【发布时间】:2017-09-07 09:38:00
【问题描述】:
我有三个表,我想计算每个表中每个度假村的记录数。我得到了一个我无法解释的意外结果。
我的表格如下:
CREATE TABLE `game_items` (
`id_items` int(11) NOT NULL,
`id_resort` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `game_items` (`id_items`, `id_resort`) VALUES
(36, 81),
(38, 81),
(39, 67);
CREATE TABLE `game_slopes` (
`id_slopes` int(11) NOT NULL,
`id_resort` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `game_slopes` (`id_slopes`, `id_resort`) VALUES
(16, 81);
CREATE TABLE `game_staff` (
`id_staff` int(11) NOT NULL,
`id_resort` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `game_staff` (`id_staff`, `id_resort`) VALUES
(1, 69),
(3, 67),
(5, 81),
(7, 81),
(8, 81),
(12, 81);
CREATE TABLE `game_resorts` (
`id_resort` int(11) NOT NULL,
`id_player` int(11) DEFAULT NULL,
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `game_resorts` (`id_resort`, `id_player`) VALUES
(66, 59),
(67, 60),
(68, 61),
(69, 62),
(70, 63),
(81, 67),
(82, 68);
我的查询:
SELECT `game_players_tbl`.`id_player`, `game_resorts`.`id_resort`,
COUNT(game_items_tbl.id_items) as item_count,
COUNT(game_slopes_tbl.id_slopes) as slope_count,
COUNT(game_staff_tbl.id_staff) as staff_count
FROM `game_resorts`
INNER JOIN `game_players` as `game_players_tbl` ON `game_resorts`.`id_player` = `game_players_tbl`.`id_player`
LEFT OUTER JOIN `game_items` as `game_items_tbl` ON `game_resorts`.`id_resort` = `game_items_tbl`.`id_resort`
LEFT OUTER JOIN `game_slopes` as `game_slopes_tbl` ON `game_resorts`.`id_resort` = `game_slopes_tbl`.`id_resort`
LEFT OUTER JOIN `game_staff` as `game_staff_tbl` ON `game_resorts`.`id_resort` =`game_staff_tbl`.`id_resort`
GROUP BY `game_resorts`.`id_resort`
ORDER BY `game_resorts`.`reputation` DESC
结果是:
id_player id_resort item_count slope_count staff_count
61 68 0 0 0
63 70 0 0 0
67 81 8 8 8
68 82 0 0 0
62 69 0 0 1
59 66 0 0 0
60 67 1 0 1
但我希望:
id_player id_resort item_count slope_count staff_count
61 68 0 0 0
63 70 0 0 0
67 81 2 1 4
68 82 0 0 0
62 69 0 0 1
59 66 0 0 0
60 67 1 0 1
我不明白为什么我在度假村 ID 81 的每个计数中都得到 8。我尝试了不同的替代方案,但从未得到正确的结果。
编辑:添加了 game_resorts
【问题讨论】:
-
可以添加game_resorts的数据吗?
-
测试它,我会添加它作为答案。当然,解释原因
-
不幸的是,我得到与您的更改相同的结果。我在原始帖子中添加了我的表格
-
现在我看到数据我知道问题了。我现在很忙...如果没有人回答你我会...底线,你必须计算每张桌子,你现在的做法是只计算度假村的注册表
-
@JorgeCampos 感谢您的提示!
标签: mysql codeigniter join