MySQL difficulty grammar - Connection

Benpian involved data tables can be found on its own chapter "MySQL difficulty grammar - subqueries"

There are three connections between MySQL data table: equivalent connection, external connection, since the connection. The following describes by way of example three connections

　　1, the equivalent connection

　　　　　　Equijoins generally applicable to the connection of the main table and an outer (primary key and the foreign key table is equal to the outer primary)

　　　　　　　　Requirements: 1 View the name of the department where all employees

the SELECT s_emp.id staff, last_name employee name, s_dept.id department number, name department name
 from s_dept, s_emp
 the WHERE s_dept.id = s_emp.dept_id;

 

　　　　　　　　Requirements: 2 query average wage of more than 1200 department, and displays the names of those sectors

select s_dept.id, name
from s_emp, s_dept
where s_emp.dept_id=s_dept.id
group by dept_id
    having avg(salary) > 1200;

 

　　　　　　　　Requirements: 3 viewing area where the names of all employees

select last_name, s_dept.id, s_region.name
from s_emp, s_dept, s_region
where s_emp.dept_id=s_dept.id and s_dept.region_id=s_region.id

 

　　　　　　　　Requirements: 4 to view the employee id, last_name, salary, department name, the name of the region, these employees have the following conditions: salary greater than the average wage in the area where or with Chang Chang employees not in the same departments

the SELECT s_emp.id, last_name, salary, s_dept.name, s_region.name
 from s_emp, s_dept, s_region
 the WHERE s_emp.dept_id = s_dept.id and s_dept.region_id = s_region.id
 and (salary > sector (# chang in your area the average wage
     select  AVG (salary) 
     from s_emp
     the wHERE dept_id in (# of areas where the department chang
         select the above mentioned id
         from s_dept
         the wHERE region_id = (# Chang area where
             select region_id
            from s_emp, s_dept
            where s_emp.dept_id=s_dept.id and last_name="chang"
        )
    ) or last_name != "chang")
);

 

　　　　　　　　Requirements: 5 View wages higher than the average wage of employees where the department id and name

the SELECT s_emp.id, last_name
 from s_emp, s_dept, (the average salary for each department table #
     the SELECT dept_id, AVG (salary) AS avg_salary 
     from s_emp 
     Group  by dept_id) AS avg_table
 the WHERE s_emp.dept_id = s_dept.id
     and s_dept.id = avg_table.dept_id # payroll department number and department table joins
     and salary > avg_salary;

 

　　　　　　　　Requirements: 6 Query salary greater than the average wage of the employee's department number 41, and the average wage in the sector where employees have more than the average wage of 41 departments, showing the name of the current average wage sector where employees and department where the employee

SELECT e.dept_id, n.avg_salary, d.name
 from s_emp AS E, s_dept AS D, (# sector pay table
             SELECT the dept_id, 
                 AVG (the salary) AS avg_salary 
             from s_emp Group  by the dept_id) AS n-
 WHERE e.dept_id = D. ID and d.id = n.dept_id 
 and n.dept_id ! =  41    # 41 is greater than the sector does not exist
 and the salary > (# 41 greater than the average salary department
     SELECT avg_salary 
    from (     # 部门薪资表
        select dept_id, 
            avg(salary) as avg_salary 
        from s_emp group by dept_id) as n 
    where dept_id =41)

　　　2, the outer connector

　　　　　　External connections can be divided into two connections: connect the left and right connections

　　　　　　　　Left connection: Table 1 left join table relationships and table 2 ... on the table, on the ground behind is a condition, not necessarily the equivalence relationship between two tables (Table 1 data are complete)

　　　　　　　　　　Requirements: sales staff to view customer names and their corresponding names

select c.name, e.last_name
from s_customer as c left join s_emp as e    # 左连接
on e.id=c.sales_rep_id;

　　　　　　　　Right connection: Table 1 right jion Table 2 .... ON association table and the table, followed by a contact ON condition, the equivalence relation is not necessarily two tables (Data Table 2 is intact)

　　　　　　　　　　Requirements: sales staff to view customer names and their corresponding names

select c.name, e.last_name
from s_emp as e right join s_customer as c    # 右连接
on e.id=c.sales_rep_id;

　　　　3, since the connection

　　　　　　　　　　As the name suggests it is to form their own association - an association table a foreign key to its primary key to the employee table have their own leadership, but leadership itself is an employee: examples are as follows

　　　　　　　　　　　　Demand: See all employee names and their corresponding names manager

select id, last_name
from s_emp
where manager_id = id