Analysis of connection query examples in MySQL
1. Cartesian product
Table 1 has m rows of data, Table 2 has n rows of data, and the query result has m*n rows of data.
2. Classification
(1) Classification by age
sql92 standard: only supports inner joins
sql99 standard (recommended): supports inner joins, Outer joins (left outer joins and right outer joins), cross joins
(2) Classification by function
Inner joins: equivalent joins, non-equivalent joins, self-joins
Outer join: left outer join, right outer join, full outer join
Cross join
3. Equivalent join
(1) Query the girl’s name and its corresponding Boyfriend’s name
SELECT girlname, boyname FROM boys, girls WHERE girls.boyfriend_id=boys.girlfriend_id;
(2) Query the employee name (last_name) and the corresponding department name (department_name)------------- (query based on the associated id)
SELECT last_name, department_name FROM JDSC, SNSC WHERE JDSC.`id`=SNSC.`id`;
(3) Query employee name (last_name), job type number (job_id), job type name (job_title) (employee table: JDSC work table: JOBSC) ------ Alias the table to improve simplicity , to avoid ambiguity
#"e.job_id"是为了避免歧义 SELECT last_name, e.job_id, job_title FROM ESC e, JOBSC j WHERE e.`job_id`=j.`job_id`;
Note:
If an alias is given to the table, the queried fields cannot be qualified by the original table name.
The order of the above two tables can be interchanged
(4) Query the names of employees and departments with bonuses
#员工名:last_name #部门名:department_name #奖金率:commissiom_pct SELECT last_name, department_name, commissiom_pct FROM employees e, department d WHERE e.`department_id`=d.`department_id` AND e.`commissiom_pct` IS NOT NULL;
(5) Group query-mdash;Query Number of departments in each city
#城市分组表名:city #部门分组表名:departments SELECT COUNT(*) 个数, city FROM departments d, locations l WHERE d.`location_id`=l.`location_id` GROUP BY city;
(6) Sorting - Query the job name and number of employees for each type of work, and sort them in descending order by the number of employees
SELECT job_title, COUNT(*) FROM emloyees e, jobs j WHERE e.`job_id`=j.`job_id` GROUP BY job_title ORDER BY COUNT(*) DESC;
(7) Three Table connection-query employee name, department name and city
SELECT last_name, department_name, city FROM employees e, departments d, locations l WHERE e.`department_id`=d.`department_id` AND d.`location_id`=l.`location_id` AND city LIKE 's%';
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