我有這個查詢,當我添加一個OR,返回多個行,我該如何解決這個問題?我需要另一個左連接或右連接?在選擇與多個行或
SELECT `images`.`id` as `id_image`, `images_lang`.`title` as `title`, `images_lang`.`autor` as `author`, `media_lang`.`file`, `category_lang`.`title` as `category`
FROM `images`
JOIN `images_lang` ON `images_lang`.`id_images` = `images`.`id`
JOIN `category_lang` ON `images`.`id_category` = `category_lang`.`id_category`
JOIN `media` ON `images`.`id` = `media`.`id_item`
JOIN `media_lang` ON `media`.`id` = `media_lang`.`id_media`
JOIN `relation` ON `relation`.`from_id` = `images`.`id`
JOIN `tag` ON `tag`.`id` = `relation`.`to_id`
JOIN `tag_lang` ON `tag`.`id` = `tag_lang`.`id_lang`
WHERE `media`.`table` = 'images'
AND `media_lang`.`id_lang` = '1'
AND `images_lang`.`id_lang` = '1'
AND `category_lang`.`id_lang` = '1'
AND `images`.`active` = 1
AND `relation`.`type` = 2
AND `tag_lang`.`id_lang` = '1'
AND `tag_lang`.`slug` = 'pa-am-oli'
OR `tag_lang`.`slug` = 'playa'
LIMIT 9
和笨的代碼,但我不如何添加或之後與($關鍵字是過濾數組中的元素)
$this->db->select('images.id as id_image, images_lang.title as title, images_lang.autor as author, media_lang.file, category_lang.title as category');
$this->db->from($this->table);
$this->db->join($this->table.'_lang',$this->table.'_lang.id_images = '.$this->table.'.id');
$this->db->join('category_lang',$this->table.'.id_category = category_lang.id_category');
$this->db->join('media',$this->table.'.id = media.id_item');
$this->db->join('media_lang','media.id = media_lang.id_media');
$this->db->where('media.table',$this->table);
$this->db->where('media_lang.id_lang',$id_lang);
$this->db->where($this->table.'_lang.id_lang', $id_lang);
$this->db->where('category_lang.id_lang', $id_lang);
$this->db->where('images.active', 1);
if($category){
$this->db->where('category_lang.title', $category);
}
if($keywords){
$this->db->join('relation', 'relation.from_id = '.$this->table.'.id');
$this->db->join('tag', 'tag.id = relation.to_id');
$this->db->join('tag_lang', 'tag.id = tag_lang.id_lang');
$this->db->where('relation.type', _IMAGES_2_TAGS_);
$this->db->where('tag_lang.id_lang', $id_lang);
foreach($keywords as $tag){
$this->db->or_where('tag_lang.slug', $tag);
}
}
if($from || $limit)
{
$this->db->limit((int)$limit, (int)$from);
}
$query = $this->db->get();
return $query->result_array();
如果更改條件(例如,僅「cesta」),但不是真實的,圖像不具有這些標記,則始終返回相同的圖像。用這種方法和我試過的其他方法。 – NARTONIC
對不起。我不明白這一點。 – Strawberry
我有此(實施例) ID名稱標籤 1表格中的PA-AM-OLI 2 B CESTA 圖3C灘 但是,選擇總是返回第一行 – NARTONIC