8. The sql databawe ss is required.
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- 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).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.Create a table “books” in the database (name:ebookshop) with the following fieldsTable 1.• Write a code to establish a connection with the data source you want to use.• Create a webform request for the user to check the book information by id.• Write a SQL query and execute the query via the SQL SELECT Statement and SQL UPDATEStatement.• Process the query result.
- 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.SQL: Dog DataIn each question below, you will define a new table based on the following tables.CREATE TABLE parents ASSELECT "abraham" AS parent, "barack" AS child UNIONSELECT "abraham" , "clinton" UNIONSELECT "delano" , "herbert" UNIONSELECT "fillmore" , "abraham" UNIONSELECT "fillmore" , "delano" UNIONSELECT "fillmore" , "grover" UNIONSELECT "eisenhower" , "fillmore";CREATE TABLE dogs ASSELECT "abraham" AS name, "long" AS fur, 26 AS height UNIONSELECT "barack" , "short" , 52 UNIONSELECT "clinton" , "long" , 47 UNIONSELECT "delano" , "long" , 46 UNIONSELECT "eisenhower" , "short" , 35 UNIONSELECT "fillmore" , "curly" , 32 UNIONSELECT "grover" , "short" , 28 UNIONSELECT "herbert" , "curly" , 31;CREATE TABLE sizes ASSELECT "toy" AS size, 24 AS min, 28 AS max UNIONSELECT "mini" , 28 , 35 UNIONSELECT "medium" , 35 , 45 UNIONSELECT "standard" , 45 , 60; Q3: SentencesThere are two pairs of siblings that have the same size. Create a table that contains a row with a string foreach of…SQL: Dog DataIn each question below, you will define a new table based on the following tables.CREATE TABLE parents ASSELECT "abraham" AS parent, "barack" AS child UNIONSELECT "abraham" , "clinton" UNIONSELECT "delano" , "herbert" UNIONSELECT "fillmore" , "abraham" UNIONSELECT "fillmore" , "delano" UNIONSELECT "fillmore" , "grover" UNIONSELECT "eisenhower" , "fillmore";CREATE TABLE dogs ASSELECT "abraham" AS name, "long" AS fur, 26 AS height UNIONSELECT "barack" , "short" , 52 UNIONSELECT "clinton" , "long" , 47 UNIONSELECT "delano" , "long" , 46 UNIONSELECT "eisenhower" , "short" , 35 UNIONSELECT "fillmore" , "curly" , 32 UNIONSELECT "grover" , "short" , 28 UNIONSELECT "herbert" , "curly" , 31;CREATE TABLE sizes ASSELECT "toy" AS size, 24 AS min, 28 AS max UNIONSELECT "mini" , 28 , 35 UNIONSELECT "medium" , 35 , 45 UNIONSELECT "standard" , 45 , 60;Q1: Size of DogsThe Fédération Cynologique Internationale classifies a standard poodle as over 45 cm and up to 60 cm.The sizes table…
- Create an anonymous block that returns the number of students in a section. Prompt for section id.Name the file P2.1.sql. Create an anonymous block that return the average numeric grade for a section. Prompt for sectionid and return the average grade. Name the file P2.2.sqlSQL: Dog DataIn each question below, you will define a new table based on the following tables.CREATE TABLE parents ASSELECT "abraham" AS parent, "barack" AS child UNIONSELECT "abraham" , "clinton" UNIONSELECT "delano" , "herbert" UNIONSELECT "fillmore" , "abraham" UNIONSELECT "fillmore" , "delano" UNIONSELECT "fillmore" , "grover" UNIONSELECT "eisenhower" , "fillmore";CREATE TABLE dogs ASSELECT "abraham" AS name, "long" AS fur, 26 AS height UNIONSELECT "barack" , "short" , 52 UNIONSELECT "clinton" , "long" , 47 UNIONSELECT "delano" , "long" , 46 UNIONSELECT "eisenhower" , "short" , 35 UNIONSELECT "fillmore" , "curly" , 32 UNIONSELECT "grover" , "short" , 28 UNIONSELECT "herbert" , "curly" , 31;CREATE TABLE sizes ASSELECT "toy" AS size, 24 AS min, 28 AS max UNIONSELECT "mini" , 28 , 35 UNIONSELECT "medium" , 35 , 45 UNIONSELECT "standard" , 45 , 60; Q2: By Parent HeightCreate a table by_parent_height that has a column of the names of all dogs that have a parent,ordered by the height…Using php and sql: create a query and any supporting PHP code that will output only messages from the user named “Abla”.
- Create a CRUD API to manage recipes. Design the database and create your createDB.sql file. Seed your database with some initial records. Your application should store these recipe fields: id name ingredient_1 ingredient_2 ingredient_3 recipe_author_first_name recipe_author_last_name recipe_category_name Follow RESTful conventions to set up the five standard routes for a CRUD api. Make sure your API responds appropriately to various situations: Errors Invalid requests Etc.The primary keys are identified below. The following data types are defined in the SQL Server. tblLevels Level – Identity PKClassName – text 20 – nulls are not allowed tblPoolPool – Identity PKPoolName – text 20 – nulls are not allowedLocation – text 30 tblStaffStaffID – Identity PKFirstName – text 20MiddleInitial – text 3LastName – text 30Suffix – text 3Salaried – BitPayAmount – money tblClassesLessonIndex – Identity PKLevel – Integer FKSectionID – IntegerSemester – TinyIntDays – text 20Time – datetime (formatted for time)Pool – Integer FKInstructor – Integer FKLimit – TinyIntEnrolled – TinyIntPrice – money tblEnrollmentLessonIndex – Integer FKSID – Integer FK (LessonIndex and SID) Primary KeyStatus – text 30Charged – bitAmountPaid – moneyDateEnrolled – datetime tblStudentsSID – Identity PKFirstName – text 20MiddleInitial – text 3LastName – text 30Suffix – text 3Birthday – datetimeLocalStreet – text 30LocalCity – text 20LocalPostalCode – text 6LocalPhone – text 10 Implement this…Please help with the following: Using oracle sql live or sql developer what sql statements could be use to find the following: A simple database composed of the following tables: PATIENT, DOCTOR, DRUG, and PRESCRIPTION, is now loaded with this assignment and you are ready to start. Please explore the database and then write the appropriate SQL command. List doctor’s details where speciality is either Dermatology or Neurology. (Use INoperator). List the results by the first name in ascending order and the last name indescending order sql doc below Beginfor c in (select table_name from user_tables) loopexecute immediate ('drop table '||c.table_name||' cascade constraints');end loop;End;//* */ ALTER SESSION SET NLS_DATE_FORMAT = 'MM/DD/YYYY';CREATE TABLE DOCTOR (DOC_ID varchar2(5) PRIMARY KEY,DOC_LNAME varchar2(15),DOC_FNAME varchar2(15),DOC_INITIAL varchar2(1),DOC_SPECIALTY varchar2(15));INSERT INTO DOCTOR VALUES('29827','Sanchez','Julio','J','Dermatology');INSERT INTO DOCTOR…