CakePHP2虚拟场问题

CakePHP2虚拟场问题,cakephp,cakephp-2.0,Cakephp,Cakephp 2.0,我正在进行我的第一个CakePHP博客项目: 如您所见,有一个页脚部分显示了前25个最流行的标记 我使用了三个相关的表来实现这些常用标记: POSTS: id, title, content, slug TAGS: id, name, slug POST_TAG_LINK: id, post_id, tag_id 我试图通过$this->tag->find进行CakePHP查询,但出现了一个无法修复的持久性SQL错误。 因此,我尝试了“$this->tag->querySQL方式: debug

我正在进行我的第一个CakePHP博客项目: 如您所见,有一个页脚部分显示了前25个最流行的标记

我使用了三个相关的表来实现这些常用标记:

POSTS: id, title, content, slug
TAGS: id, name, slug
POST_TAG_LINK: id, post_id, tag_id
我试图通过
$this->tag->find
进行CakePHP查询,但出现了一个无法修复的持久性SQL错误。 因此,我尝试了“
$this->tag->query
SQL方式:

debug($this->Tag->query(
    "SELECT
        Tag.name,
        COUNT(PostTagLink.id) AS count
    FROM
        tags AS Tag
    INNER JOIN
        post_tag_links AS PostTagLink
    ON
        tag.id = PostTagLink.tag_id
    WHERE
        Tag.show = 'Y'
    GROUP BY
        Tag.name
    ORDER BY
        Tag.name ASC"
));
问题是输出数组不是很好:

array(
    (int) 0 => array(
        'Tag' => array(
            'name' => 'Beauty'
        ),
        (int) 0 => array(
            'count' => '2'
        )
    ),
    (int) 1 => array(
        'Tag' => array(
            'name' => 'Koken'
        ),
        (int) 0 => array(
            'count' => '1'
        )
    ),
    (int) 2 => array(
        'Tag' => array(
            'name' => 'Lente'
        ),
        (int) 0 => array(
            'count' => '2'
        )
    ),
    (int) 3 => array(
        'Tag' => array(
            'name' => 'Wonen'
        ),
        (int) 0 => array(
            'count' => '4'
        )
    )
)
我想要这样的东西:

array(
    (int) 0 => array(
        'Tag' => array(
            'name' => 'Beauty',
            'count' => '2'
        )
    ),
    (int) 1 => array(
        'Tag' => array(
            'name' => 'Koken',
            'count' => '1'
        )
    ),
    (int) 2 => array(
        'Tag' => array(
            'name' => 'Lente',
            'count' => '2'
        )
    ),
    (int) 3 => array(
        'Tag' => array(
            'name' => 'Wonen',
            'count' => '4'
        )
    )
)

是否有人对此提出了解决方案?

如果您坚持CakePHP惯例,会容易得多。您可以尝试以下方法:

  $this->Tag->virtualFields['count'] = "SELECT COUNT(PostTagLink.id) FROM
  post_tag_links AS PostTagLink WHERE PostTagLink.tag_id = Tag.id";

  $this->Tag->recursive = -1; // If you're using join, keep the recursive to minimum in order to keep it optimized.

  $tags = $this->Tag->find("all", 
            array(
                "fields"     =>  array("Tag.name", "Tag.count"),
                "conditions" =>  array("Tag.show" => 'Y'),
                "order"      =>  array("Tag.name ASC"),
                "group"      =>  array("Tag.name")
            )
        );
当您想要创建自定义字段(例如“count”)时,virtualFields的概念就应运而生

希望这有帮助


和平!xD

也祝你和平:D.我会试试的,谢谢你的帮助;)