If a table SALES has 2 columns namely SALESID, SALES_PERSONNAME and a table PERSON has 2 columns namely PERSONID, PERSONNAME. Is it possible to attempt join on the tables SALES and PERSON? 1 O a. Common columns are always and are only needed to perform join O b. Possible O c. Not Possible O d. Cannot say What is the correct query to display all the matching and non-matching rows from the tables INVOICE and SHIPMENT? O a. Select * from INVOICE i LEFT OUTER JOIN SHIPMENT s on(i.invoiceid=s.invoiceid); O b. Select * from INVOICE i JOIN SHIPMENT s on(i.invoiceid=s.invoiceid); O c. Select * from INVOICE i RIGHT OUTER JOIN SHIPMENT s on(i.invoiceid=s.invoiceid); O d. Select * from INVOICE i FULL OUTER JOIN SHIPMENT s on(i.invoiceid=s.invoiceid); Assume that ItemlD is the primary key in a Product table . Select price from Product where Itemld =(select ItemlD from Product where ProductName='screws'); The above given query is a, O a. Can't say O b. Multi-row subquery O c. Single-row subquery O d. Simple Query

If a table SALES has 2 columns namely SALESID, SALES_PERSONNAME and a table PERSON has 2 columns namely PERSONID, PERSONNAME. Is it possible to attempt join on the tables SALES and PERSON? 1 O a. Common columns are always and are only needed to perform join O b. Possible O c. Not Possible O d. Cannot say What is the correct query to display all the matching and non-matching rows from the tables INVOICE and SHIPMENT? O a. Select * from INVOICE i LEFT OUTER JOIN SHIPMENT s on(i.invoiceid=s.invoiceid); O b. Select * from INVOICE i JOIN SHIPMENT s on(i.invoiceid=s.invoiceid); O c. Select * from INVOICE i RIGHT OUTER JOIN SHIPMENT s on(i.invoiceid=s.invoiceid); O d. Select * from INVOICE i FULL OUTER JOIN SHIPMENT s on(i.invoiceid=s.invoiceid); Assume that ItemlD is the primary key in a Product table . Select price from Product where Itemld =(select ItemlD from Product where ProductName='screws'); The above given query is a, O a. Can't say O b. Multi-row subquery O c. Single-row subquery O d. Simple Query

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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Write a query to show all record of job-history of those employees whose department number 60? SELECT first_name from employees where first_name like ‘S%’; SELECT first_name from employees where first_name like ‘_e%’; table Name :- EMP EDIT EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO 7839 KING PRESIDENT - 11/17/1981 5000 - 10 7698 BLAKE MANAGER 7839 05/01/1981 2850 - 30 7782 CLARK MANAGER 7839 06/09/1981 2450 - 10 7566 JONES MANAGER 7839 04/02/1981 2975 - 20 7788 SCOTT ANALYST 7566 12/09/1982 3000 - 20 7902 FORD ANALYST 7566 12/03/1981 3000 - 20 7369 SMITH CLERK 7902 12/17/1980 800 - 20 7499 ALLEN SALESMAN 7698 02/20/1981 1600 300 30 7521 WARD SALESMAN 7698 02/22/1981 1250 500 30 7654 MARTIN SALESMAN 7698 09/28/1981 1250 1400 30 7844 TURNER SALESMAN 7698 09/08/1981 1500 0 30 7876 ADAMS CLERK 7788 01/12/1983 1100 - 20 7900 JAMES CLERK 7698 12/03/1981 950 - 30 7934 MILLER CLERK 7782 01/23/1982 1300 - 10
Some rows of the table STUDENT are shown below: CODE NAME DEPARTMENT GPA YEAR 291 ALEX PSI 3.1 1992 938 MICHELE PHY 2.3 1992 931 JHON MD 3.3 2001 182 JOE MD 3.4 2002 190 REY PHY 2.0 2001 330 RON PSI 3.9 2001 You want the name all departments that have a department GPA average of 2004 greater than the global GPA average. Which query gives you the desired result? a. SELECT DEPARTMENT FROM STUDENT WHERE YEAR = 2004 GROUP BY DEPARTMENT HAVING AVG(GPA) > (SELECT AVG(GPA) FROM STUDENT); b. SELECT DEPARTMENT FROM STUDENT WHERE YEAR = 2004 AND AVG(GPA) > (SELECT AVG(GPA) FROM STUDENT) GROUP BY DEPARTMENT; c. SELECT DEPARTMENT FROM STUDENT WHERE YEAR = 2004 GROUP BY DEPARTMENT HAVING AVG(GPA) > (SELECT GPA FROM STUDENT GROUP BY DEPARTMENT); d. SELECT * FROM STUDENT WHERE YEAR = 2004 GROUP BY DEPARTMENT HAVING AVG(GPA) > (SELECT AVG(GPA) FROM STUDENT);
Write a query that will show all attributes and field in the Movie table that do not contain a value.
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i need to create a query that joins these 3 tables, without the duplicate column which would be the shop_id. i cant get them to join without that column duplicating. help would be appreciated thanks! CREATE TABLE COFFEE_SHOP ( shop_id int, shop_name VARCHAR(50), city VARCHAR(50),state CHAR(2), PRIMARY KEY (shop_id)); CREATE TABLE EMPLOYEE ( employee_id int, first_name VARCHAR(30), last_name VARCHAR(30),hire_date Date,job_title VARCHAR(30),shop_id int, PRIMARY KEY (employee_id),FOREIGN KEY (shop_id) REFERENCES COFFEE_SHOP(shop_id)); CREATE TABLE COFFEE ( coffee_id int, shop_id int, supplier_id int,coffee_name VARCHAR(30),price_per_pound NUMERIC(5,2), PRIMARY KEY (coffee_id),FOREIGN KEY (shop_id) REFERENCES COFFEE_SHOP(shop_id),FOREIGN KEY (supplier_id) REFERENCES SUPPLIER(supplier_id));
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JOIN TABLES One row in the PAT_ENC table represents one patient encounter. One row in the ORDER MED table represents one medication order. One patient encounter can have many medication orders but one medication order can only have one patient encounter. In other words, the cardinality of this PAT_ENC to ORDER_MED relationship is one-to- many. You start a query with PAT_ENC. You then add ORDER_MED using an inner join. What is true about the granularity of the result set before and after adding the ORDER_MED table? SELECT ONE OF THE FOLLOWING A. The granularity stays at one row per patient encounter. B. The granularity stays at one row per medication order. C. The granularity changes from one row per patient encounter to one row per medication order on an encounter. D. The granularity changes from one row per medication order to one row per patient encounter.
The PA T_ENC_HSP table has 100 rows each, representing the hospital and counter. The ADMISSION _PROV_ID column identifies encounters admitting provider. There are five unique known values in ADMISSION _PROV_ID and separately, there are encounters with a null admitting provider. Use the following query to answer the question below: Select ADMISSION _PROV_ID FROM PAT_ENC_HSP GROUP BY ADMISSION_PROV_ID How many rows will be in the results? Select one of the following:A. 0 B. 5 C . 6 D. 100 E. 500
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