id'), [
    'contain' => [
        'Articles' => ['Comments']
    ]
]);

如果尝试了一些回调方法，但这似乎只能在第一级工作。像这样，例如：

$user = $this->Users->get($this->Auth->user('id'), [
    'contain' => [
        'Articles' => function ($q) {
            return $q->where(['Articles.published' => true]);
        }
    ]
]);

有什么想法吗？

Answer 1

你应该能够做到：

$user = $this->Users->get($this->Auth->user('id'), [
     'contain' => [
        'Articles' => function ($q) {
            return $q->where(['Articles.published' => true]);
        },
        'Articles.Comments' => function ($q) {
            return $q->where(['Comments.deleted' => false]);
        }
    ]
]);

或

$user = $this->Users->get($this->Auth->user('id'), [
     'contain' => [
        'Articles' => function ($q) {
            $q->contain([
                 'Comments' => function ($q) {
                    return $q->where(['Comments.deleted' => false]);
                }
            ]);
            return $q->where(['Articles.published' => true]);
        }            
    ]
]);

Answer 2

@ arilia的选项应该可以正常工作。我的偏好是

$user = $this->Users->get($this->Auth->user('id'), [
    'contain' => [
        'Articles' => [
            'queryBuilder' => function ($q) {
                return $q->where(['Articles.published' => true]);
            },
            'Comments' => [
                'querybuilder' => function ($q) {
                    return $q->where(['Comments.deleted' => false]);
                }
            ],
        ],
    ],
]);

（试图将此作为评论发布，而不是单独的答案，但它不会在那里正确格式化代码。）

如何在get（）调用中将条件传递给“contains”？

2 个答案: