我正在做我的第一个CakePHP博客项目:http://kattenbelletjes.be/正如你所看到的,有一个页脚部分显示了前25个最受欢迎的标签。
我使用了三个相关的表格来实现那些流行的标签:
POSTS: id, title, content, slug
TAGS: id, name, slug
POST_TAG_LINK: id, post_id, tag_id我试图通过$this->tag->find进行CakePHP查询,但是出现了一个我无法修复的持久性错误。因此,我在"$this->tag->query SQL way:
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'
)
)
)有没有人能解决这个问题?
发布于 2016-03-11 11:06:12
如果你坚持CakePHP的约定,那就容易多了。您可以尝试如下所示:
$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")
)
);当你想创建像"count“这样的自定义字段时,virtualFields的概念就应运而生了。
希望这能有所帮助。
安息吧!xD
https://stackoverflow.com/questions/35925551
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