在 Codeigniter 中使用 $this->db->like(） 会返回错误/缺失的结果

我有一个现有的SQL查询，可以按照我想要的方式完美运行：

$this->db->select('places.*, category.*')
            ->select('COUNT(places_reviews.place_id) AS num_reviews')
            ->select('(places_popularity.rating_1 + 2*places_popularity.rating_2 + 3*places_popularity.rating_3 + 4*places_popularity.rating_4 + 5*places_popularity.rating_5)/(places_popularity.rating_1 + places_popularity.rating_2 + places_popularity.rating_3 + places_popularity.rating_4 + places_popularity.rating_5) AS average_rating')
            ->from('places')
            ->join('category', 'places.category_id = category.category_id')
            ->join('places_reviews', 'places_reviews.place_id = places.id', 'left')
            ->join('places_popularity', 'places_popularity.place_id = places.id', 'left')
            ->where('places.category_id', $category_id)
            ->group_by('places.id')
            ->limit($limit, $offset)
            ->order_by($sort_by, $sort_order);

但是现在我想通过在上面再添加一行来向查询添加一个 LIKE 子句以获得：

$this->db->select('places.*, category.*')
            ->select('COUNT(places_reviews.place_id) AS num_reviews')
            ->select('(places_popularity.rating_1 + 2*places_popularity.rating_2 + 3*places_popularity.rating_3 + 4*places_popularity.rating_4 + 5*places_popularity.rating_5)/(places_popularity.rating_1 + places_popularity.rating_2 + places_popularity.rating_3 + places_popularity.rating_4 + places_popularity.rating_5) AS average_rating')
            ->from('places')
            ->join('category', 'places.category_id = category.category_id')
            ->join('places_reviews', 'places_reviews.place_id = places.id', 'left')
            ->join('places_popularity', 'places_popularity.place_id = places.id', 'left')
            ->where('places.category_id', $category_id)
                            ->like('places.name', $term)
            ->group_by('places.id')
            ->limit($limit, $offset)
            ->order_by($sort_by, $sort_order);

但是，它给了我不准确的结果。例如，当我让正在搜索的字符串$term = "hong" 并且我有 3 行，其中"name"列与"hong"匹配，即。（香港咖啡厅，香港咖啡厅，拉面行），我只会得到（香港咖啡厅，香港咖啡馆）退货。现在，如果$term="香港"，我只返回一个"香港咖啡馆"，而不是两个。

另一个更让我困惑！有一行叫做"Dozo"。当 $term = 'dozo' 时，不返回任何结果！

知道为什么会这样吗？

实际生成的 SQL抱歉，它出现在 1 行中

SELECT `places`.*, `category`.*, COUNT(places_reviews.place_id) AS num_reviews, (places_popularity.rating_1 + 2*places_popularity.rating_2 + 3*places_popularity.rating_3 + 4*places_popularity.rating_4 + 5*places_popularity.rating_5)/(places_popularity.rating_1 + places_popularity.rating_2 + places_popularity.rating_3 + places_popularity.rating_4 + places_popularity.rating_5) AS average_rating FROM (`places`) JOIN `category` ON `places`.`category_id` = `category`.`category_id` LEFT JOIN `places_reviews` ON `places_reviews`.`place_id` = `places`.`id` LEFT JOIN `places_popularity` ON `places_popularity`.`place_id` = `places`.`id` WHERE `places`.`category_id` = 1 AND `places`.`name` LIKE '%Dozo%' GROUP BY `places`.`id` ORDER BY `average_rating` desc LIMIT 1, 3

更新

解决。这是一个将错误的变量传递给 LIMIT 子句的分页问题。谢谢！