Re: Virtual fields problem

How about this?

$array = Hash::combine($array, '{n}Tag.name,{n}0.aantai,{n}');

On Fri, Mar 11, 2016 at 4:14 AM, Sam Clauw <sam.clauw@gmail.com> wrote:

Hi 2 all,

I'm working on my first CakePHP blog project: http://kattenbelletjes.be/

As you can see, there's a footer section that shows the top 25 most popular tags.

There are three relevant tables I use the implement those popular tags:

POSTS: id, title, content, slug

TAGS: id, name, slug

POST_TAG_LINK: id, post_id, tag_id

I tried to make a CakePHP query via $this->tag->find, but there was a persistent SQL error that I couldn't fix.

So, I tried it on the "$this->tag->query" SQL way:

debug($this->Tag->query(
    "SELECT
        Tag.name,
        COUNT(PostTagLink.id) AS aantal
    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"
));

The problem is that the output array isn't very nice:

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

I wan't something like this instead:

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'
)
)
)

Is there somebody with a solution on this?

Thanks 4 helping ;) 

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