Question

正如您在此sqlfiddle上看到的，我有这个架构：

CREATE TABLE reviews
    (`id` int(11) NOT NULL AUTO_INCREMENT, 
     `shop_id` int(11), 
     `order_id` char(255), 
     `product_id` char(32), 
     `review_time` int(11),  
     PRIMARY KEY (`id`)
    )
;
INSERT INTO reviews
    (`shop_id`, `order_id`, `product_id`, `review_time`)
VALUES
    ('10', '100', '1000', '1466190000'),
    ('10', '100', '1000', '1466276400'),
    ('10', '100', '1000', '1466462800'),
    ('20', '800', '8000', '1466249200')
;

CREATE TABLE tags
    (`id` int(11) NOT NULL AUTO_INCREMENT, 
     `shop_id` int(11), 
     `order_id` char(255), 
     `product_id` char(32), 
     `tag_time` INT(11) NULL,  
     PRIMARY KEY (`id`)
    )
;
INSERT INTO tags
    (`shop_id`, `order_id`, `product_id`, `tag_time`)
VALUES
    ('10', '100', '1000', '1466449200'),
    ('10', '100', '1000', NULL),
    ('10', '100', '3000', NULL),
    ('20', '800', '8000', '1469449200')
;

我需要按日期显示统计数据，显示每个日期有多少评论，有多少评论以及有多少评论没有。我正在使用此查询：

SELECT
  DATE_FORMAT(FROM_UNIXTIME(r.`review_time`), "%d.%m.%Y") AS review_submited_on,
  r.`shop_id`,
  COUNT(*) as total_orders,
  COUNT(*) as tagged_orders 
FROM
  reviews AS r
LEFT JOIN tags as t 
  ON r.`shop_id` = t.`shop_id` AND 
  r.`order_id` = t.`order_id` AND 
  r.`product_id` = t.`product_id` 
WHERE
    t.`tag_time` IS NOT NULL    
GROUP BY r.`shop_id`, r.`order_id`, r.`product_id`  
ORDER BY review_submited_on ASC

更新预期结果如下：

| review_submited_on | shop_id | total_orders | tagged_orders |
|--------------------|---------|--------------|---------------|
|         17.06.2016 |      10 |            3 |             1 |
|         18.06.2016 |      20 |            1 |             1 |

我为演示创建了此sqlfiddle。感谢您的帮助:)）

Answer 1

试试这个，如果你想要的话，请告诉我。

SELECT review_submited_on, shop_id, total_orders, IFNULL(tagged_orders, 0) tagged_orders
FROM
    (SELECT shop_id, COUNT(DISTINCT shop_id, order_id, product_id) total_orders, DATE_FORMAT(FROM_UNIXTIME(review_time), "%d.%m.%Y") AS review_submited_on
    FROM reviews
    GROUP BY shop_id) review_counter
    LEFT JOIN
    (SELECT shop_id, COUNT(DISTINCT shop_id, order_id, product_id) tagged_orders
    FROM tags 
    WHERE tag_time IS NOT NULL
    GROUP BY shop_id) tag_counter
USING (shop_id)

结果

| review_submited_on | shop_id | total_orders | tagged_orders |
|--------------------|---------|--------------|---------------|
|         17.06.2016 |      10 |            1 |             1 |
|         18.06.2016 |      20 |            1 |             1 |

MySQL左连接分组和日期

1 个答案: