ch
Feedback

不要被骗子欺骗!Telemetrio 会找到并标记这些频道 👉 如果想查看标记,请订阅

SQL Interview Questions

SQL Interview Questions

前往频道在 Telegram

Channel for SQL learning: @sql_res Chennal for SQL syntax: @sql_syntax Chennal for SQL interview questions: @etl_sql_interview

显示更多
未指定国家未指定类别
2 405
订阅者
无数据24 小时
-27 天
-1730 天
帖子存档
Click below links To Refer three tables for writing queries👇 https://t.me/etl_sql_interview/8 To Refer questions for writing queries👇 https://t.me/etl_sql_interview/34 Answers: 26. SELECT * FROM Worker WHERE MOD (WORKER_ID, 2) <> 0; 27. SELECT * FROM Worker WHERE MOD (WORKER_ID, 2) = 0; 28. SELECT * INTO WorkerClone FROM Worker; 29. (SELECT * FROM Worker) INTERSECT (SELECT * FROM WorkerClone); 30. SELECT * FROM Worker MINUS SELECT * FROM Title;

Click below link to Refer three tables here for writing queries👇 https://t.me/etl_sql_interview/8 Q-26. Write an SQL query to show only odd rows from a table. Q-27. Write an SQL query to show only even rows from a table. Q-28. Write an SQL query to clone a new table from another table. Q-29. Write an SQL query to fetch intersecting records of two tables. Q-30. Write an SQL query to show records from one table that another table does not have.

Click below link to Refer three Tables here for writing queries👇 https://t.me/etl_sql_interview/8 Q-21. Write an SQL query to fetch the count of employees working in the department ‘Admin’. Q-22. Write an SQL query to fetch worker names with salaries >= 50000 and <= 100000. Q-23. Write an SQL query to fetch the no. of workers for each department in the descending order. Q-24. Write an SQL query to print details of the Workers who are also Managers. Q-25. Write an SQL query to fetch duplicate records having matching data in some fields of a table. For Answers 🤔 click here 👇 https://t.me/etl_sql_interview/32

Click below link to Refer three Tables here for writing queries👇 https://t.me/etl_sql_interview/8 Answer: 1. Select FIRST_NAME AS WORKER_NAME from Worker; 2. Select upper(FIRST_NAME) from Worker; 3. Select distinct DEPARTMENT from Worker; 4. Select substring(FIRST_NAME,1,3) from Worker; 5.Select INSTR(FIRST_NAME, BINARY'a') from Worker where FIRST_NAME = 'Amitabh'; 6.Select RTRIM(FIRST_NAME) from Worker; 7.Select LTRIM(DEPARTMENT) from Worker; 8.Select distinct length(DEPARTMENT) from Worker; 9.Select REPLACE(FIRST_NAME,'a','A') from Worker; 10.Select CONCAT(FIRST_NAME, ' ', LAST_NAME) AS 'COMPLETE_NAME' from Worker; 11.Select * from Worker order by FIRST_NAME asc; 12.Select * from Worker order by FIRST_NAME asc,DEPARTMENT desc; 13.Select * from Worker where FIRST_NAME in ('Vipul','Satish'); 14.Select * from Worker where FIRST_NAME not in ('Vipul','Satish'); 15. Select * from Worker where DEPARTMENT like 'Admin%'; 16. Select * from Worker where FIRST_NAME like '%a%'; 17. Select * from Worker where FIRST_NAME like '%i'; 18. Select * from Worker where FIRST_NAME like '_h'; 19. Select * from Worker where SALARY between 100000 and 500000; 20.Select * from Worker where year(JOINING_DATE) = 2014 and month(JOINING_DATE) = 2; 21. SELECT COUNT(*) FROM worker WHERE DEPARTMENT = 'Admin'; 22. SELECT CONCAT(FIRST_NAME, ' ', LAST_NAME) As Worker_Name, Salary FROM worker WHERE WORKER_ID IN (SELECT WORKER_ID FROM worker WHERE Salary BETWEEN 50000 AND 100000); 23. SELECT DEPARTMENT, count(WORKER_ID) No_Of_Workers FROM worker GROUP BY DEPARTMENT ORDER BY No_Of_Workers DESC; 24. SELECT DISTINCT W.FIRST_NAME, T.WORKER_TITLE FROM Worker W INNER JOIN Title T ON W.WORKER_ID = T.WORKER_REF_ID AND T.WORKER_TITLE in ('Manager'); 25. SELECT WORKER_TITLE, AFFECTED_FROM, COUNT(*) FROM Title GROUP BY WORKER_TITLE, AFFECTED_FROM HAVING COUNT(*) > 1;