我在使用select语句时遇到问题。到目前为止我所拥有的是这个 -
SELECT COUNT(booked.desk_id),
name,
desk.desk_id,
phone,
fax,
dock,
pc
FROM desk, booked
WHERE desk.desk_id = booked.desk_id
AND booking_id >=1
AND location = "Cheltenham"
哪个输出
"12" "Desk 1" "1" "1" "0" "0" "1"
这与我想要的很接近,但桌面上还有另一张名为Desk 2的桌子,它完全无视。事实上,如果有第二台预订,那么它包含了对第一台桌子的计数......
整个表格结构如下:
table "booked"
INSERT INTO `booked` (`id`, `booking_id`, `desk_id`, `member_id`, `date_booked`) VALUES
(246, 1358121601, 1, 1, 'Monday 14th January at 4:40pm'),
(247, 1358121602, 1, 1, 'Monday 14th January at 4:40pm'),
(248, 1358121604, 1, 1, 'Monday 14th January at 4:40pm'),
(249, 1358121603, 1, 1, 'Monday 14th January at 4:40pm'),
(250, 1358121606, 1, 1, 'Monday 14th January at 4:40pm'),
(251, 1358121605, 1, 1, 'Monday 14th January at 4:40pm'),
(252, 1358121607, 2, 1, 'Monday 14th January at 4:40pm'),
(253, 1358121609, 2, 1, 'Monday 14th January at 4:40pm'),
(254, 1358121608, 2, 1, 'Monday 14th January at 4:40pm'),
(255, 1358121610, 2, 1, 'Monday 14th January at 4:40pm'),
(256, 1358121612, 2, 1, 'Monday 14th January at 4:40pm'),
(257, 1358121611, 2, 1, 'Monday 14th January at 4:40pm');
table "desk"
INSERT INTO `desk` (`location`, `desk_id`, `name`, `phone`, `fax`, `dock`, `pc`) VALUES
('Cheltenham', 1, 'Desk 1', 1, 0, 0, 1),
('Cheltenham', 2, 'Desk 2', 1, 1, 0, 1);
我需要帮助的是如何正确地构造语句,以便为每个桌面输出单独的行及其相关信息。
答案 0 :(得分:2)
您遗漏了GROUP BY
以及您的汇总功能:
SELECT COUNT(booked.desk_id),
name,
desk.desk_id,
phone,
fax,
dock,
pc
FROM desk
INNER JOIN booked
ON desk.desk_id = booked.desk_id
WHERE booking_id >=1
AND location = "Cheltenham"
GROUP BY name;
在MySQL中,您不必在选择列表中GROUP BY
所有字段,但在其他RDBMS中您必须使用:
SELECT COUNT(booked.desk_id),
name,
desk.desk_id,
phone,
fax,
dock,
pc
FROM desk
INNER JOIN booked
ON desk.desk_id = booked.desk_id
WHERE booking_id >=1
AND location = "Cheltenham"
GROUP BY name, desk.desk_id, phone, fax, dock, pc
根据您的示例数据和评论,您可以使用:
SELECT coalesce(CountDesk, 0) Total,
name,
d.desk_id,
phone,
fax,
dock,
pc
FROM desk d
LEFT JOIN
(
select COUNT(booked.desk_id) CountDesk,
desk_id
from booked
WHERE booking_id >=1
GROUP BY desk_id
) b
ON d.desk_id = b.desk_id
WHERE location = "Cheltenham"
如果你想在没有子查询的情况下这样做:
SELECT
Coalesce(count(b.desk_id), 0) Total,
name,
d.desk_id,
phone,
fax,
dock,
pc
FROM desk d
LEFT JOIN booked b
ON d.desk_id = b.desk_id
WHERE booking_id >=1
AND location = "Cheltenham"
GROUP BY name, d.desk_id, phone, fax, dock, pc ;