Write the SQL code that will create only the table structure for a table named EMP_1.This table will be a subset of the EMPLOYEE table. The basic EMP_1 table structure issummarized in the following table. Use EMP_NUM as the primary key. Note that theJOB_CODE is the FK to JOB so be certain to enforce referential integrity. Your codeshould also prevent null entries in EMP_LNAME and EMP_FNAME.TABLE P8.1 CONSTRUCTCO DATABASE STRUCTUREATTRIBUTE (FIELD) NAME DATA DECLARATIONEMP_NUM CHAR(3)EMP_LNAME VARCHAR(15)EMP_FNAME VARCHAR(15)EMP_INITIAL CHAR(1)EMP_HIREDATE DATEJOB_CODE CHAR(3)

Write an SQL query that will select OrderID and OrderDate from theOrders table and OrderDetailID from the OrderDetails table. The queryshould join on the Orders table using the OrderID attribute.

Create a View to show the city name of any location that has a department. Only show the city name once in the results. You must use a non-correlated subquery.Call this view: Department_LocationsInclude a second query to display from the View after it is created. Refer to the attached for the table.

The BLOOD_ADMIN_INFO are ORDER ID and LINE. The MAR_ADMIN_INFO MAR_BLOOD_INFO_LN table has one row per blood unit associated with an order. Its primary key columns table has one row per medication administration. The columns ORDER_MED_ID and form the foreign key to the BLOOD_ADMIN_INFO table. Use the followin*query to answer the question below it- SELECT mar.ORDER_MED_ID, mar.SCHEDULED_TIME, mar.TAKEN TIME, bld.BLOOD_ADMIN_TYPE FROM MAR_ADMIN_INFO as mar INNER JOIN BLOOD_ADMIN_INFO as bld What join condition is appropriate in ? AS [Blood Type] Select on of the following: A. mar.ORDER_MED_ID = bld.ORDER_ID B. mar.MAR_BLOOD_INFO_LN = bld.LINE C. mar.ORDER_MED_ID = bld.ORDER_ID AND mar.MAR_BLOOD_INFO_LN = bld.LINE D. It is not possible to join these tables

Emp:- Id is Primary keyProject:- EmpId foreign key.Find out the output of the left outer join, right outer join and inner join.Take the below columns in output:-Emp table ID, name.Project table projectID, NAME

For this problem create a (temporary) table called instructor_course_nums. Write a procedure that accepts an instructor ID as input. The procedure calculates the total number of course sections taught by that instructor, and adds a tuple to the temporary table consisting of the instructors ID number, name, and total courses taught - call these attributes: ID, name, and tot_courses. If the instructor already has an entry in the table,then the procedure makes sure the total number of courses taught in the temporary table is up-to-date. You must name your procedure: Written in PostgreSQL

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,…

The given SQL creates a Song table and inserts some songs. The first SELECT statement selects songs released after 1992. The second SELECT statement selects the release year for song with ID 800. Create a third query that combines the two existing queries. The first SELECT should be the outer query, and the second SELECT should be the subquery. The ORDER BY clause should appear after the subquery. Run your solution and verify the new query returns a result table with five rows, all with release years after 1992. CREATE TABLE Song (ID INT,Title VARCHAR(60),Artist VARCHAR(60),ReleaseYear INT,Genre VARCHAR(20),PRIMARY KEY (ID)); INSERT INTO Song VALUES(100, 'Hey Jude', 'Beatles', 1968, 'pop rock'),(200, 'You Belong With Me', 'Taylor Swift', 2008, 'country pop'),(300, 'You\'re Still the One', 'Shania Twain', 1998, 'country pop'),(400, 'Need You Now', 'Lady Antebellum', 2011, 'country pop'),(500, 'You\'ve Lost That Lovin\' Feeling', 'The Righteous Brothers', 1964, 'R&B'),(600, 'That\'s…

Task 3: The InstantRide Finance team wants to collect the price and discount information with the driver names for each travel in the system. You need to return the TRAVEL_ID, DRIVER_FIRST_NAME, DRIVER_LAST_NAME, TRAVEL_PRICE, and TRAVEL_DISCOUNT information from the TRAVELS and DRIVERS tables combined over DRIVER_ID field with the ON keyword. Task: Calculate each user's price and discount information. (SQL Database Test)

SQL Code for  Extend the code in (Perform a LEFT OUTER JOIN between the tables Wicketkeeping and Batting. You may select any column(s). Look at the result and state how many wicketkeepers have not played as batsmen (no need to write code for this part ) To show the name of the fielders who have played 7 times as many matches as the wicketkeeper(s) in 5).