The command DESC DEPARTMENTS shows the following information Name Null? Type DEPARTMENT ID NOT NULL NUMBER (4) DEPARTMENT NAME NOT NULL VARCHAR2 (30) MANAGERID NUMBER (6) LOCATION ID NUMBER (4) Which of the following is correct? This table has a primary key. This table has a foreign key. This table has both primary and foreign keys. This table has neither primary nor foreign keys. Not enough information to answer.

The command DESC DEPARTMENTS shows the following information Name Null? Type DEPARTMENT ID NOT NULL NUMBER (4) DEPARTMENT NAME NOT NULL VARCHAR2 (30) MANAGERID NUMBER (6) LOCATION ID NUMBER (4) Which of the following is correct? This table has a primary key. This table has a foreign key. This table has both primary and foreign keys. This table has neither primary nor foreign keys. Not enough information to answer.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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In Structure Query Information Select all columns from the jobs table using an asterisk.2. Select only job title, min salary, and max salary columns from the jobs table.3. Select all columns from the d_partners table by writing out all of the columns inthe select statement.4. Using the d_partners table, display the id, and then concatenate the first_nameand last_name together with a space in between the two columns. Give theconcatenated column an alias of NAME. Be sure to use the optional ASkeyword.5. Using the job_grades table, show the grade level, and then subtract thelowest_sal from the highest_sal to show the amount of variance there is for eachgrade level. Give the calculated column an alias of Salary_Variance. Do notuse the optional AS keyword.6. Using the employees table, show an unduplicated list of department_id’s.7. Change the SELECT statement you built for question 4, adding a single columncalled Partner_Information that concatenates the partner’s…
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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…
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
Lab #9 – Joining Data From Multiple Tables List the order number and order date for every order that was placed by Mary Nelson and that contains an order line for an iron. Use an INTERSECT operator. Lab #10 – Sub Queries Find the sales rep number, last name, and first name for every sales rep who represents at least one customer with a credit limit of $2000. List the order number and order date for every order that was placed by Mary Nelson and that contains an order line for an iron. Down below are the tables ORDER_LINE ORLN_ORDER_NUMBER ORLN_PART_NUMBER ORLN_NUMBER_ORDERED ORLN_QUOTED_PRICE ORDER_TABLE ORDR_ORDER_NUMBER ORDR_ORDER_DATE ORDR_CUSTOMER_NUMBER CUSTOMER CUST_NUMBER CUST_LAST CUST_FIRST CUST_STREET CUST_CITY CUST_STATE CUST_ZIP_CODE CUST_BALANCE CUST_CREDIT_LIMIT CUST_SALES_REP_NUM PART PART_NUMBER PART_DESCRIPTION UNITS_ON_HAND…
help please Create a table named BRANCH. The table has the following structureBRANCH_NUM decimal(2,0) NOT NULL default '0'BRANCH_NAME char(50) default NULLBRANCH_LOCATION char(50) default NULLNUM_EMPLOYEES decimal(2,0) deault NULL PRIMARY KEY (`BRANCH_NUM`) (Please type answer no write by hand)
Create a VIEW named MAJOR_CUSTOMER. It consists of the customer ID, first name, last name, balance, credit limit, and rep ID for every customer whose credit limit is $500 or less.
SQL: For every invoice, display the invoice number, invoice date and the total dollar amount for all products purchased in the invoice, ordered by invoice number in descending order and then by invoice date in ascending order. Here you will use an aggregate SUM(x,y) function in your SELECT clause to calculate the total dollar amount (a calculated field – name it TOTAL). Don’t forget to join on the tables as well. [hint: you should have 8 rows of output ordered properly]. select inv_number, inv_date sum(list_price) as 'total' from invoice, line order by desc inv_num, asc inv_date. I am getting the error: ORA-00923: FROM keyword not found where expected Am I formulating the query incorrectly?
Write a SELECT statement that returns three columns: EmailAddress, OrderID, and the order total for each customer. To do this, you can group the result set by the EmailAddress and OrderID columns. In addition, you must calculate the order total from the columns in the OrderItems table. Write a second SELECT statement that uses the first SELECT statement in its FROM clause. The main query should return two columns: the customer�s email address and the largest order for that customer. To do this, you can group the result set by the EmailAddress column.
Lab #9 – Joining Data From Multiple Tables List the order number and order date for every order that was placed by Mary Nelson and that contains an order line for an iron. Use an INTERSECT operator. Lab #10 – Sub Queries Find the sales rep number, last name, and first name for every sales rep who represents at least one customer with a credit limit of $2000. List the order number and order date for every order that was placed by Mary Nelson and that contains an order line for an iron. Repeat the exercise from the prior step using the ANY operator in the query instead of the ALL operator.
PLZ HELP WITH THE FOLLOWING: 1. Fix the sql file so that the values can properly be inserted. 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(VID_NUM CHAR(5)…
1) Add the following record to the OWNER table: INSERT INTO OWNER VALUES('SA100', 'Sam', 'Afyouni', '100 Hello St', 'Anytown', 'MA', '55555'); 2) Create the UPD_OWNER_LAST_NAMEprocedure to change the last name of the owner whose number is stored in I_OWNER_NUM (provided as a parameter) to the value currently found in I_LAST_NAME.