Php SQL搜索城市id中的用户和用户状态

我不知道这个标题是否有意义，但情况如下：见下表4

用户表（有200多个用户）

城市表

+---------+-------+----------+
| idcity  | city  | idstate  |
+---------+-------+----------+
|    123  | city1 |     1    |
|    542  | city2 |     2    |
|    623  | city3 |     3    |
+---------+-------+----------+

+---------+--------+------------+
| idstate | state  | idcountry  |
+---------+--------+------------+
|     1   | state1 |      4     |
|     2   | state2 |      5     |
|     3   | state3 |      6     |
+---------+--------+------------+

    +-----------+----------+
    | idcountry |  country |
    +-----------+----------+
    |     4     | country1 |
    |     5     | country2 |
    |     6     | country3 |
    +-----------+----------+

状态表

+---------+-------+----------+
| idcity  | city  | idstate  |
+---------+-------+----------+
|    123  | city1 |     1    |
|    542  | city2 |     2    |
|    623  | city3 |     3    |
+---------+-------+----------+

+---------+--------+------------+
| idstate | state  | idcountry  |
+---------+--------+------------+
|     1   | state1 |      4     |
|     2   | state2 |      5     |
|     3   | state3 |      6     |
+---------+--------+------------+

    +-----------+----------+
    | idcountry |  country |
    +-----------+----------+
    |     4     | country1 |
    |     5     | country2 |
    |     6     | country3 |
    +-----------+----------+

国家/地区表

+---------+-------+----------+
| idcity  | city  | idstate  |
+---------+-------+----------+
|    123  | city1 |     1    |
|    542  | city2 |     2    |
|    623  | city3 |     3    |
+---------+-------+----------+

+---------+--------+------------+
| idstate | state  | idcountry  |
+---------+--------+------------+
|     1   | state1 |      4     |
|     2   | state2 |      5     |
|     3   | state3 |      6     |
+---------+--------+------------+

    +-----------+----------+
    | idcountry |  country |
    +-----------+----------+
    |     4     | country1 |
    |     5     | country2 |
    |     6     | country3 |
    +-----------+----------+

注意=我需要找出属于

州的所有人

并通过

idcity

输出

+--------+----------+-------------------+---------+---------+
| iduser |   user   |      address      |  idcity |  state  |
+--------+----------+-------------------+---------+---------+
|   11   | person11 | some address      |     123 |  state1 |
|    2   | person2  | another address   |     123 |  state1 |
|   13   | person13 | different address |     123 |  state1 |
+--------+----------+-------------------+---------+---------+

或

如何通过一个连接实现这一点？我尝试在一个连接（在查询中）中组合这两个表，但感觉不太对劲。请帮助我如何修复此pblm。谢谢

像这样试试

select u.iduser,u.user,u.address,u.idcity,s.state 
From usertable u
inner join city c on c.idcity=u.idcity
inner join state s on s.idstate=c.idstate

获取城市身份证

select 
 c.idcity 
from city c 
where c.idcity = <someidcity>

选择
c、 城市
来自c市
其中c.idcity=

加入州名

select 
  c.idcity, 
  s.state
from city c
join state s on s.idstate = c.idstate
where c.idcity = <someidcity>

选择
c、 艾德城，
s、 陈述
来自c市
在s.idstate=c.idstate上连接状态s
其中c.idcity=

然后加入用户

select 
  u.iduser,
  u.user,
  u.address,
  c.idcity, 
  s.state
from city c
join state s on s.idstate = c.idstate
join user u on u.idcity = c.idcity
where c.idcity = <someidcity>

选择
u、 iduser，
u、 用户，
u、 地址:，
c、 艾德城，
s、 陈述
来自c市
在s.idstate=c.idstate上连接状态s
在u.idcity=c.idcity上加入用户u
其中c.idcity=

您不能在单个联接中执行此操作，因为您需要来自表用户和表状态的信息（ID除外），它们之间没有直接关系。

试试这个

SELECT
    a.iduser,
    a.user,
    a.address,
    b.idcity,
    c.state
FROM
    users a
INNER JOIN
    cities b
ON
    a.idcity=b.idcity
INNER JOIN
    states c
ON
    b.idstate=c.idstate
WHERE
    c.state='myState'
AND b.city='myCity'

---

您必须将城市和州表与用户表连接，使用单个连接这是不可能的，但与按idcity搜索类似，例如：-从用户表u内部连接中选择u.iduser、u.user、u.address、u.idcity、s.state。。。。。。。。。。。。。。。。其中，idcity=“542”操作请求结果为单个joinOP请求结果为单个joinOP请求join@SandipPatel是的，但他也显示了期望的输出。。。