Question

我有一个现有的SQL查询，可以完全按照我的意愿运行：

$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添加一行来为查询添加一个LIKE子句来获取：

$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);

然而，它给我的结果不准确。例如，当我让字符串被搜索时$ term =“hong”而我有3行，其中'name'列匹配“hong”即。（香港咖啡厅，香港咖啡厅，拉面红），我只会（香港咖啡厅，香港咖啡厅）退货。现在，如果$ term =“hong kong”，我只回来了一个'香港咖啡馆'而不是两个。

另一个让我更进一步困惑！有一排叫做“Dozo”。当$ term ='dozo'时，不会返回任何结果！

为什么会发生这种情况？

生成实际SQL 对不起，它出现在第1行

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。它是一个将错误的变量传递给LIMIT子句的分页问题。谢谢！

Answer 1

从您的实际查询中，offset从1而不是0开始，这样就会忽略第一条记录（偏移0）。

因此：

另一个让我更进一步困惑！有一排叫做“Dozo”。什么时候 $ term ='dozo'，没有结果返回！

显然不会返回任何内容。

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

1 个答案: