Q1. Please write SQL codes to query data from member tables based on below criteria  Please write a SELECT statement to display each member data that street must NOT be ‘STREET’ and ‘LANE’, and valid_date CANNOT be ‘FEB’, ‘APR’, and ‘JUN’. Please display the result in descending order of the last_name. The result should be identical to below results: [SCREENSHOT]   I try to use this, but it gives me this error: SELECT MEMBER_ID, FIRST_NAME, LAST_NAME, STREET, FORMAT(VALID_DATE, 'dd-MM--yy') FROM MEMBER WHERE STREET NOT IN ('STREET', 'LANE') AND MONTH(VALID_DATE) NOT IN (2, 4, 6) ORDER BY LAST_NAME DESC;   ====== error ===== ORA-00904: "MONTH": invalid identifier ORA-00904: "FORMAT": invalid identifier

how is duplicate column prefixed when duplicate column appear in Join Query? a) Table name followed by period b) Table name followed by plus c) Table name followed by equal d) Table name followed by space abd then column name

Task 4: The InstantRide Finance team also wants to analyze travels where more than the average discount rate is applied. They want to look for any correlation between higher discount amounts against other travel characteristics. You need to create a SELECT statement which is filtered with a subquery to calculate the AVG of the TRAVEL_DISCOUNT column. Task: Find all travels where an above average discount was applied. (SQL Database Test)

P7

Create a new table, EMP_PROMOTION to store the employees information. You must use a subquery to create the new table. Make the salary column invisible. Display the structure of the table in SQL Developer. Also use a SELECT statement to display all records from the new table EMP_PROMOTION.

Q. Please write SQL codes to query data from member tables based on below criteria  Please write a SELECT statement to display each member’sdata that member_id is an odd number. Please display the result in ascending order of the last_name. The result should be identical to below results: [Screenshot]

Use the code below to write a query to display the movie title, movie year, and movie genre for all movies CREATE TABLE MOVIE ( MOVIE_NUM    NUMBER(8,0)  PRIMARY KEY, MOVIE_TITLE  VARCHAR2(75) NOT NULL, MOVIE_YEAR  NUMBER(4,0)  CHECK (MOVIE_YEAR > 1900), MOVIE_COST   NUMBER(5,2), MOVIE_GENRE VARCHAR2(50), PRICE_CODE    NUMBER(2,0) CONSTRAINT MOVIE_PRICE_CODE_FK REFERENCES PRICE ); INSERT INTO MOVIE VALUES (1234, 'The Cesar Family Christmas', 2011, 39.95, 'FAMILY', 2); INSERT INTO MOVIE VALUES (1235, 'Smokey Mountain Wildlife', 2008, 59.95, 'ACTION', 1); INSERT INTO MOVIE VALUES (1236, 'Richard Goodhope', 2012, 59.95, 'DRAMA', 2); INSERT INTO MOVIE VALUES (1237, 'Beatnik Fever', 2011, 29.95, 'COMEDY', 2); INSERT INTO MOVIE VALUES (1238, 'Constant Companion', 2012, 89.95, 'DRAMA', NULL); INSERT INTO MOVIE VALUES (1239, 'Where Hope Dies', 2002, 25.49, 'DRAMA', 3); INSERT INTO MOVIE VALUES (1245, 'Time to Burn', 2009, 45.49, 'ACTION', 1);

2. Write a SQL statement to insert a new user with userId ‘XXXXX’ into the users table and populate at least 5 other columns including column userName. Note: userName column is varchar(20) and its value must be unique in the table users. DO NOT insert into the “created” column but pay attention to it’s value after the insert Paste the SQL below.

CREATE TABLE Employee( firstName char(30) , lastName char(30) , hireDate date ,empNo int , empInitial char(30) , years int );   Write the SQL syntax to create an insert statement for the employee table you created above.

PLZ help with the following using ORACLE command sql Use subqueries    CREATE TABLE MEMBERSHIP ( MEM_NUM CHAR(3) CONSTRAINT MEMBER_MEM_NUM_PK PRIMARY KEY, MEM_FNAME   VARCHAR(30) NOT NULL, MEM_LNAME   VARCHAR(30) NOT NULL, MEM_STREET  VARCHAR(15), 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_RENT_NUM_PK  PRIMARY KEY, RENT_DATE DATE, MEM_NUM CHAR(3), CONSTRAINT RENTAL_MEM_NUM_FK FOREIGN KEY (MEM_NUM) REFERENCES MEMBERSHIP);   CREATE TABLE PRICE (PRICE_CODE CHAR(1) CONSTRAINT PRICE_PRICE_CODE_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_MOVIE_NUM_PK PRIMARY KEY, MOVIE_NAME  VARCHAR(30) NOT NULL, MOVIE_YEAR CHAR(4), MOVIE_COST NUMBER(5,2), MOVIE_GENRE VARCHAR(15), PRICE_CODE CHAR(1), CONSTRAINT MOVIE_PRICE_CODE_FK FOREIGN KEY (PRICE_CODE) REFERENCES PRICE);     CREATE TABLE VIDEO…