选择所有连接的行都匹配的记录

Question

给定以下架构：

employees
id | name

employee_attributes
id | employee_id | key | value

我想选择所有具有提供的属性的员工。

以下语句起作用：

SELECT employees.* FROM employees
INNER JOIN employee_attributes ON employee_attributes.employee_id = employees.id
WHERE employee_attributes.key = 'foo' AND employee_attributes.value = 'bar'

但只允许我通过一个属性来找到一名雇员。 我该如何调整它以通过多个属性来检索员工？

需要明确的是，如果我提供了两组属性来匹配，查询应该只返回至少有这两个属性的员工。

例如，如果Bob仅有一个属性：

key | value
===========
foo | bar

但我为查询提供了两个属性（ foo和bar ， bin和baz ），不应返回Bob。

Answer 1

使用条件聚合：

SELECT employees.*
FROM employees
INNER JOIN employee_attributes
    ON employee_attributes.employee_id = employees.id
GROUP BY employee_attributes.employee_id
HAVING SUM(CASE WHEN employee_attributes.key = 'foo' AND
                     employee_attributes.value = 'bar' THEN 1 ELSE 0 END) > 0 AND
       SUM(CASE WHEN employee_attributes.key = 'bin' AND
                     employee_attributes.value = 'baz' THEN 1 ELSE 0 END) > 0

Answer 2

以下应该工作：

SELECT employees.id, employees.name, count(employee_attributes.id) as attribute_count FROM employees
INNER JOIN employee_attributes ON employee_attributes.employee_id = employees.id
WHERE (employee_attributes.key = 'foo' AND employee_attributes.value = 'bar') OR (employee_attributes.key = 'bin' AND employee_attributes.value = 'baz')
group by employees.id, employees.name
having attribute_count >= 2;

Answer 3

您可以使用聚合获取员工ID：

SELECT ea.employee_id
FROM employee_attributes.employee_id 
WHERE (ea.key = 'foo' AND ea.value = 'bar') OR
      (ea.key = 'bin' AND ea.value = 'baz')
GROUP BY ea.employee_id
HAVING COUNT(DISTINCT ea.key) = 2;

有关完整信息，可以使用JOIN ：

SELECT e.*
FROM employee e JOIN
     (SELECT ea.employee_id
      FROM employee_attributes.employee_id 
      WHERE (ea.key = 'foo' AND ea.value = 'bar') OR
            (ea.key = 'bin' AND ea.value = 'baz')
      GROUP BY ea.employee_id
      HAVING COUNT(DISTINCT ea.key) = 2
     ) ea
     ON ea.employee_id = e.id;