Build the Students Table that is shown in the picture using python SQLhint:The SQL command to build the table iscreate table student( SID int primary key not null, Name varchar(20), GPA char(3), Major varchar(25)) Q2) Build the Students Table that is shown below.SIDNameGPAMajor1Mike3.4Accounting2Mary4.0 Information Systems3 William2.3Finance4Joan4.0MarketingHint: The SQL command to build the table iscreate table student( SID int primary key not null, Name varchar(20), GPA char(3), Major varchar(25))in [ ]:

Since the table is stored in db, there is no output. I'll directly post python code.

Build the Students Table that is shown in the picture using python SQL hint: The SQL command to build the table is create table student( SID int primary key not null, Name varchar(20), GPA char(3), Major varchar(25))

Build the Students Table that is shown in the picture using python SQL hint: The SQL command to build the table is create table student( SID int primary key not null, Name varchar(20), GPA char(3), Major varchar(25))

Database Systems: Design, Implementation, & Management

12th Edition

ISBN:9781305627482

Author:Carlos Coronel, Steven Morris

Publisher:Carlos Coronel, Steven Morris

Chapter11: Database Performance Tuning And Query Optimization

Section: Chapter Questions

Problem 7RQ: How is the processing of SQL DDL statements (such as CREATE TABLE) different from the processing...

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/* Created with SQL Script Builder v.1.5 */ /* Type of SQL : SQL Server */ CREATE TABLE EMPLOYEE ( EMP_CODE int, EMP_TITLE varchar(4), EMP_LNAME varchar(15), EMP_FNAME varchar(15), EMP_INITIAL varchar(1), EMP_DOB datetime, JOB_CODE varchar(5), STORE_CODE int ); INSERT INTO EMPLOYEE VALUES('1','Mr.','Williamson','John','W','5/21/1964','SEC','3'); INSERT INTO EMPLOYEE VALUES('2','Ms.','Ratula','Nancy','','2/9/1969','MGR','2'); INSERT INTO EMPLOYEE VALUES('3','Ms.','Greenboro','Lottie','R','10/2/1961','GEN','4'); INSERT INTO EMPLOYEE VALUES('4','Mrs.','Rumpersfro','Jennie','S','6/1/1971','GEN','5'); INSERT INTO EMPLOYEE VALUES('5','Mr.','Smith','Robert','L','11/23/1959','GEN','3'); INSERT INTO EMPLOYEE VALUES('6','Mr.','Renselaer','Cary','A','12/25/1965','GEN','1'); INSERT INTO EMPLOYEE VALUES('7','Mr.','Ogallo','Roberto','S','7/31/1962','MGR','3'); INSERT INTO EMPLOYEE VALUES('8','Ms.','Johnsson','Elizabeth','I','9/10/1968','SEC','1'); INSERT INTO EMPLOYEE…
/* Created with SQL Script Builder v.1.5 */ /* Type of SQL : SQL Server */ CREATE TABLE EMPLOYEE ( EMP_CODE int, EMP_TITLE varchar(4), EMP_LNAME varchar(15), EMP_FNAME varchar(15), EMP_INITIAL varchar(1), EMP_DOB datetime, JOB_CODE varchar(5), STORE_CODE int ); INSERT INTO EMPLOYEE VALUES('1','Mr.','Williamson','John','W','5/21/1964','SEC','3'); INSERT INTO EMPLOYEE VALUES('2','Ms.','Ratula','Nancy','','2/9/1969','MGR','2'); INSERT INTO EMPLOYEE VALUES('3','Ms.','Greenboro','Lottie','R','10/2/1961','GEN','4'); INSERT INTO EMPLOYEE VALUES('4','Mrs.','Rumpersfro','Jennie','S','6/1/1971','GEN','5'); INSERT INTO EMPLOYEE VALUES('5','Mr.','Smith','Robert','L','11/23/1959','GEN','3'); INSERT INTO EMPLOYEE VALUES('6','Mr.','Renselaer','Cary','A','12/25/1965','GEN','1'); INSERT INTO EMPLOYEE VALUES('7','Mr.','Ogallo','Roberto','S','7/31/1962','MGR','3'); INSERT INTO EMPLOYEE VALUES('8','Ms.','Johnsson','Elizabeth','I','9/10/1968','SEC','1'); INSERT INTO EMPLOYEE…
/* Created with SQL Script Builder v.1.5 */ /* Type of SQL : SQL Server */ CREATE TABLE EMPLOYEE ( EMP_CODE int, EMP_TITLE varchar(4), EMP_LNAME varchar(15), EMP_FNAME varchar(15), EMP_INITIAL varchar(1), EMP_DOB datetime, JOB_CODE varchar(5), STORE_CODE int ); INSERT INTO EMPLOYEE VALUES('1','Mr.','Williamson','John','W','5/21/1964','SEC','3'); INSERT INTO EMPLOYEE VALUES('2','Ms.','Ratula','Nancy','','2/9/1969','MGR','2'); INSERT INTO EMPLOYEE VALUES('3','Ms.','Greenboro','Lottie','R','10/2/1961','GEN','4'); INSERT INTO EMPLOYEE VALUES('4','Mrs.','Rumpersfro','Jennie','S','6/1/1971','GEN','5'); INSERT INTO EMPLOYEE VALUES('5','Mr.','Smith','Robert','L','11/23/1959','GEN','3'); INSERT INTO EMPLOYEE VALUES('6','Mr.','Renselaer','Cary','A','12/25/1965','GEN','1'); INSERT INTO EMPLOYEE VALUES('7','Mr.','Ogallo','Roberto','S','7/31/1962','MGR','3'); INSERT INTO EMPLOYEE VALUES('8','Ms.','Johnsson','Elizabeth','I','9/10/1968','SEC','1'); INSERT INTO EMPLOYEE…
can you help me iwth these two queries in mySQL. Write a SQL statement to create a table job_history including columns employee_id, start_date, end_date, job_id and department_id and make sure that, the employee_id column does not contain any duplicate value at the time of insertion and the foreign key column job_id contain only those values which are exists in the jobs table. Write a SQL statement to create a table employees including columns employee_id, first_name, last_name, email, phone_number hire_date, job_id, salary, commission, manager_id and department_id and make sure that, the employee_id column does not contain any duplicate value at the time of insertion and the foreign key columns combined by department_id and manager_id columns contain only those unique combination values, which combinations are exists in the departments table.
51. The given SQL statement is intended to either (1) list the customer numbers, customers names, and headquarter cities of customers headquartered in Los Angeles or (2) satisfy the two conditions of having a customer number higher than 1000 and being headquartered in Denver. SELECT CUSTNUM, CUSTNAME, HQCITY FROM CUSTOMER WHERE HQCITY= 'Los Angeles" Which set of lines should be added to the end of the statement to make the statement work as intended? A: (CUSTNUM>1000 AND HQCITY-'Denver'); B: AND (CUSTNUM>1000 AND HQCITY='Denver*); C:OR CUSTNUM>1000 OR HQCITY='Denver": D:OR (CUSTNUM> 1000 AND HQCITY='Denver*);
Below is a schema for an HR database:employee(empid, fname, lname, managerid, departmentid, employee_rank)It's an employee table, which has employee id, first name, last name, manager id (which is an employee id), department id, and employee_rank, such as VP, CEO, SVP, etc.Using SQL, answer this question (write a SQL query that answers this question) [tip: use a recursive query]. 11. For employee 42, find the path-of-managers directly to the CEO?
Given the following SQL query: Student (sid, name, age, address) Book (bid, title, author) Checkout (sid, bid, date) SELECT S.name FROM Student S, Book B, Checkout C WHERE S.sid = C.sid AND B.bid = C.bid AND B.author = ’Olden Fames’ AND S.age > 12 AND S.age < 20 Write relational algebra query for the above SQL query. Using a drawing tool such as Visio or others (ex: https://www.draw.io/ is a free one) to draw initial relational algebra query trees for the query generated in question 1. Transform the initial query trees created above into a more efficient form (use the heuristic rules and show your steps (.
You are given the schema of a table with its columns inside the bracket as follows: STUDENT(Name, Student_id, Major, Marks, Email) (The Primary Key is underlined) Write an SQL query for the following tasks: (Hint: the queries are nested) 1. Write an SQL query to retrieve the name, student id and the maximum marks of students of each major where that maximum marks is greater than all students with CS major. 2. Write an SQL query to retrieve the name, student id and the maximum marks of students with ECE major where that maximum marks is greater than all students with CS major.
Consider the following tables. Primary keys are identified after each table. Hotel(Hnum,HName,Location) PK = HNum Room(RNum,HNum,Type,View,Price) PK = RNUM , HNum Booking(id,HNum,GNum,RNum,DateFrom,DateTo) PK = id Guest(GNum,Fname,Lname,Gender,DOB,City) PK = GNum Write SQL queries AND Relational algebra expressions (using the relational algebra operators: ρ, σ, π, , υ, ∩, sum, avg, count) for each of the following queries a) List all information about the rooms that have a price below the average price of all single rooms b) For each Hotel, List the total number of rooms c) List the names of hotels that have no booking in them d) List the first name and last name of Parisian guests who book a double room in Paris PLEASE ANSWER THE QUESTION BY WRITING SQL QUERIES AND RELATIONAL ALGEBRA EXPRESSION. Thank you.
2) Write SQL query that returns the Name of ALL employees (based on EMP_LNAME and EMP_FNAME in EMPLOYEE table, call this EMP_NAME), Gender of ALL employees (based on EMP_TITLE in EMPLOYEE table, call this EMP_NAME), and Pilot ratings if any (PIL_RATINGS in PILOT table). Output: EMP_NAME EMP_GENDER Rhett VanDam M John Lange M Jeanine Duzak F
Consider the following schema and write SQL query for given statements. Student ( regNo, sName, sCity, sCGPA, sAge)Course ( cCode, cName, credit_hours)Enrollment (regNo, cCode, semester,quality_points,Grade) 1- Display 2nd row of Student table using Cursors. 2- Display all records of Enrollment table which are placed at even rows. 3- Execute given query in SQL and do as directed.a) UPDATE STUDENT SET ECITY = ‘ISB’ WHERE EID>101b) Count how many records have been updated in above query using cursor. 4- Write a mechanism which must not allow user to delete more than 1 records using cursor and transaction.
Question 1 Write SQL statements to answer questions 1.1, 1.2, 1.3 and 1.4 based on the following schema. Primary keys are( BOLD and italicized) and attributes with the same names in different tables are foreign keys. Doctors(doctorId, doctorLastName, doctorFirstName,specialtyID, numberOfPatientsSeen) Patients(patientId, patientLastName, patientFirstName, sex, patientBalance, numberOfVisits, patientDateOfBirth ) Records(doctorId, patientId, date, diagnosis, visitCost) Specialty(specialtyId, specialtyName, specialtyCost) 1.1) Create a view, called PatientSmithDoctors, to list all the doctors a patient with the last name of Smith has seen (be careful on this one). 1.2) Write a trigger, each time a row is inserted in the table Records, the numberOfVisits in the corresponding Patient row is updated by 1 (i.e. add 1 to the numberOfVisits) and the numberOfPatientsSeen in the Doctors table is also updated by 1. 1.3) Find the Patient’s full name with the…

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