See the following table CREATE TABLE product ( product_id NUMBER(3), product_name VARCHAR(30) NOT NULL, price NUMBER (3,2) CHECK (price > 0), quantity NUMBER(3)
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- Repeat Exercise 6, but this time use the EXISTS operator in your query.Your company needs to contact the 1st author of each textbook and to discuss reprint options. Pull out the 1st author for all the books. Please note that the order of authors has been marked as integer values 1, 2, 3, etc. in the field Auth_Order in the AuthorshipLog table. The query result should show the following information in the given order: Book title, 1st author’s last name, 1st author’s email, and 1st author’s phone. Save and name the query 2 First Author Contacts. How would I go about this in Access's query design?Q1. Please write a query statement from emp table to display employee's empno, ename, length of ename, number of 'S',and number of 'L'. SQL> SELECT empno AS "EmpNo", ename AS "Emp Name", ... AS "Name Length", .... AS "# Of S", .... AS "# Of L"FROM emp; [IMAGE below] Material to answer the question above: Material: https://drive.google.com/file/d/14OwWDIybjAIeuHCO7McupV08gJs9nB-8/view?usp=sharing https://drive.google.com/file/d/1npRH8JoJ6GYBzPU3iNjLXluvCyhr2tMu/view?usp=sharing
- Some rows of the STUDENT table are shown below: CODE NAME GPA YEAR 291 ALEX 3.1 2 938 MICHELE 2.3 1 931 JHON 3.3 1 182 JOE 3.4 2 190 REY 2.0 2 330 RON 3.9 3 Which best describes the result of the query below? SELECT YEAR, AVG(GPA) FROM STUDENT WHERE GPA > 2.0 GROUP BY YEAR a. The average maximum GPA of all the students in each year. b. The year of the student with the maximum GPA. c. The average GPA of students with a GPA higher than 2.0 each year. d. The average GPA of the students with a GPA higher than 2.0.In Structure Query Information Select all columns from the jobs table using an asterisk.2. Select only job title, min salary, and max salary columns from the jobs table.3. Select all columns from the d_partners table by writing out all of the columns inthe select statement.4. Using the d_partners table, display the id, and then concatenate the first_nameand last_name together with a space in between the two columns. Give theconcatenated column an alias of NAME. Be sure to use the optional ASkeyword.5. Using the job_grades table, show the grade level, and then subtract thelowest_sal from the highest_sal to show the amount of variance there is for eachgrade level. Give the calculated column an alias of Salary_Variance. Do notuse the optional AS keyword.6. Using the employees table, show an unduplicated list of department_id’s.7. Change the SELECT statement you built for question 4, adding a single columncalled Partner_Information that concatenates the partner’s…Use this info to answer the questionTABLE NAME: students COLUMNS:student-no NUMBER(6) fname VARCHAR2(12) lname VARCHAR(20) sex CHAR(1)major VARCHAR2(24) 7. Write a SQL statement that will display the student number(studentno),firstname(fname),and last name (lname) for all students who are female (F) in the table named students.
- Assume you have these tables.. Q) Write in SQL queries to populate the above tables by inserting minimum 4 tuples in each one. CREATE TABLE Student (studid Char(9) Primary Key,FirstName Varchar2(50) NOT NULL,LastName Varchar2(50) NOT NULL,Email Varchar2(50) NOT NULL UNIQUE,PhoneNumber Number(8) NOT NULL UNIQUE,DateOfBirth Date NOT NULL,GPA Number(1,2) NOT NULL); CREATE TABLE Club (clubID Number(3) Primary Key,ClubName Varchar2(20) NOT NULL,studid Char(9) NOT NULL UNIQUE, Foreign key(studid) references Student(studid)); CREATE TABLE MemberOf (ClubID Number(3) Primary Key, Studid char(9), JoiningDate date, Foreign Key(ClubID) references club(clubID), Foreign Key(Studid) references Student(studid)); CREATE TABLE Activities (Actid Number(3) Primary Key,Acdt DATE NOT NULL,Place Varchar2(50) NOT NULL,durationNbHour Number NOT NULL); CREATE TABLE Organize (actid Number(3) Primary Key, clubID Number(3), Fee Number(4,2), Foreign Key(actid) references Activities(Actid),Foreign Key(clubID)…PLZ HELP WITH THE FOLLOWING: make the following work select MOVIE_NAME, MOVIE_YEAR from MOVIE where MOVIE_COST > (select a.MOVIE_COST from MOVIE a, PRICE b where a.PRICE_CODE = b.PRICE_CODE and b.PRICE_DESC ="weekly Special"); CREATE TABLE MEMBERSHIP ( MEM_NUM CHAR(3) CONSTRAINT MEMBER_MEMNUM_PK PRIMARY KEY, MEM_FNAME VARCHAR(30) NOT NULL, MEM_LNAME VARCHAR(30) NOT NULL, MEM_STREET VARCHAR(30), MEM_CITY VARCHAR(10), MEM_STATE CHAR(2), MEM_ZIP CHAR(5), MEM_BALANCE NUMBER(2) ); CREATE TABLE RENTAL ( RENT_NUM CHAR(4) CONSTRAINT RENTAL_RENTNUM_PK PRIMARY KEY, RENT_DATE DATE, MEM_NUM CHAR(3), CONSTRAINT RENTAL_MEMNUM_FK FOREIGN KEY (MEM_NUM) REFERENCES MEMBERSHIP ); CREATE TABLE PRICE ( PRICE_CODE CHAR(1) CONSTRAINT PRICE_PRICECODE_PK PRIMARY KEY, PRICE_DESC VARCHAR(20), PRICE_RENTFEE NUMBER (3, 1), PRICE_DAILYATFEE NUMBER(3, 1) ); CREATE TABLE MOVIE ( MOVIE_NUM CHAR(4) CONSTRAINT MOVIE_MOVIENUM_PK PRIMARY KEY, MOVIE_NAME VARCHAR(30) NOT NULL,…Janice Jones bought three books in May 2018. It was assigned a sale id of 2 in the volume table. Use PARTITION BY to list the ISBNs of those books, the individual selling price, and the total (do this in one query). CREATE TABLE `volume` ( `inventory_id` int(11) NOT NULL,`ISBN` char(17) DEFAULT NULL,`condition_code` int(11) DEFAULT NULL,`date_acquired` date DEFAULT NULL,`asking_price` decimal(7,2) DEFAULT NULL,`selling_price` decimal(7,2) DEFAULT NULL,`sale_id` int(11) DEFAULT NULL);
- WEEK SQL ASSIGNMENT Complete the following exercise: Alter your student table adding a new column called enroll_date using the datetime data type. Populate the new column (enroll_date) with the current date and time. Using your student table return the student first and last names concatenated with a space and then the enroll_date minus one month. Here is an example of the output: Name Date Jane Smith 2019-05-03 13:14:12 Tom Jones 2020-08-07 05:23:56 Using your student table return the month name for the enroll_date column. For example, if the enroll_date was 11-6-2019 then the month name would be June.Write the SQL code required to list the employee number, last name, first name, and middle initial of all employees whose last names start with Smith. In other words, the rows for both Smith and Smithfield should be included in the listing. Sort the results by employee number. Assume case sensitivity. In the SQL Query do the following Select clause: Emp_Num, Emp_LName, Emp_FName, Emp_Initial From: EMPLOYEE Where clause: last name starts with Smith. (You will be using Like operator) Order by clause: Emp_Num
