Saturday, 7 December 2013

Set of A-Z SQL Command 2



  1. Single row Function
  2. Conversion Function (to_char with date)
  3.  
  4. 1.   select to_char(sysdate,’dd-mm-yyyy’) from dual;
  5. 2.   select to_char(sysdate,’dd-month-year’) from dual;
  6. 3.   select to_char(sysdate,’dd-mon-year’) from dual;
  7. 4.   select to_char(sysdate,’dd-DAY-month-year’) from dual;
  8. 5.   select to_char(sysdate,’dd-DY-month-year’) from dual;
  9. 6.   select to_char(sysdate,’dd-mon-yyyy, HH:MI:SS’) from dual;
  10. 7.   select to_char(sysdate,’dd-mon-yyyy, HH24:MI:SS’) from dual;
  11. 8.   select to_char(sysdate,’dd-mon-yyyy, HH:MI:SS AM’) from dual;
  12. 9.   select to_char(sysdate,’DDth-mon-yyyy’) from dual;
  13. 10.        select to_char(sysdate,’DDth “of” mon-yyyy, HH:MI:SS’) from dual;
  14. 11.        select to_char(sysdate,’DDsp-mon-yyyy, HH:MI:SS’) from dual;
  15. 12.        select to_char(sysdate,’DDspth-mon-yyyy, HH:MI:SS’) from dual;
  16. 13.        select to_char(hire_date,’DDspth-mon-yyyy, HH:MI:SS’) from employees;
  17.  
  18.  
  19. Conversion Function (to_char with Number)
  20.  
  21. 14.        select to_char(salary,’99999.99’) from employees;
  22. 15.        select to_char(salary,’$99999.99’) from employees;
  23. 16.        select to_char(salary,’09999.99’) from employees;
  24. 17.  select to_char(salary,’L99999.99’) from employees; //L use to local currency //
  25. 18.        select to_char(salary,’99,999.99’) from employees;
  26. 19.  select to_char(salary-3000,’099999MI’) from employees; // MI use for minus//
  27. 20.  select to_char(salary-3000,’099999PR’) from employees; //PR to use the paraththice//
  28. 21.  select to_char(salary,’99999EEEE’) from employees; //EEEE to see the exponencial  //
  29. 22.  select to_char(salary,’99999V99’) from employees; //after salary 00 use to V//
  30.  
  31.  
  32. Conversion Function (to char with Characters)
  33.  
  34. 23.        select to date(’05-09.2010’,’dd-mm-yyyy’) from dual;
  35. 24.        select to date(’05-09.2010’,’fxdd-mm-yyyy’) from dual;  //fix to fix the date //
  36. 25.        select to date(’05-09.2010’,’fxdd-mm.yyyy’) from dual;
  37. 26.        select sysdate-to_date(’05-09-2010’,’dd-mm-yyyy’) from dual;
  38. 27.        alter session set nls_currency=’Rs’;
  39. 28.        select to_char(salary,’L99999.99’) from employees;
  40. 29.        select (sysdate-hire_date) from employees;
  41. 30.        select (sysdate-12-may-10’) from employees;
  42.  
  43.  
  44.  
  45.  
  46. Displaying Data from Multiple Table
  47.  
  48.  
  49. 1.   select * from employees;
  50. 2.   desc employees;
  51. 3.   select * from departments;
  52. 4.   desc departments;
  53. 5.   select last_name,department_name from employees,departments; (Cartesion product)
  54. 6.   select e.employee_id,e.last_name,d.department_name from employees e,departments d;
  55. 7.   select e.employee_id,e.last_name,d.department_name from employees e,departments d where e.department_id=d.department_id;
  56. 8.   select last_name,department_name from employees,departments  where employees.department_id=departments.department_id;
  57. 9.   select e.employee_id,e.last_name,d.department_name from employees e,departments d where e.department_id=d.department_id and last_name='Smith';
  58. 10.        select e.last_name,e.employee_id,d.department_name,d.department_id,l.location_id,l.city from employees e,departments d,locations l where e.department_id=d.department_id and d.location_id=l.location_id;
  59.  
  60. Non Equi join (Cartesion Product)
  61.  
  62. 11.        select e.last_name,e.hire_date,e.job_id,e.department_id,j.employee_id from employees e,job_history j where e.hire_date between j.start_date and j.end_date;   
  63. 12.        select e.last_name,e.hire_date,e.job_id,e.department_id,j.employee_id from employees e,job_history j where e.hire_date between j.start_date and j.end_date and last_name='Gee';          
  64.  
  65. Ooter Join
  66.  
  67. 13.        select e.employee_id,e.last_name,d.department_name from employees e,departments d where e.department_id=d.department_id;
  68. 14.        select e.employee_id,e.last_name,d.department_name from employees e,departments d where e.department_id(+)=d.department_id;
  69. 15.        select e.employee_id,e.last_name,d.department_name from employees e,departments d where e.department_id=d.department_id(+);
  70.  
  71. Self Join
  72.  
  73. 16.        select e.employee_id,e.last_name,m.last_name "Manager" from employees e,employees m where e.manager_id=m.employee_id;
  74. 17.        select e.last_name||'  work for  '||m.last_name "Manager" from employees e,employees m where e.manager_id=m.employee_id;
  75.  
  76. Displaying Data from Multiple Table
  77.  
  78. Join with Oracle’s 1999 Syntax
  79.  
  80. Cross join (cartesion Product)
  81.  
  82. 18.    select last_name,salary,department_name from employees cross join departments;
  83.  
  84. Natural Join
  85.  
  86. 19.    select last_name,salary,department_name from employees natural join departments;
  87.  
  88. With the on clause
  89.  
  90. 20.    select e.employee_id,e.last_name,d.department_name from employees e join departments d on e.department_id=d.department_id;
  91. 21.    select e.last_name,d.department_name,d.department_id,l.city from employees e join departments d on e.department_id=d.department_id join locations l on d.location_id=l.location_id;
  92. Left outer Join
  93.  
  94. 22.    select e.employee_id,e.last_name,d.department_name from employees e left outer join departments d on e.department_id=d.department_id;
  95.  
  96. Right outer Join
  97.  
  98. 23.    select e.employee_id,e.last_name,d.department_name from employees e right outer join departments d on e.department_id=d.department_id;
  99.  
  100. Full outer Join
  101.  
  102. 24.    select e.employee_id,e.last_name,d.department_name from employees e full outer join departments d on e.department_id=d.department_id;
  103. 25.    select e.employee_id,e.last_name,d.department_name from employees e  join departments d on e.department_id=d.department_id and last_name=’Smith’;
  104.  
  105. Aggregation Data using Group Functions
  106.  
  107. Group Function (operate on set of row to give one result per group)
  108.  
  109. 1.   select max(salary) from employees;
  110. 2.   select max(salary),min(salary),avg(salary),sum(salary) from employees;
  111. 3.   select sum(salary+salary+salary) from employees;
  112. 4.   select max(salary) max_sal,min(salary) min_sal,avg(salary) avg_sal,sum(salary) sum_sal from employees;
  113. 5.   select min(hire_date),max(hire_date) from employees;
  114. 6.   select min(last_name),max(last_name) from employees;
  115. 7.   select count(*) from employees;
  116. 8.   select count(commission_pct),count(job_id),count(*) from employees;
  117. 9.   select count(*) from employees where department_id=50;
  118. 10.        select count(department_id) from employees;
  119. 11.        select count(distinct department_id) from employees;
  120. 12.        select avg(commission_pct) from employees;
  121. 13.        select avg(nvl(commission_pct,0)) from employees;
  122.  
  123. Creating Group of data
  124.  
  125. 14.        select department_id,avg(salary) from employees group by department_id;
  126. 15.        select avg(salary) from employees group by department_id;
  127. 16.        select job_id,avg(salary),sum(salary),count(*) from employees group by job_id;
  128.  
  129. Group by more then one column
  130.  
  131. 17.        select department_id,job_id,avg(salary) from employees group by department_id,job_id;
  132. 18.        select department_id,job_id,avg(salary) from employees; (illegal Query)
  133.  
  134. Having Clause
  135.  
  136. 19.        select department_id,avg(salary) from employees where avg(salary)>5000 group by department_id; (illegal Query)
  137. 20.        select department_id,avg(salary) from employees where department_id<60 group by department_id;
  138. 21.        select department_id,avg(salary) from employees  group by department_id having avg(salary) >5000;
  139. 22.        select department_id,avg(salary) from employees  group by department_id having avg(salary)>5000 order by avg(salary);
  140.  
  141. Nesting Group Function
  142.  
  143. 23.        select max(avg(salary)) from employees  group by department_id;



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