我正在开发我的第一个CakePHP博客项目:http://kattenbelletjes.be/ 正如您所看到的,有一个页脚部分显示了前25个最受欢迎的标签。
我使用三个相关表来实现那些流行的标签:
void add(float*, float*, float*);
int main()
{
float arr1[SIZE] = {1,2,3,4};
float arr2[SIZE] = {4,3,2,1};
float arr3[SIZE];
add(arr3, arr2, arr1);
for(int q=0;q<SIZE;q++){
cout<<*(arr3+q);
}
return 0;
}
void add(float* ptr, float* ptr1, float* ptr2){
*ptr++ = *(ptr1++) + *(ptr2++);
}
我试图通过POSTS: id, title, content, slug
TAGS: id, name, slug
POST_TAG_LINK: id, post_id, tag_id
进行CakePHP查询,但是有一个持续的SQL错误我无法解决。
所以,我尝试了#34; $this->tag->find
SQL方式:
$this->tag->query
问题是输出数组不是很好:
debug($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"
));
我想要这样的东西:
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'
)
)
)
是否有人对此有解决方案?
答案 0 :(得分:0)
如果你坚持CakePHP惯例,那就容易多了。你可以尝试这样的事情:
$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的概念就会受到拯救。例如。
希望这有帮助。
和平!的xD