这是我的mysql数据库架构:
training。*,[type.type],[voivodeship.name] 如果没有与给定训练相关的类型或省级,则应在列值中返回null。
例如:
{
"id": 1,
"name": "Example training 2",
"description": "This is a description 2",
"status": "status 2",
"registerFrom": null,
"registerTo": null,
"DateOfInsert": "2018-02-06T12:00:57.000Z",
"training types": "Example type 1,Example type 3,Example type 2"
"localizations": "loc 1, loc 3"
},
...
{
"id": 99,
"name": "Example training 99",
"description": "This is a description 99",
"status": "status 2",
"registerFrom": null,
"registerTo": null,
"DateOfInsert": "2018-02-06T12:00:57.000Z",
"training types": null,
"localizations": null
},
{
"id": 99,
"name": "Example training 99",
"description": "This is a description 99",
"status": "status 2",
"registerFrom": null,
"registerTo": null,
"DateOfInsert": "2018-02-06T12:00:57.000Z",
"training types": "Example type 9,Example type 4,Example type 2",
"localizations": "loc 56, loc 32"
},
这是我当前的查询,返回有关其本地化的信息的训练(遗憾的是它没有返回没有它们的那些),我也不知道如何修改它以返回所有类型:
SELECT `training`.*, GROUP_CONCAT(`voivodeship`.`name`) AS `Localizations`
FROM `training_localization`
INNER JOIN `training` ON (`training_localization`.`training_id` = `training`.`id`)
INNER JOIN `voivodeship` ON (`training_localization`.`voivodeship_id` = `voivodeship`.`id`)
GROUP BY `training`.`id`
我对sql不是很有经验。甚至可以使用一个查询吗?
根据戈登的回答,我提出了新的查询,看起来它正在起作用:):
SELECT t.*, GROUP_CONCAT(vs.name) AS Localizations, tw.types AS types
FROM training t
LEFT JOIN training_localization tl ON tl.training_id = t.id
LEFT JOIN voivodeship vs ON tl.voivodeship_id = vs.id
LEFT JOIN(
SELECT t.*, GROUP_CONCAT(ty.type) AS Types
FROM training t
LEFT JOIN training_type tt ON tt.training_id = t.id
LEFT JOIN type ty ON tt.type_id = ty.id
GROUP BY t.id
) tw ON tw.id = t.id
GROUP BY t.id;
答案 0 :(得分:1)
如果你想要所有的东西,请考虑“外部联接”。如果您想要所有培训,那应该是left join
s系列中的第一个表:
SELECT t.*, GROUP_CONCAT(vs.name) AS Localizations
FROM training t LEFT JOIN
training_localization tl
ON tl.training_id = t.id LEFT JOIN
voivodeship vs
ON tl.voivodeship_id = vs.id
GROUP BY t.id;
注意:
LEFT JOIN
将所有内容保存在第一个表中并匹配后续表中的行(除非使用内部联接或where子句撤消它)。