让我们一步一步来...
首先，您选择的实体在leave_request表中。所以让我们从那里开始：
SELECT leave_request.* FROM leave_request
现在，您需要知道applied_by所需结果中列的数据。所以你加入staff表格：
SELECT
applied_staff.name AS applied_by
FROM
leave_request
INNER JOIN staff AS applied_staff ON leave_request.staff_id = applied_staff.staff_id
（请注意，我使用别名作为表名。稍后会派上用场。）
现在您需要知道applied_fromand applied_to，您已经拥有了：
SELECT
applied_staff.name AS applied_by,
leave_request.applied_from,
leave_request.applied_to
FROM
leave_request
INNER JOIN staff AS applied_staff ON leave_request.staff_id = applied_staff.staff_id
现在您需要知道approved_from和approved_to，它们在leave_approval表中：
SELECT
applied_staff.name AS applied_by,
leave_request.applied_from,
leave_request.applied_to,
admin_approval.approved_from,
admin_approval.approved_to
FROM
leave_request
INNER JOIN staff AS applied_staff ON leave_request.staff_id = applied_staff.staff_id
INNER JOIN leave_approval AS admin_approval ON leave_request.request_id = admin_approval.request_id
哦，现在我们有一个问题。存在一对多的关系，所以现在我们在结果中重复了请假请求。我们需要以某种方式过滤掉它。你没有具体说明如何，所以我要做几个假设：你想知道“管理员”批准的approved_from和approved_to只有一个“管理员”批准。
让我们在表连接中反映这些假设：
SELECT
applied_staff.name AS applied_by,
leave_request.applied_from,
leave_request.applied_to,
admin_approval.approved_from,
admin_approval.approved_to
FROM
leave_request
INNER JOIN staff AS applied_staff ON leave_request.staff_id = applied_staff.staff_id
INNER JOIN leave_approval AS admin_approval ON leave_request.request_id = admin_approval.request_id
INNER JOIN staff AS approved_staff ON admin_approval.approved_by = approved_staff.staff_id
INNER JOIN group AS approved_staff_group on approved_staff.group_id = approved_staff_group.group_id
WHERE
approved_staff_group.group_name = 'admin'
那应该更好。请注意，表别名在这里派上用场，因为我们现在staff在同一个查询中有两个用于两个不同目的的表实例。所以我们需要区分它们。（请记住，我在这里盲目飞行，实际上无法测试任何这些。如果在此过程中遇到任何问题，请纠正我。我也是徒手编写这段代码，因为我没有 MySQL方便，所以如果还有语法错误，请告诉我。）
现在让我们将approved_admin字段添加到结果中，该字段已经可用：
SELECT
applied_staff.name AS applied_by,
leave_request.applied_from,
leave_request.applied_to,
admin_approval.approved_from,
admin_approval.approved_to,
approved_staff.name AS approved_admin
FROM
leave_request
INNER JOIN staff AS applied_staff ON leave_request.staff_id = applied_staff.staff_id
INNER JOIN leave_approval AS admin_approval ON leave_request.request_id = admin_approval.request_id
INNER JOIN staff AS approved_staff ON admin_approval.approved_by = approved_staff.staff_id
INNER JOIN group AS approved_staff_group on approved_staff.group_id = approved_staff_group.group_id
WHERE
approved_staff_group.group_name = 'admin'
最后，我们需要知道approved_hr. 并且null被允许？那么，我们将为此使用不同的连接。我也在做出与上述类似的假设。让我们试试这个：
SELECT
applied_staff.name AS applied_by,
leave_request.applied_from,
leave_request.applied_to,
admin_approval.approved_from,
admin_approval.approved_to,
approved_staff.name AS approved_admin,
hr_staff.name AS approved_hr
FROM
leave_request
INNER JOIN staff AS applied_staff ON leave_request.staff_id = applied_staff.staff_id
INNER JOIN leave_approval AS admin_approval ON leave_request.request_id = admin_approval.request_id
INNER JOIN staff AS approved_staff ON admin_approval.approved_by = approved_staff.staff_id
INNER JOIN group AS approved_staff_group on approved_staff.group_id = approved_staff_group.group_id
LEFT OUTER JOIN leave_approval AS hr_approval ON leave_request.request_id = hr_approval.request_id
LEFT OUTER JOIN staff AS hr_staff ON hr_approval.approved_by = hr_staff.staff_id
LEFT OUTER JOIN group AS hr_staff_group ON hr_staff.group_id = hr_staff_group.group_id
WHERE
approved_staff_group.group_name = 'admin'
AND hr_staff_group.group_name = 'HR'
我不完全确定后者LEFT OUTER JOIN。第一个肯定需要是允许null值的连接，但我不确定查询引擎如何处理除此之外的连接。我希望它们INNER JOIN在 initial 的范围内LEFT OUTER JOIN。但我想所有这一切也确实取决于数据的完整性，我无法保证。
还值得注意的是，"Jack"当值为"jack". 我没有在这段代码中进行任何字符串操作来实现这一点。如果值应该在数据中大写，则在数据中将其大写。
同样，我不能保证这个代码。但作为一个演练，它应该让你朝着正确的方向前进。正如我在对该问题的评论中提到的，如果您要编写 MySQL 代码，我真的建议您拿起一本关于 MySQL 的书。
编辑：我可以给出的一项建议是数据本身的结构。具体来说，leave_approval那张桌子感觉有点乱，只有那张桌子引起了混乱。我有几个建议：
approval_type在leave_approval表中添加一个。至少这将表明它是管理员批准、人力资源批准还是任何其他类型的批准。（甚至还有其他类型吗？会有吗？）然后您还可以使用request_idandapproval_type作为组合主键，或者至少是组合唯一约束，以加强数据完整性并防止重复批准。- 如果只有两种批准并且可能不会改变，请将它们都反映在
leave_approval表格中。有一组用于 的列admin_approval_*，一组用于hr_approval_*。（每组都将包括staff_id批准的相关日期。）然后request_id它本身可能是leave_approval使其与 . 一对一的主键leave_request。这将极大地简化关系数据，实质上将记录转换为leave_approval记录的一组可选附加信息leave_request。连接会变得更简单，数据会更清楚地表达自己。