Question

表人：

ID，NAME，AGE，DESCRIPTION，STATUS

值：

1, Casey, 18, A fresh grad, Probation
2, Wedy, 25, Middle career, Approve
3, Stacy, 33, A working professional, Confirmed
4, Steve, 45, Senior Management, Confirmed
5, Goat, 55, Retire, Rejected

表格许可：

PermID，TYPE，VALUE

值：

1, normaluser, can view training 1 and 2
1, probationuser, can view training 3 part 1
2, normaluser, can view training 1 and 2
2, traininguser, can view training 3
2, extradinoaryuser, Wendy can view full training
3, extradinoaryuser, Stacy can view full training
3, superuser, Stacy also can edit full training
3, supertrainer, can create full training

努力实现：

显示具有＆＃39;状态＆＃39;的完整个人表格数据=已批准或＆＃39;状态＆＃39; =已确认

输出 - 显示Person row id 2,3,4

如果将人员ID与权限PermID匹配，请输入＆＃39;输入＆＃39; = extradinoaryuser或＆＃39; type＆＃39; =超级用户，忽略其余类型，与Person表数据合并。

输出 - 附加权限＆＃39;输入＆＃39;和＆＃39;价值＆＃39;人行id 2,3

如果条目同时具有超级用户和extradinoaryuser类型，请复制条目

输出 - 人员ID行3作为超级用户和extradinoaryuser

期望的结果：

2, Wedy, 25, Middle career, Approve, extradinoaryuser, Wendy can view full training<br/>
3, Stacy, 33, A working professional, Confirmed, extradinoaryuser, Stacy can view full training<br/>
3, Stacy, 33, A working professional, Confirmed, superuser, Stacy also can edit full training<br/>
4, Steve, 45, Senior Management, Confirmed

Sql命令（不工作）：

select Person.id, Person.name, Person.age, Person.description, Person.status 
from Person  
inner join Permission ON Person.id = Permission.PermID 
where Person.status like “Approve”  
   or Person.status like “Confirmed” 
  AND (permission.name = 'superuser' or permission.name = ‘extradinoaryuser’);

Answer 1

对字符串变量使用正确的引号（''）：

select p.id, p.name, p.age, p.description, p.status 
  from Person p inner join Permission m ON ( p.id = m.PermID )
 where p.status like 'Approve' or p.status like 'Confirmed'
   and (m.name = 'superuser' or m.name = 'extradinoaryuser');

P.S。确保表中字符串值的（下/上）情况。

Answer 2

问题是使用正确的引号（“”）和permission.name列不存在。使用permission.type代替

构建架构：

create table Person
(ID int primary key Auto_Increment
, `NAME` varchar(25)
, AGE int
, DESCRIPTION varchar(255)
, STATUS varchar(25) );

insert into person values 
( 1, "Casey", 18, "A fresh grad", "Probation" ), 
(2, "Wedy", 25, "Middle career", "Approve"),
(3, "Stacy", 33, "A working professional", "Confirmed"), 
(4, "Steve", 45, "Senior Management", "Confirmed"), 
(5, "Goat", 55, "Retire", "Rejected");

create table Permission 
(PermID int
, `TYPE` varchar(25)
, `VALUE` varchar(255)
);

insert into Permission values
( 1, "normaluser","can view training 1 and 2"), 
(1, "probation user", "can view training 3 part 1"), 
(2, "normaluser", "can view training 1 and 2"), 
(2, "traininguser", "can view training 3"), 
(2, "extradinoaryuser", "Wendy can view full training"), 
(3, "extradinoaryuser", "Stacy can view full training"),
(3, "superuser", "Stacy also can edit full training"), 
(3,"supertrainer", "can create full training");

将您的查询更正为

首先获取状态为Approve或Confirmed的所有用户，然后将用户权限类型过滤为超级用户或extradinoaryuser或null以显示用户没有权限

select * 
  from ( select Person.id
          , Person.name
          , Person.age
          , Person.description
          , Person.status
          , Permission.type
          , Permission.value
      from Person 
      left 
      join Permission 
        ON Person.id = Permission.PermID
     where (Person.status like "Approve" or Person.status like "Confirmed")
        ) t 
 where (t.type = 'superuser' or t.type = "extradinoaryuser" or t.type IS NULL) ;

查看结果 http://sqlfiddle.com/#!9/f5a6be/27

Answer 3

问题似乎是你的报价。但是，我建议查询两件事：

表别名。
IN

查询看起来像：

select p.id, p.name, p.age, p.description, p.status 
from Person p inner join
     Permission pn
     ON p.id = pn.PermID 
where p.status in ('Approve', 'Confirmed') and
      pn.name in ('superuser', 'extradinoaryuser');

在SQL中，标准字符串分隔符是单引号。这些应该用于定义字符串和日期常量（而不是用于其他目的）。

如果ID匹配，则内连接附加

3 个答案: