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- a. What is the result of this query:SELECT AVG (S.rating)FROM Sailors S b. What is the result of this query:SELECT COUNT (S.rating)FROM Sailors S c. Show the full outer join of S with itself, with the join condition being sid=sid .The OR_CASE table has one bro per surgical case. The OR_LOG table has one row per surgical log. It's LOG_ID column identifies each log. If a case has an associated log, OR_CASE.LOG_ID will store the logs ID. Otherwise, it will be null. A query uses the following join: FROM OR_CASE INNER JOIN OR_LOG ON OR_CASE.LOG_ID=OR_LOG.LOG_ID which of the following would be the results? A. A log with no associated case.B. A log with an associated case. C. A case with no associated log.D. Rows with no case or log.(A) Write a query which shows the employee IDs that are unique to the employee table. Order the employee IDs in descending order. An employee ID Is the same in both tables if the integer value of the ID matches. (B) We want to add a new column to the employee table. We want to provide a new column with a more complete phone number. Right now the PHONENO column only shows the last 4 digits. We want a new column which is called PHONE and consists of ###-###-####. The last 4-digits are already in the PHONENO column. The first three digits should be 416 and the next three should be 123. To improve clarity in the table, we also want to rename the PHONENO column to PHONEEXT. Show all the commands used to accomplish this, then, select all data for employees who have the last name of 'smith' (case insensitive). (C) Show a list of employee id, names, department, years and job of any employee in the staff table who makes a total amount more than their manager or has more years of service than…
- reate a query that SELECT all of the EMPLOYEES from the EMPLOYEE TABLE that are Mangers. Include Employee ID first name, Lastname and Salary (Hint: Self Join) /* Write a query the gives the Employee ID first name, Lastname and Salary and Project Number of EMPLOYEES who aren't currently assigned to a project (hint outer join) Create a query that lists the lastnme, edlevel , job, the number of years they've worked as of Jan 01/2002 ( hint : year function Jan 01/2002 minus hiredate), and their salary. Get the employees that have the same Job as the employee named starts with J (hint subquery from employee) and hiredate < Jan 01/2002 Sort the listing by highest salary first. */How many inner joins are possible in the attached scenario? write a query for inner join, left join, and right join?Add two columns to the EMPLOYEES table. One column, named EmpDate, contains thedate of employment for each employee, and its default value should be the system date. Thesecond column, named EndDate, contains employees’ date of termination. When I put this formal it says, line 2 has invalid identifier! Can you help me what I need to fix from line 2? alter table EMPLOYEES2add column EmpDate date = getdate(),add column EndDate;
- The 'customer' table contains ten columns but no rows. The table is then updated with 10 new rows and 3 rows discarded. What is the table'customer's degree and cardinality?Hello, can you please assit me with this. I am struggling to get the code to run correctly without errors: 1. Write a query using a join that displays Department names (not Department_IDs) Employees’ IDs, last & first names Sort your results by department name and employee last and first name. NOTE: INNER JOIN is OK here. NOTE: I suggest you use table aliases.*/ /*2. Write a query using a join that displays Department names (no DEPARTMENT_IDs). Job titles (not Job_IDs) Employees’ IDs, last & first names, Sort your results by department name, job title, and employee last and first name. NOTE: INNER JOIN is OK here. NOTE: I suggest you use table aliases.*/ /*3. Write a query using a self-join to display Employees’ IDs and names along with Their manager’s ID and name. Use column aliases so that it is clear which columns are for the Manager and which columns are for the Employee.…Use the following quert to answer the question below it: SELECT *FROM HSP_ACCT_CVG_LISTLEFT OUTER JOIN CLARITY_EMPON COVERAGE.PAYOR_ID = CLARITY_EMP.PAYOR_IDLEFT OUTER JOIN COVERAGEON HSP_ACCT_CVG_LIST.COVERAGE_ID = COVERAGE.COVERAGE_ID Why does this query fail to run?
- Q: This is an SQL question, which is correct: Which of the following statements about JOINs are true? a. If NATURAL JOIN and FULL OUTER JOIN have the same number of rows, then LEFT OUTER JOIN AND RIGHT OUTER JOIN will not have the same number of rows. b. Number of rows in a NATURAL JOIN is always less than the number of rows in a FULL OUTER JOIN c. LEFT OUTER JOIN and RIGHT OUTER JOIN will always have the same number of rows d. Number of rows in a NATURAL JOIN can be equal to the number of rows in a FULL OUTER JOIN e. LEFT OUTER JOIN can produce more rows than FULL OUTER JOINFROM customer WHERE city IN (SELECT city FROM employee WHERE reports_to= 2); 2. SELECT customer.* FROM customer INNER JOIN employee ON customer.city = employee.city WHERE reports_to = 2; Query #1 would be more efficient as it is based on primary and foreign keys. O Query #1 would be more efficient as it is not using indexed columns. Both would be the same as both use the same indices for the join and filter. Query #2 would be more efficient as it is based on primary and foreign keys. 70°F Cloudy QLCreate a query for displaying e1.first_name with e1.last_name of employee under column alias “Employee Full Name” by using concatenation from employees table with a table alias e1 and e2.first_name with e2.last_name of employee under column alias “Manager Full Name” self joining with employees table with a table alias e2 on the basis of equality of manager_id from e1 with employee_id from e2. Use order by clause to display list according to first name of manager e2.first_name in ascending order.