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7月28

spa 詳細資料 strong rst add subst 工資 and distinct

選擇部門30中的雇員

Select last_name, first_name from employees where department_id=30;

列出所有職員(CLERK)的姓名、編號和部門

Select last_name, first_name, employee_id, job_id from employees;

找出薪金大於5000的雇員

Select last_name, first_name from employees where salary>5000;

找出獎金高於0.1的雇員

Select last_name, first_name from employees where commission_pct>0.1;

找出部門50中的所有員工和部門30中的經理的詳細資料

Select * from employees where department_id=50 or (department_id=30 and job_id like ‘%\_MAN’);

找出收取獎金的雇員的不同工作職位 每種職位顯示一次 distinct

Sesect distinct job_id from employees where comission_pct is null;

找出不收取獎金或收取的工資低於5000的雇員

Select last_name, first_name from employees where commission_pct is null or salary*12*(commission_pct)<5000;

顯示last_name不帶有‘R‘的雇員姓名

Select last_name, first_name from employees where last_name not like ‘%\_R’;

Order by排序 默認是升序 加上desc表示降序。

多行函數:

單行函數:

Character functions(字符串)函數:

Trunc截斷,如 select last_name, trunc((sysdate - hire_date) /30) from employees;

Conversion 轉換函數 implicat隱含類型轉換 explicit明確類型轉換 類型必須一致。

To_char做格式轉換的使用。

Rr代表離當前年份最近的。Yy代表和當前年份在同一個世紀中。年份必須四個y

排序:

SQL> select last_name, salary from employees order by salary;

SQL> select last_name, salary from employees order by salary desc;

SQL> select last_name, salary from employees order by last_name;

SQL> select last_name, hire_date from employees order by hire_date;

SQL> select last_name, salary, commission_pct from employees order by salary desc, commission_pct desc;

SQL> select last_name, salary*12*(1+commission_pct) from employees order by 2;2表示第幾列

SQL> select last_name, salary*12*(1+commission_pct) total_salary from employees order by total_salary;

單行函數

SQL> select upper(first_name), lower(last_name), length(last_name) from employees;

SQL> select (sysdate-hire_date)/7 from employees;

SQL> select trunc((sysdate-hire_date)/30, 0) from employees;

SQL> select trunc(months_between(sysdate,hire_date), 0) from employees;

SQL> select sysdate+3650 from dual;

SQL> select add_months(sysdate, 120) from dual;

SQL> select next_day(‘2015-09-01‘, ‘friday‘) from dual;

SQL> select next_day(‘2015-10-01‘, 6) from dual;

SQL> select last_day(sysdate) from dual;

SQL> select round(to_date(‘2015-10-10‘,‘yyyy-mm-dd‘), ‘MONTH‘) from dual;

SQL> select round(to_date(‘2015-10-16‘,‘yyyy-mm-dd‘), ‘MONTH‘) from dual;

SQL> select round(to_date(‘2015-10-10‘,‘yyyy-mm-dd‘), ‘YEAR‘) from dual;

SQL> select round(sysdate, ‘DAY‘) from dual;

練習:

找出各月最後三天內受雇的所有雇員

extract(month from hire_date+4) != extract(month from hire_date)

找出早於25年之前受雇的雇員

months_between(sysdate, hire_date)/300>=25

顯示正好為6個字符的雇員姓名

length(last_name)=6

顯示所有雇員的姓名的前三個字符

substr(last_name, 1, 3)

顯示所有雇員的姓名,用a替換所有‘A‘

replace(last_name, ‘A‘, ‘a‘)

類型轉換和其他函數

SQL> select to_char(salary, ‘$999,999.00‘) from employees;

SQL> select last_name, to_char(hire_date, ‘dd-Mon-RR‘) from employees;

SQL> select to_char(sysdate, ‘yyyy-mm-dd hh24:mi:ss‘) from dual;

SQL> select to_char(sysdate, ‘yyyy-mm-dd hh:mi:ss AM‘) from dual;

SQL> select last_name from employees where hire_date=to_date(‘2006-05-23‘, ‘yyyy-mm-dd‘);

SQL> select to_number(‘$123,456.78‘, ‘$999,999.00‘) from dual;

練習:

查詢2006年入職員工:

select last_name

from employees

where hire_date between to_date(‘2006-01-01‘, ‘yyyy-mm-dd‘)

and to_date(‘2006-12-31‘, ‘yyyy-mm-dd‘);

select last_name

from employees

where to_char(hire_date, ‘yyyy‘)=‘2006‘;

select last_name

from employees

where extract(year from hire_date)=2006;

--不推薦

select last_name

from employees

where hire_date like ‘2006%‘;

查詢歷年9月份入職的員工:

select last_name

from employees

where to_char(hire_date, ‘mm‘)=‘09‘;

select last_name

from employees

where extract(month from hire_date)=9;

其他函數:

nvl

nvl(val1, val2)

if val1 is not null

then

return val1;

else

return val2;

SQL> select last_name, salary*12*(1+nvl(commission_pct, 0)) total_salary from employees;

練習:

顯示所有員工部門編號,沒有部門的顯示未分配部門

Select last_name, nvl(to_char(department_id), ‘N/A’) from employees;

casedecode

IT_PROG +1000

SA_REP +1500

ST_CLERK +2000

其他人工資不變 select salary from employees where job_id not in=‘IT_PROG‘; SA_REP ST_CLERK

select salary+1000 from employees where job_id=‘IT_PROG‘;

select last_name, job_id, salary,

case job_id

when ‘IT_PROG‘ then salary+1000

when ‘SA_REP‘ then salary+1500

when ‘ST_CLERK‘ then salary+2000

else salary

end new_salary

from employees;

select last_name, job_id, salary,

decode( job_id,

‘IT_PROG‘, salary+1000,

‘SA_REP‘, salary+1500,

‘ST_CLERK‘, salary+2000,

salary) new_salary

from employees;

練習:

按照員工工資,對員工分級顯示:

A 20001-25000

B 15001-20000

C 10001-15000

D 5001-10000

E 0-5000

select last_name, salary, (case when salary>=20001 and salary<=25000 then ‘A‘ when salary>=15001 and salary<=20000 then ‘B‘ when salary>=10001 and salary<=15000 then ‘C‘ when salary>=5001 and salary<=10000 then ‘D‘ when salary>=0 and salary<=5000 then ‘E‘ end) qualified_salary from employees;

7月28