A Guide to SQL
9th Edition
ISBN: 9781111527273
Author: Philip J. Pratt
Publisher: Course Technology Ptr
expand_more
expand_more
format_list_bulleted
Expert Solution & Answer
Chapter 3, Problem 12RQ
Explanation of Solution
Difference between “CHAR” and “VARCHAR”:
| CHAR | VARCHAR |
| “CHAR” is fixed length data type. | “VARCHAR” is the variable length data type. |
| The storage size is equal to the maximum size of the column. | The storage size is equal to the actual length of the data. |
| The keyword “CHAR” is used to declare the variable... |
Explanation of Solution
Appropriate place to use “CHAR” data type:
The “CHAR” data type is a good choice when storing the initial of the name as it takes up the same amount of bytes. It is also suitable to specify the name of the individual as it hold only the list of characters.
Example:
CREATE TABLE EMPLOYEE (
EMP_NU...
Explanation of Solution
URL:
The following given URL provides the examples as referen...
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
example inn 100 sqlmstatement
Need to create two store procedures with help.
1-Write the syntax to insert into a relational table called students the address column references an object table called addresses that was created using an address_type.
Other columns exist in the students table, but you are only inserting into the ones below. Aliases used should be the first letter of the table name, eg students would be s, addresses would be a
Relational Table Name
students
attribute
student_id
attribute
surname
attribute
address
Chapter 3 Solutions
A Guide to SQL
Ch. 3 - Prob. 1RQCh. 3 - How do you delete a table using SQL?Ch. 3 - Prob. 3RQCh. 3 - Prob. 4RQCh. 3 - Prob. 5RQCh. 3 - Prob. 6RQCh. 3 - Prob. 7RQCh. 3 - Prob. 8RQCh. 3 - Prob. 9RQCh. 3 - Prob. 10RQ
Ch. 3 - Prob. 11RQCh. 3 - Prob. 12RQCh. 3 - Prob. 13RQCh. 3 - Use SQL to complete the following exercises....Ch. 3 - Prob. 2TDCh. 3 - Prob. 3TDCh. 3 - Prob. 4TDCh. 3 - Prob. 5TDCh. 3 - Prob. 6TDCh. 3 - Prob. 7TDCh. 3 - Prob. 1CATCh. 3 - Add the following row to the ADVENTURE_TRIP table:...Ch. 3 - Prob. 3CATCh. 3 - Prob. 4CATCh. 3 - Prob. 5CATCh. 3 - Prob. 6CATCh. 3 - Review the data for the TRIP table in Figure 1-5...Ch. 3 - Colonial Adventure Tours would like to increase...Ch. 3 - Prob. 1SCGCh. 3 - Add the following record to the VACATION_UNIT...Ch. 3 - Delete the VACATION_UNIT table.Ch. 3 - Prob. 4SCGCh. 3 - Prob. 5SCGCh. 3 - Prob. 6SCGCh. 3 - The SERVICE_REQUEST table uses the CHAR data type...
Knowledge Booster
Similar questions
- The SERVICE_REQUEST table uses the CHAR data type for the DESCRIPTION and STATUS fields. Is there an alternate data type that could be used to store the values in these fields? Justify your reason for choosing an alternate data type or for leaving the data type as CHAR.arrow_forwardWhat is a NULL value?arrow_forwardDelete the index named ITEM_INDEX3.arrow_forward
- What happens if you try to decrease the scale or precision of a NUMBER column to a value less than the data already stored in the field?arrow_forwardPATHS is a table that contains information about paths on a number line. The structure of PATHS is as follows, where x1 and x2 represent that there is a path from Coordinate x1 to Coordinate x2 (You can't move from Coordinate x2 to Coordinate x1 ). NAME ΤΥΡE NULLABLE X1 INT FALSE X2 INT FALSE Problem Please write an SQL statement that returns the beginning and end point of each path in PATHS . Sort them by the value of the beginning point in ascending order. Constraints • In the case where a direct path from Coordinate a to Coordinate b is available, no path from Coordinate b to Coordinate a will be given. |x1-x2| = 1 • No path will overlap with one another. Example Suppose that PATHS is as follows: x1 x2 1 2 2 3 4 7 7 6 This table can be visualized as follows: START END END START 7 8. 9 Therefore, your SQL statement must return the following: start end 1 4 8 6arrow_forwardUsing php and sql: create a query and any supporting PHP code that will output only messages from the user named “Abla”.arrow_forward
- SBN Title Author 12345678 The Hobbit J.R.R. Tolkien 45678912 DaVinci Code Dan Brown Your student ID DBS311 Your Name use the following statement to Write a PL/SQL Function that accepts 4 parameters, 2 strings representing students names, and 2 integers representing the marks they received in DBS311. The function will determine which student had the higher mark and return the name of the student. If the marks were the same, then return the word "same" and return "error" if an error occurs or the calculation can not be determined.arrow_forwardDo this in MySQL please: Create a stored procedure named prc_inv_amounts to update the INV_SUBTOTAL, INV_TAX, and INV_TOTAL. The procedure takes the invoice number as a parameter. The INV_SUBTOTAL is the sum of the LINE_TOTAL amounts for the invoice, the INV_TAX is the product of the INV_SUBTOTAL and the tax rate (8 percent), and the INV_TOTAL is the sum of the INV_SUBTOTAL and the INV_TAX.arrow_forwardThe 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).arrow_forward
- You will be using the Colonial Adventure Tours database. List the trip IDs and trip names for each pair of trips that have the same start location.(You will need to assign an alias). The first trip id should be the major sort key, and the second trip id should be the minor sort key.arrow_forward1) Write the SQL code that will create the table structure for a table named EMP. This table is a subset of the EMPLOYEE table. The basic EMP table structure is summarized in the table below. (Note that the EMP_NUM is the PK)? ATTRIBUTE(FIELD) NAME DATADECLARATION Constraints EMP_NUM INT Auto generated Identifier, start from 101 EMP_LNAME VARCHAR(15) Not null EMP_FNAME VARCHAR(15) Not Null EMP_INITIAL CHAR(1) EMP_HIREDATE DATE Default value is system date JOB_CODE CHAR(3) Make sure input is in the range500,501,502arrow_forwardAssignment : PL/SQL Practice Note: PL/SQL can be executed in SQL*Plus or SQL Developer or Oracle Live SQL. Write an anonymous PL/SQL block that will insert a new doctor into the DOCTOR Verify that insert has been done successfully by issuing a select * from doctor.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
A Guide to SQLComputer ScienceISBN:9781111527273Author:Philip J. PrattPublisher:Course Technology Ptr
Np Ms Office 365/Excel 2016 I NtermedComputer ScienceISBN:9781337508841Author:CareyPublisher:Cengage
- COMPREHENSIVE MICROSOFT OFFICE 365 EXCEComputer ScienceISBN:9780357392676Author:FREUND, StevenPublisher:CENGAGE L
A Guide to SQL
Computer Science
ISBN:9781111527273
Author:Philip J. Pratt
Publisher:Course Technology Ptr
Np Ms Office 365/Excel 2016 I Ntermed
Computer Science
ISBN:9781337508841
Author:Carey
Publisher:Cengage
COMPREHENSIVE MICROSOFT OFFICE 365 EXCE
Computer Science
ISBN:9780357392676
Author:FREUND, Steven
Publisher:CENGAGE L