我正在开发REST API。尝试访问资源时:我们要么提供403(禁止)或404(未找到)错误。我们的表格是:
CREATE TABLE `Action` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`created_By_Id` int(10) unsigned NOT NULL,
`name` varchar(60) NOT NULL,
`updated_action_at` datetime(3) DEFAULT NULL,
`created_At` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,
`notes` varchar(400) DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `action_empId_fk` (`created_By_Id`),
CONSTRAINT `action_empId_fk` FOREIGN KEY (`created_By_Id`)
REFERENCES `Employee` (`id`) ON DELETE CASCADE,
) ENGINE=InnoDB AUTO_INCREMENT=502004 DEFAULT CHARSET=latin1
CREATE TABLE `ActionAssignedTo` (
`action_Id` int(10) unsigned DEFAULT NULL,
`assignee_Id` int(10) unsigned DEFAULT NULL,
KEY `actionassignedto_emp_id_foreign` (`emp_Id`),
KEY `actionassignedto_action_id_foreign` (`action_Id`),
CONSTRAINT `ActionAssignedTo_ibfk_1` FOREIGN KEY (`assignee_Id`)
REFERENCES `Employee` (`id`) ON DELETE CASCADE,
CONSTRAINT `ActionAssignedTo_ibfk_2` FOREIGN KEY (`action_Id`)
REFERENCES `Action` (`id`) ON DELETE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=latin1
CREATE TABLE `Employee` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`vendor_Id` int(10) unsigned DEFAULT NULL,
`name` varchar(40) NOT NULL,
`mobile_Number` varchar(15) NOT NULL,
`active` tinyint(1) DEFAULT '1',
`updated_At` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,
`created_At` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
KEY `employee_vendor_id_foreign` (`vendor_Id`),
CONSTRAINT `employee_vendor_id_foreign` FOREIGN KEY (`vendor_Id`)
REFERENCES `Vendor` (`vendor_Id`)
) ENGINE=InnoDB AUTO_INCREMENT=511 DEFAULT CHARSET=latin1
我们正在运行一个查询来获取ID为17的Action,其创建者员工的身份编号为9,供应商ID为1,并且该员工已创建该操作,以便他可以查看该业务(业务规则)。可以将操作分配给多个员工。
select Action.name,
group_concat(AssigneeNameTable.name) as assignedTo,
group_concat(AssigneeNameTable.id) as assignedToId,
ActionAssignedTo.action_Id as actionId
from Action
inner join Employee
on Action.created_By_Id = Employee.id
and Employee.vendor_Id = 1
inner join ActionAssignedTo
on Action.id = ActionAssignedTo.action_Id
and ActionAssignedTo.action_Id = 17
inner join Employee as AssigneeNameTable
on ActionAssignedTo.assignee_Id = AssigneeNameTable.Id
where Action.created_By_Id = 9
and Action.deleted_at is null
group by Action.id
limit 2
现在,让我们说一下,如果在DB中根本不存在Action - >在这种情况下,上面的查询返回空结果集
the problem is we can not differentiate the query return empty set because
1. either the action with id:17 did not exist(404- Not Found)
2. or the business rule failed (as in the person requested the action was not
at all related to the action(403 - Forbidden).
我能想到的解决方案之一是: 首先运行一个小查询,如:
select * from Action where id = 17
如果此查询返回空集,则表示该操作在db中不存在。
在此之后我运行更大的查询
结果集的不同组合(数组中的数字表示返回的记录):
Small Query | Big Query | Interpretation
---------------------------------------
[0] | [0] | Resource Not Found(404)
[1] | [0] | Forbidden (403)
如果小查询返回0结果 - >我们可以直接发送404错误;否则我们执行Big Query。
答案 0 :(得分:1)
我使用了左外连接的概念,正如我的朋友建议的那样。请在下面找到新的查询:
select *
from
(select id
from Action
where id = 17) AS act1
left Outer Join
(select Action.name,
group_concat(AssigneeNameTable.name) as assignedTo,
group_concat(AssigneeNameTable.id) as assignedToId,
ActionAssignedTo.action_Id as actionId
from Action
inner join Employee
on Action.created_By_Id = Employee.id
and Employee.vendor_Id = 1
inner join ActionAssignedTo
on Action.id = ActionAssignedTo.action_Id
and ActionAssignedTo.action_Id = 17
inner join Employee as AssigneeNameTable
on ActionAssignedTo.assignee_Id = AssigneeNameTable.Id
where Action.created_By_Id = 9
and Action.deleted_at is null
group by Action.id
limit 2) AS act2
on act1.id = act2.actionId
概念很简单
如果输出中没有结果 - >找不到对象(404)
如果输出包含id
字段但不包含第二个子查询中的任何单个字段,则表示该实体存在于db中,但业务规则不允许,因此禁止(403)。< / p>