Php 多个内部连接情况

我的情况：

表“\u客户”

表“_项目”

多对多表“\u项目\u节”

就我而言，我现在需要所有“活跃”的客户。此外，它们必须来自特定的部分。因此，作为一个例子，我希望所有活跃客户都来自“1”部分。我还找到了获取正确数据的方法，但是——这是一件奇怪的事情——只有当需要id大于1的部分时。如果我试图在第1节中获取所有活动客户，查询仍会显示所有具有第2、3、4节id的项目

质疑

而错误的结果是：

Array
(
[customer_id] => 1
[name] => customer I
[section_id] => 1
)

---

也许有人能帮我，因为我不明白。非常感谢你

根据您的描述，查询应该是：

SELECT c.customer_id, c.name, ps.section_id
FROM _customers c
  INNER JOIN _projects p
     ON p.customer_id = c.customer_id 
  INNER JOIN _project_sections ps
    ON ps.section_id = p.project_ID 
  WHERE c.active = 1 AND ps.section_id = 1
GROUP BY c.name
ORDER BY LOWER(c.name)

---

特定部分的约束在

WHERE

子句中，而不在

JOIN

中。编写此查询不需要GROUP BY子句

首先，让我们看看活跃的客户及其项目

-- All active customers 
SELECT c.customer_id, c.name
FROM customers c
WHERE c.active = 1;

customer_id    name
--
1              customer I

-- All active customers and their projects
SELECT c.customer_id, c.name, p.project_id, p.project_name
FROM customers c
INNER JOIN projects p ON (p.customer_id = c.customer_id)
WHERE c.active = 1;

customer_id  name          project_id  project_name  
--
1            customer I    1           project I
1            customer I    2           project II

-- All projects from section 1
SELECT ps.project_id, ps.section_id
FROM project_sections ps
WHERE ps.section_id = 1;

project_id    section_id
--
2             1

-- All projects from section 1, including project name
SELECT ps.project_id, p.project_name, ps.section_id
FROM project_sections ps
INNER JOIN projects p ON (p.project_id = ps.project_id)
WHERE ps.section_id = 1;

project_id    project_name    section_id
--
2             project II      1

只有一个活跃的客户，他有两个项目

-- All active customers 
SELECT c.customer_id, c.name
FROM customers c
WHERE c.active = 1;

customer_id    name
--
1              customer I

-- All active customers and their projects
SELECT c.customer_id, c.name, p.project_id, p.project_name
FROM customers c
INNER JOIN projects p ON (p.customer_id = c.customer_id)
WHERE c.active = 1;

customer_id  name          project_id  project_name  
--
1            customer I    1           project I
1            customer I    2           project II

-- All projects from section 1
SELECT ps.project_id, ps.section_id
FROM project_sections ps
WHERE ps.section_id = 1;

project_id    section_id
--
2             1

-- All projects from section 1, including project name
SELECT ps.project_id, p.project_name, ps.section_id
FROM project_sections ps
INNER JOIN projects p ON (p.project_id = ps.project_id)
WHERE ps.section_id = 1;

project_id    project_name    section_id
--
2             project II      1

现在让我们看看第1节的项目

-- All active customers 
SELECT c.customer_id, c.name
FROM customers c
WHERE c.active = 1;

customer_id    name
--
1              customer I

-- All active customers and their projects
SELECT c.customer_id, c.name, p.project_id, p.project_name
FROM customers c
INNER JOIN projects p ON (p.customer_id = c.customer_id)
WHERE c.active = 1;

customer_id  name          project_id  project_name  
--
1            customer I    1           project I
1            customer I    2           project II

-- All projects from section 1
SELECT ps.project_id, ps.section_id
FROM project_sections ps
WHERE ps.section_id = 1;

project_id    section_id
--
2             1

-- All projects from section 1, including project name
SELECT ps.project_id, p.project_name, ps.section_id
FROM project_sections ps
INNER JOIN projects p ON (p.project_id = ps.project_id)
WHERE ps.section_id = 1;

project_id    project_name    section_id
--
2             project II      1

只有一个第1部分的项目。现在把这些碎片放在一起

-- All active customers and their projects from section 1
SELECT c.customer_id, c.name, p.project_id, p.project_name, ps.section_id
FROM customers c
INNER JOIN projects p ON (p.customer_id = c.customer_id)
INNER JOIN project_sections ps ON (p.project_id = ps.project_id)
WHERE c.active = 1 AND ps.section_id = 1;

customer_id  name          project_id    project_name    section_id
-- 
1            customer I    2             project II      1

这就是我所期望的

稍后

---

我看到你改变了数据。如果我在这里做了相同的更改，那么上面的最后一个查询不会返回任何行。我想这就是你所说的你所期待的。（而且，随着您对数据的更改，这也是我所期望的。）

对于您显示的数据，该结果是否正确？第1部分只有一个项目，而该项目只有一个客户：客户I。事实上，您的两个项目都只有一个客户，客户I，所以无论您指定哪个部分，您都会将该客户取回。很抱歉，您的回答完全正确，我已更正。因此，在上面的例子中，结果应该是空的。但不幸的是，事实并非如此。非常感谢。谢谢你，我更新了查询，不幸的是仍然有相同的错误结果。@user707907正如nnnnnn已经说过的，结果似乎适合查询。。。不是吗？只有一个活跃的客户也在第1部分。嘿，夸斯邓克，非常感谢你的帮助。是的，没错。我已经纠正了，问题仍然存在。