Show the list of students (all attributes) who were born after 2000-01-01 and are currently in Active status in the alphabetical order of the last and the first name. Hint: To calculate easily, always use the date format YYYY-MM-DD
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- Show the list of students (all attributes) who were born after 2000-01-01 and are currently in Active status in the alphabetical order of the last and the first name. Hint: To calculate easily, always use the date format YYYY-MM-DD
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- Which of the following data dictionary objects should be used to view information about the constraints in a database? USER_TABLES USER_RULES USER_COLUMNS USER_CONSTRAINTS None of the above objects should be used.In the initial creation of a table, if a UNIQUE constraint is included for a composite column that requires the combination of entries in the specified columns to be unique, which of the following statements is correct? a. The constraint can be created only with the ALTER TABLE command. b. The constraint can be created only with the table-level approach. c. The constraint can be created only with the column-level approach. d. The constraint can be created only with the ALTER TABLE MODIFY command.What is the purpose of a column alias?
- 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).PL/SQL Help I need to create a PL/SQL trigger called aprove_rental that will prevent an INSERT if a clients rating (client_credit_rtg) is less than 1.5 from the rental table. Im using sqlplus to test this trigger. Tables : create table client (client_num char(4), client_lname varchar2(15) not null, client_fname varchar2(15), client_phone char(8), client_credit_rtg decimal(2,1) check(client_credit_rtg between 0.0 and 5.0), client_fave_cat char(3), primary key (client_num), foreign key (client_fave_cat) references movie_category(category_code) ); create table rental (rental_num char(7), client_num char(4), vid_id char(6), date_out date not null, date_due date not null, date_returned date, primary key (rental_num), foreign key (client_num) references client, foreign key (vid_id) references video ); PLEASE WRITE THIS FROM SCRATCH AND NOT JUST COPY AND PASTE FROM ANOTHER ANSWER, BECAUSE THEY DONT WORK FOR MELX SQLLuery1.sql - ML-RefVm-820127.AdventureWorks2017 (ML-RefVm-820127student (57))* - Microsoft SQL Server Management Studio (Administrator) File Edit View Query Project Tools Window Help
- Task 4: The Driver Relationship team wants to have quick search options for the active drivers. The team specifically mentioned that they are using first name, last name and driving license ID to search the drivers. Create an index called NameSearch on the ACTIVE_DRIVERS table created in task 3. SQL Database TestIncomplete Create an INDEX to search the ACTIVE_DRIVERS table Test Query SHOW INDEX FROM ACTIVE_DRIVERS Table Non_unique Key_name Seq_in_index Column_name Collation Cardinality Sub_part Packed Null Index_type Comment Index_comment Visible Expression ACTIVE_DRIVERS 0 PRIMARY 1 DRIVER_ID A 3 NULL NULL BTREE YES NULL ACTIVE_DRIVERS 1 NameSearch 1 DRIVER_FIRST_NAME A 3 NULL NULL YES BTREE YES NULL ACTIVE_DRIVERS 1 NameSearch 2 DRIVER_LAST_NAME A 3 NULL NULL YES BTREE YES NULL ACTIVE_DRIVERS 1 NameSearch 3 DRIVER_DRIVING_LICENSE_ID A 3 NULL NULL YES BTREE YES NULLCreate an ERD in SQL with primary keys. ReservationID HotelID HotelAddress CheckInDate LengthOfStay RoomNumber RoomCapacity PoolInHotel HotelManager RatePerNightExclTaxExclDiscount RateType PercentDiscountOnRoomRate RoomTaxPercent RoomService CustomerID CustomerName RewardsID RewardsPointsEarned DateOfLastStay CustomeRatingOfStaySubmit Result Key Here.. Subm SQL Injection Escaping Challenge To complete this challenge, you must exploit SQL injection flaw in the following form to find the result key. The developer of this level has attempted to stop SQL Injection attacks by escaping apostrophes so the database interpreter will know not to pay attention to user submitted apostrophes Challenge Hint This is the query you are injecting code into! Be aware that your apostrophes are being escaped with a leading backslash before being sent to the interpreter SELECT FROM customers WHERE customerld =" Please enter the Customer Id of the user that you want to look up Get user There were no results found in your search
- CREATE 3 tables as listed below in your own database. */ CREATE TABLE Customer (CustomerID VARCHAR(20) PRIMARY KEY, CustomerLName VARCHAR(30), CustomerFName VARCHAR(30), CustomerStatus VARCHAR(10)); CREATE TABLE SaleOrder (OrderID INT IDENTITY PRIMARY KEY, CustomerID VARCHAR(20) REFERENCES Customer(CustomerID), OrderDate DATE, OrderAmountBeforeTax INT); CREATE TABLE SaleOrderDetail (OrderID INT REFERENCES SaleOrder(OrderID), ProductID INT, Quantity INT, UnitPrice INT, PRIMARY KEY (OrderID, ProductID)); /* Write a trigger to put the total sale order amount before tax (unit price * quantity for all items included in an order) in the OrderAmountBeforeTax column of SaleOrder. */A simple blog database A sql-5-7.sql file has been opened for you. In the sql-5-7.sql file: 1. Use the blog database 2. Write a SQL RIGHT OUTER JOIN statement that joins the user_id column from the blog.posts table, the name column of the blog.users table and the body column of the blog posts table together.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.