Assistance with SQL statements Previous code; CREATE TABLE Lab4program_piershed ( code VARCHAR(5) PRIMARY KEY, name VARCHAR(20), tel VARCHAR(20) ); CREATE TABLE Lab4course_xxxx ( cid INT PRIMARY KEY, name VARCHAR(20), credits INT, mycode VARCHAR(5), FOREIGN KEY (mycode) REFERENCES Lab4program_xxxx(code) ); 1. Insert 3 records into your Lab4program_xxxx table. The code should be "CPS", "TECH", "MATH", or "BIO". Please copy/paste your SQL statements with the output message in the text that shows you successfully insert 3 records to your table. 2. Insert 4 records into your Lab4course_xxxx table with at least 2 different mycode. Please copy/paste your SQL statements with the output message in the text that shows you successfully insert 4 records to your table.

Problem 92 Write a query to display the author last name, first name, book title, and replacement cost for each book. Sort the results by book number and then author ID (Figure P7.92). (25 rows)

The database contains a Horse table, with columns: ID integer, primary key Registered Name variable-length string. The database contains a Student table, with columns: • ID integer, primary key First Name - variable-length string LastName variable-length string Write a SQL Query to create a Schedule table, with columns: HorseID - integer with range 0 to 65535, not NULL • Student ID integer with range 0 to 65535 Lesson DateTime - date/time, not NULL (HorseID, Lesson DateTime) is the primary key Also, create the following foreign key constraints on Schedule columns: HorseID references Horse. When an ID is deleted from Horse matching Lesson Schedule rows are deleted. Student ID references Student. When an ID is deleted from Student, matching Student ID 's are set to NULL).

1-Write the syntax to insert into a relational table called students the address column references an object table called addresses that was created using an address_type. Other columns exist in the students table, but you are only inserting into the ones below. Aliases used should be the first letter of the table name, eg students would be s, addresses would be a Relational Table Name students attribute student_id attribute surname attribute address

101. The basic operations that can be performed on relations are a. deletion b. insertion c. modification d. all of above

Task 2: The Car Maintenance team also wants to store the actual maintenance operations in the database. The team wants to start with a table to store CAR_ID (CHAR(5)), MAINTENANCE_TYPE_ID (CHAR(5)) and MAINTENANCE_DUE (DATE) date for the operation. Create a new table named MAINTENANCES. The PRIMARY_KEY should be the combination of the three fields. The CAR_ID and MAINTENANCE_TYPE_ID should be foreign keys to their original tables. Cascade update and cascade delete the foreign keys. SQL DataBase Test: Create a new table to store maintenance operations Test Query: DESCRIBE MAINTENANCES Expected Results Field Type Null Key Default Extra CAR_ID char(5) NO PRI NULL MAINTENANCE_TYPE_ID char(5) NO PRI NULL MAINTENANCE_DUE date NO PRI NULL

What are some common uses for the STRING_AGG function in SQL?

SQL ( ORACLE SQL DEVELOPER ) : THIS IS FIRST , SECOND AND THIRD TABLE for the Task 1 to solve : create table author_45( auth_name char(50), gender char(1), age number(3), nationality char(50), PRIMARY KEY (auth_name) ); create table book_45( isbn number(10), title char(50), ver_no number(2), publisher char(50), PRIMARY KEY (isbn) ); create table writtenBy_45( isbn number(10), auth_name char(50), yr_published number(4), price number(3), no_copy number(4), PRIMARY KEY (isbn,auth_name), foreign key (isbn) references book_45, foreign key (auth_name) references author_45 ); insert into author_45 values ('Benoit','F',42,'French'); insert into author_ 45 values ('Antonini','M',33,'Italian'); insert into author_ 45 values ('McAven','F',31,'Australian'); insert into author_ 45 values ('Williams','M',90,'British'); insert into book_45 values (101,'atabases',1,'Possum'); insert into book_45 values (202,'SQL Primer',2,'Hall'); insert into book_45 values...

Task 2: The Car Maintenance team also wants to store the actual maintenance operations in the database. The team wants to start with a table to store CAR_ID (CHAR(5)), MAINTENANCE_TYPE_ID (CHAR(5)) and MAINTENANCE_DUE (DATE) date for the operation. Create a new table named MAINTENANCES. The PRIMARY_KEY should be the combination of the three fields. The CAR_ID and MAINTENANCE_TYPE_ID should be foreign keys to their original tables. Cascade update and cascade delete the foreign keys. Create a new table to store maintenance operations Test Query DESCRIBE MAINTENANCES Expected Results Field Type Null Key Default Extra CAR_ID char(5) NO PRI NULL MAINTENANCE_TYPE_ID char(5) NO PRI NULL MAINTENANCE_DUE date NO PRI NULL

Task 2: The Car Maintenance team also wants to store the actual maintenance operations in the database. The team wants to start with a table to store CAR_ID (CHAR(5)), MAINTENANCE_TYPE_ID (CHAR(5)) and MAINTENANCE_DUE (DATE) date for the operation. Create a new table named MAINTENANCES. The PRIMARY_KEY should be the combination of the three fields. The CAR_ID and MAINTENACNE_TYPE_ID should be foreign keys to their original tables. Cascade update and cascade delete the foreign keys. Answer in MYSQL please

MySQL CREATE TABLE students ( id INT PRIMARY KEY, first_name VARCHAR(50), last_name VARCHAR(50), age INT, major VARCHAR(50), faculty VARCHAR(50)); CREATE TABLE location ( id INT PRIMARY KEY, name VARCHAR(50), rooms INT); CREATE TABLE faculty ( id INT PRIMARY KEY, name VARCHAR(50), department_id INT); 4. Find the number of employees in each department who get no commission or have salary less than5000.5. Find the maximum salary of employees in each department that the employee was hired 15 yearsbefore now. *hint: user TIMESTAMPDIFF(<unit type>,<Date_value 1>,<Date_value 2>), the unitcan be YEAR, MONTH, DAY, HOUR, etc...6. Find the last name of all employees that were not hired on Tuesday *hint: (UseDATE_FORMAT() function amd the format you need is %W, also use UPPER()7. Find the number of employees in each department who have a manager.8. Find the number of employees for each manager whose employees' minimum salary is greaterthan 5000.9. Find the number of...

C sharp Table: Student (the headers are the field names in the Students table) StudentID Name Age Gender ProgramID 791 Stephanie Brown 19 Female BCS 236 Shannon Dawn 25 Female BA 618 Geoff Berg 24 Male ARET 256 Andrew Schilling 22 Male BSC 902 Gary Sang 23 Male DAAD Note: There is a StudentDataSet with a Student table, a StudentTableAdapter, a StudentBindingSource, and a StudentDataGridView control on the form.Note: There is an Average query, named Average, that returns the average age of the student from the Student table.Note: There is also Max query, named Highest, that returns the highest age of the student from the Student table.Write the code you would place in the AverageButton click event on your form to call the Average query and the Highest query and display in a DifferenceLabel, the difference between the highest age of a student and the average age.