(ESSN, PROJECT_NO, HOURS ), Query: SELECT DISTINCT ESSN FROM WORKS_ON WHERE PROJECT_NO IN (1, 2, 3) will: a. Retrieve the SSN of employees who work in project 1, 2, 3 b. Retrieve the SSN of employees having works_on attribute = 1, 2, 3 c. Retrieve the SSN of employees having SSN attribute = 1, 2, 3 d. None of the options
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Give table WORKS_ON (ESSN, PROJECT_NO, HOURS ), Query: SELECT DISTINCT ESSN FROM WORKS_ON WHERE PROJECT_NO IN (1, 2, 3) will:
a. Retrieve the SSN of employees who work in project 1, 2, 3
b. Retrieve the SSN of employees having works_on attribute = 1, 2, 3
c. Retrieve the SSN of employees having SSN attribute = 1, 2, 3
d. None of the options
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- Answer questions 2.1 to 2.4 based on the following schema. Note that primary keys are in BOLD. Note also that attributes with the same name in different tables are foreign keys. For example, doctorID in the Treatment table is a foreign key that refers to the Doctor table. Doctor(doctorId:integer, specialtyID:integer, firstName:varchar(20), lastName:varchar(20), salary:numeric(11,2)) Specialty(specialtyID:integer, specialtyName:varchar(20), specialtyCost:numeric(11,2)) Treatment(doctorId:integer, patientSSN:integer, dateTreated:date, costOfTreatment:numeric(11,2)) Patient(patientSSN:integer, firstName:varchar(20), lastName:varchar(20), street:varchar(30), city:varchar(15), state:char(2),zipCode:integer(5)) 1. Write an SQL statement to increase the cost of all Specialty treatment by 2.5% 2. Write a SQL statement to delete all treatment rows that were done before 2015 and cost less than $500.Answer questions 2.1 to 2.4 based on the following schema. Note that primary keys are in BOLD. Note also that attributes with the same name in different tables are foreign keys. For example, doctorID in the Treatment table is a foreign key that refers to the Doctor table. Doctor(doctorId:integer, specialtyID:integer, firstName:varchar(20), lastName:varchar(20), salary:numeric(11,2)) Specialty(specialtyID:integer, specialtyName:varchar(20), specialtyCost:numeric(11,2)) Treatment(doctorId:integer, patientSSN:integer, dateTreated:date, costOfTreatment:numeric(11,2)) Patient(patientSSN:integer, firstName:varchar(20), lastName:varchar(20), street:varchar(30), city:varchar(15), state:char(2),zipCode:integer(5)) 2.3) Write an SQL statement to increase the cost of all Specialty treatment by 2.5% 2.4) Write a SQL statement to delete all treatment rows that were done before 2015 and cost less than $500. 2.5) Write an SQL statement to select all Specialties where the specialtytName has a null…Question 1 : Consider the following schema: Suppliers(sid: integer, sname: string, address: string) Parts(pid: integer, pname: string, color: string) Catalog(sid: integer, pid: integer, cost: real) The Catalog relation lists the prices charged for parts by Suppliers. Write the following queries in SQL: 1.Find the snames of suppliers who supply every part. 2.Find the sids of suppliers who supply only red parts. 3.Find the price of the least expensive red part 4.Find the sids of suppliers who charge more for some part than the average cost of that part (averaged over all the suppliers who supply that part. 5.For every supplier, print the name of the supplier and the total number of parts that he or she supplies.
- The table STUDENT has the attributes STUDENT_ID, NAME, SCHOOL_ID, and ADDR. The table SCHOOL has the attributes SCHOOL_ID, NAME, and STATE_CODE. Assume that there is a 1:N relation between SCHOOL and STUDENT. Which of the following creates a table that shows the STUDENT attributes and the corresponding SCHOOL attributes? a. SELECT * SCHOOL_ID FROM STUDENT JOIN SCHOOL b. SELECT * SCHOOL_ID FROM STUDENT JOIN SCHOOL USING(SCHOOL_ID) c. SELECT * SCHOOL_ID FROM STUDENT NATURAL JOIN SCHOOL d. SELECT * SCHOOL_ID FROM STUDENT, SCHOOL WHERE SCHOOL_ID = SCHOOL_IDQuestion 1 : Consider the following schema: Suppliers(sid: integer, sname: string, address: string) Parts(pid: integer, pname: string, color: string) Catalog(sid: integer, pid: integer, cost: real) The Catalog relation lists the prices charged for parts by Suppliers. Write the following queries in SQL: 1.Find the names of suppliers who supply every part. 2.Find the sids of suppliers who supply only red parts. 3.Find the price of the least expensive red part 4.Find the sids of suppliers who charge more for some part than the average cost of that part (averaged over all the suppliers who supply that part. 5.For every supplier, print the name of the supplier and the total number of parts that he or she supplies.SQL Queries: Given the following schema. Primary keys are underlined and highlighted in yellow and foreign keys are bold. The meaning of the relations is straightforward. Medical_Staff(MS_Id, MS_First, MS_Last, MS_PhoneNumber, MS_Occupation, Salary) Blood_Bank(Bank_Id, Bank_Name, Bank_Street, Bank_City, Bank_St, Bank_PhoneNumber) Donor(Donor_Id, Donor_First, Donor_Last, Donor_BloodType, Donor_City, Donor_St, DOB, TotalNumberOfDonations) Donation_Record(Donor_Id, Donation_Date, Donation_Amount) Employee(MS_Id, Bank_Id) Storage(Donor_Id, Donation_Date, Bank_Id) Write SQL queries to answer the following questions: a.Create a view to list of all employees for any Blood Bank that is in a city named St. Paul. (i.e., view output includes MS_Last, MS_First, and Bank_Id). Sort by Bank_Id, MS_Last, MS_First. b.Find full names of all donors who donated in 2021. (Hint: use IN) c.Create a stored procedure that takes two inputs: a Medical Staff identifier (MS_Id) and new salary. The…
- SQL Queries: Given the following schema. Primary keys are underlined and highlighted in yellow and foreign keys are bold. The meaning of the relations is straightforward. Medical_Staff(MS_Id, MS_First, MS_Last, MS_PhoneNumber, MS_Occupation, Salary) Blood_Bank(Bank_Id, Bank_Name, Bank_Street, Bank_City, Bank_St, Bank_PhoneNumber) Donor(Donor_Id, Donor_First, Donor_Last, Donor_BloodType, Donor_City, Donor_St, DOB, TotalNumberOfDonations) Donation_Record(Donor_Id, Donation_Date, Donation_Amount) Employee(MS_Id, Bank_Id) Storage(Donor_Id, Donation_Date, Bank_Id) Write SQL queries to answer the following questions: a.Find the average donation amount for Sue Thompson (Donor Id = 3329) for the year 2018. b.Sum all donations by Bank for each State (i.e. the State the blood bank is located in, not the Donor). c.Find the name of the youngest donor and all the donations made in by that donor in 2020.SELECT treatment.service_id, service.service_charge, treatment.actual_chargeFROM treatmentJOIN service ON treatment.service_id = service.service_idWHERE treatment.service_id IN (SELECT service_id FROM service WHERE service_cat_id = 'SUR')AND service.service_charge <> treatment.actual_charge; Alter the above query to only list services where the difference between the service_charge and actual_charge is greater than the average difference between these two charges for services in the ‘SUR’ category of services. The result table should only list the service_id and service_description column from the service table. Hint: Compute the average difference between these two charges for ‘SUR’ services in a subqueryThe table STUDENT has the attributes STUDENT_ID, NAME, S_ID, and ADDR. The table SCHOOL has the attributes SCHOOL_ID, NAME, and STATE_CODE. Assume that there is a 1:N relation between SCHOOL and STUDENT where the common attributes are S_ID, and SCHOOL_ID. Which of the following creates a table that shows the STUDENT attributes and the corresponding SCHOOL attributes? a. SELECT * FROM STUDENT JOIN SCHOOL USING(SCHOOL_ID) b. SELECT * FROM STUDENT JOIN SCHOOL USING(S_ID) c. SELECT * FROM STUDENT NATURAL JOIN SCHOOL ON S_ID AND SCHOOL.SCHOOL_ID d. SELECT * FROM STUDENT JOIN SCHOOL ON STUDENT.S_ID = SCHOOL.SCHOOL_ID
- Write down the relationship type and statement of the following: Ruler/Country: ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… Product/Price: ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… Zoo/Animals: ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… Recipes/ Ingredients: ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… Doctor/Patient: ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… Sim Card/Phone Number: ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… Country/Capital City: ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… Actors/Movies: …………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………Question p .Using Publisher ID as the common field, create a one-to-many relationship between the Publishers table and the Student Papers table. Enforce referential integrity and enable both cascade options. Create a relationship report with normal margins, and then save the report as Relationships Report. Close all open objects. Full explain this question and text typing work only We should answer our question within 2 hours takes more time then we will reduce Rating Dont ignore this lineCreate tables for the following, where the first 2 are entity-types and the last one is relationship. Feel free to add additional fields that you think are appropriate.Course: courseNum, department, semester, maximumEnrolled Instructor: ssn, instructorid, firstName, lastName.Teaches: instructorid, courseNumPay special attention to assigning primary key and foreign key in each table and use appropriate data types.Print out the SQL commands that you use to create the tables.Then populate your tables with at least 2 rows each and print out the contents of each table with SELECT. (Include the content of the print-out in your homework). I want answers for this part Get to know the HR database, its tables, and their contents. Show the SQL query and the output.Use INNER JOIN for multiple table query.1- Display details of jobs where the minimum salary is greater than 10000.2- Display the first name and join date of the employees who joined between 2002 and2005.3- Display first name and join…
