#第07章_单行函数的课后练习

1.显示系统时间(注：日期+时间)

SELECT NOW(),SYSDATE(),CURRENT_TIMESTAMP(),LOCALTIME(),LOCALTIMESTAMP() #大家只需要掌握一个函数就可以了
FROM DUAL;

2.查询员工号，姓名，工资，以及工资提高百分之20%后的结果（new salary）

SELECT employee_id,last_name,salary,salary * 1.2 "new salary"
FROM employees;

3.将员工的姓名按首字母排序，并写出姓名的长度（length）

SELECT last_name,LENGTH(last_name) "name_length"
FROM employees
#order by last_name asc;
ORDER BY name_length ASC;

4.查询员工id,last_name,salary，并作为一个列输出，别名为OUT_PUT

SELECT CONCAT(employee_id,',',last_name,',',salary) "OUT_PUT"
FROM employees;

5.查询公司各员工工作的年数、工作的天数，并按工作年数的降序排序

SELECT employee_id,DATEDIFF(CURDATE(),hire_date)/365 "worked_years",DATEDIFF(CURDATE(),hire_date) "worked_days",
TO_DAYS(CURDATE()) - TO_DAYS(hire_date) "worked_days1"
FROM employees
ORDER BY worked_years DESC;

6.查询员工姓名，hire_date , department_id，满足以下条件：

#雇用时间在1997年之后，department_id 为80 或 90 或110, commission_pct不为空
SELECT last_name,hire_date,department_id
FROM employees
WHERE department_id IN (80,90,110)
AND commission_pct IS NOT NULL
#and hire_date >= '1997-01-01'; #存在着隐式转换
#and date_format(hire_date,'%Y-%m-%d') >= '1997-01-01'; # 显式转换操作，格式化：日期---> 字符串
#and date_format(hire_date,'%Y') >= '1997'; # 显式转换操作，格式化
AND hire_date >= STR_TO_DATE('1997-01-01','%Y-%m-%d');# 显式转换操作，解析：字符串 ----> 日期

7.查询公司中入职超过10000天的员工姓名、入职时间

SELECT last_name,hire_date
FROM employees
WHERE DATEDIFF(CURDATE(),hire_date) >= 10000;

8.做一个查询，产生下面的结果

#<last_name> earns monthly but wants <salary*3>

SELECT CONCAT(last_name,' earns ',TRUNCATE(salary,0), ' monthly but wants ',TRUNCATE(salary * 3,0)) "Dream Salary"
FROM employees;

9.使用case-when，按照下面的条件：

/*job grade
AD_PRES A
ST_MAN B
IT_PROG C
SA_REP D
ST_CLERK E

产生下面的结果:
*/

SELECT last_name "Last_name",job_id "Job_id",CASE job_id WHEN 'AD_PRES' THEN 'A'
WHEN 'ST_MAN' THEN 'B'
WHEN 'IT_PROG' THEN 'C'
WHEN 'SA_REP' THEN 'D'
WHEN 'ST_CLERK' THEN 'E'
END "Grade"
FROM employees;