php - 使用连接从三个表中获取结果，其中一个表包含多个结果

Question

我有三个表，其中包含以下列和数据:

表一

id  | balance
100 | 10.00
101 | 5.00
102 | 8.00

表二

id | number
100 | 0890980980
100 | 7657657655
101 | 7657657656
102 | 1231231233

表三

id  | name      | active
100 | nameOne   | 1
101 | nameTwo   | 0
102 | namrThree | 1

现在我的查询将是

Query 1. SELECT * FROM table_one WHERE balance <= 8
Query 2. SELECT number(only first_matched_row) FROM table_two WHERE table_one.id = table_two.id
Query 3. SELECT name FROM table_three WHERE table_three.id = table_one.id AND table_three.active = 1

如何连接这三个查询并获得单个查询。 请注意，table_two 将获取多行，因此我想获取第一个匹配的行并忽略 table_two.id 匹配的其余行。

预期结果:

id  | name      | number
100 | nameOne   | 0890890890
102 | nameThree | 1231231233

已解决的答案:

Select onetwo.id, three.name, two.number from 
(Select two.id from
(SELECT id as id1 FROM table_one WHERE balance <= 8)one
inner join
table_two two on one.id1 = two.id
)onetwo
inner join
table_two two on two.id=onetwo.id
inner join
table_three three on three.id = onetwo.id AND three.active = 1 group by two.id

Answer 1

你会尝试吗:

      Select onetwo.id, three.name, onetwo.number from 
(Select two.id from
(SELECT id as id1 FROM table_one WHERE balance <= 8)one
inner join
table_two two on one.id1 = two.id
)onetwo
inner join
table_three three on three.id = onetwo.id AND three.active = 1 group by onetwo.id

您可以使用高级 RDBMS 进行更优雅的查询，但遗憾的是无法使用 MySQL。