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oracle查询测试样题
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1.select count(*) from employees where last_name like _A%;key:02.select count(*)from employeeswhere to_char(hire_date,YYYY)=1998;select count(*)from employeeswhere hire_date like %98;key:23select to_char(hire_date,YYYY) from employees;3.se
1. select count(*) from employees where last_name like '_A%'; key:0 2. select count(*) from employees where to_char(hire_date,'YYYY')=1998; select count(*) from employees where hire_date like '%98'; key:23 select to_char(hire_date,'YYYY') from employees; 3. select job_title, max_salary-min_salary as "SAL_DIEF" from jobs order by max_salary-min_salary desc; select job_title,(max_salary-min_salary) as "SAL_DIEF" from jobs order by 2 desc; 19行记录 4. select count(*) from employees where (salary>12000 or salary<1000) and job_id !='ST_MAN' and job_id!='SH_CLERK' ; number:6 ================ select count(*) from employees where salary not between 1000 and 12000 and job_id not in('ST_MAN','SH_CLERK') ; select salary from employees where salary<1000 and salary>12000; --当它判断1000为假时就不判断后面的大于12000了. select job_id from employees; select * from employees where job_id in('ST_MAN','SH_CLERK'); --工作岗位名称要加单引号 5. select count(*) from employees where to_char(hire_date,'YYYY')=1999 and to_char(hire_date,'mm')=02; key:3 select count(*) from employees where to_char(hire_date,'YYYY-MM')='1999-02'; 6. select last_name,salary, decode(trunc(salary/1500),0,'A', 1,'B', 2,'C', 'D' ) Grade from employees where last_name like'%s'; 7. select d.department_id,d.department_name,l.city from departments d,locations l where d.department_id in(10,40,90) and d.location_id=l.location_id; 8. select l.city,c.country_name,r.region_name from locations l,regions r,countries c where l.location_id=1000 and l.country_id=c.country_id and c.region_id=r.region_id; 9. select m.last_name "MAN_NAME",nvl(e.last_name,'NO EMPLOYEES') "EMP_NAME" from employees m,employees e where m.department_id=100 and m.employee_id=e.manager_id(+); 10行记录 select m.last_name MAN_NAME,nvl(e.last_name,'NO EMPLOYEES') EMP_NAME from employees m,employees e where m.department_id=100 and m.employee_id=e.manager_id(+); 10 select department_id,count(*) NUM from employees where salary>8000 group by department_id ; 9行记录 11 select department_id,count(*) NUM from employees where salary>5000 group by department_id having count(*)>3; 3行记录 12 select last_name,salary from employees where salary> (select salary from employees where employee_id=110) and department_id=100; 2行记录 13 select count(*) NUM from employees where commission_pct12000 and commission_pct is not null ); 24行记录 1000>
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