The course I'm taking is database management. I have attached the necessary information for question. I need the correct solution query for part A utilizing SUBQUERY. the query i have come up with isSELECT * FROM item_db WHERE unitprice > (SELECT unitprice FROM item_db WHERE item_code = '005');i get an empty set. how do i correct this so i dont receve an empty set? please provide query. thank you! Part A - Find items with a unit price greater than the unit price of the item with item code ‘005’. mysql> select * from item_db;| item_code | item_name | current_stock | unitprice |2B5 | 7100 |56 |40 |1045 |10 | 8300 |5 |33C250 |150 |855550 |650 |200 |1300 |15 |500 | 9300 |100 |7 rows in set (0.00 sec) mysql> select * from item_db;+-| item_code | item_name | current_stock | unitprice |2B5 | 710 | 833 | C855200 | 1500 | 97 rows in set (0.00 sec)100 |56 |40 |1045 |300 |5 |250 |150 |550 |650 |300 |15 |300 |100 |mysql> insert into item_db(item_code, item_name, current_stock, unitprice) values ('005', 13,52,908);ERROR 1406 (22001): Data too long for column 'item_name' at row 1mysql> select * from customer_db;+-| cust_id❘ cust_name | cust_hname | cust_street | cust_phone |2 | riya| 123abcI diablo8 | sean| 12345abc | tommy9 | riya| 123abc| diablo-+-44356798765 |4435673 rows in set (0.01 sec)mysql> SELECT * FROM item_db WHERE unitprice > (SELECT unitprice FROM item_db WHERE item_code = '005');Empty set (0.00 sec)

Subqueries, also called nested queries, are an effective feature in SQL that allows you to carry out…

The course I'm taking is database management. I have attached the necessary information for question. I need the correct solution query for part A utilizing SUBQUERY. the query i have come up with is SELECT * FROM item_db WHERE unitprice > (SELECT unitprice FROM item_db WHERE item_code = '005'); i get an empty set. how do i correct this so i dont receve an empty set? please provide query. thank you! Part A - Find items with a unit price greater than the unit price of the item with item code ‘005’.

The course I'm taking is database management. I have attached the necessary information for question. I need the correct solution query for part A utilizing SUBQUERY. the query i have come up with is SELECT * FROM item_db WHERE unitprice > (SELECT unitprice FROM item_db WHERE item_code = '005'); i get an empty set. how do i correct this so i dont receve an empty set? please provide query. thank you! Part A - Find items with a unit price greater than the unit price of the item with item code ‘005’.

Oracle 12c: SQL

3rd Edition

ISBN:9781305251038

Author:Joan Casteel

Publisher:Joan Casteel

Chapter6: Additional Database Objects

Section: Chapter Questions

Problem 6HOA: A new table has been requested to support tracking automated emails sent to customers. Create the...

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mysql> select * from order_db; | orderid | custid | itemcode | orderquantity | deliverydate 2 | 2 | 2 2 | 2 | 2 2 250 200 500 300 75 ~~~~N 2 2 B 2 5 rows in set (0.00 sec) mysql> select * from item_db; + | item_code | item_name | current_stock 10 33 | C 85 5 200 | 1 500 | 9 6 rows in set (0.00 sec) 800 600 125 200 265 100 | 300 250 | 550 | 300 | 300 2020-05-30 2021-09-21 | 2023-06-02 2020-05-29 2020-05-29 | unitprice | + 40 | 5| 150 650 15 | 100 | | | с | R | D | C | c modeofpayment | date_of_order | 2020-05-28 1998-02-15 2024-09-03 1998-02-15 1998-02-15
Use Mysql code The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key RegisteredName Breed Height BirthDate Student with columns: ID - primary key FirstName LastName Street City State Zip Phone EmailAddress LessonSchedule with columns: HorseID - partial primary key, foreign key references Horse(ID) StudentID - foreign key references Student(ID) LessonDateTime - partial primary key Write a SELECT statement to create a lesson schedule for Feb 1, 2020 with the lesson date/time, student's first and last names, and the horse's registered name. Order the results in ascending order by lesson date/time, then by the horse's registered name. Make sure unassigned lesson times (student ID is NULL) appear in the results. Hint: Perform a join on the LessonSchedule, Student, and Horse tables, matching the student IDs and horse IDs.
Write Mysql code The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key RegisteredName Breed Height BirthDate Student with columns: ID - primary key FirstName LastName Street City State Zip Phone EmailAddress LessonSchedule with columns: HorseID - partial primary key, foreign key references Horse(ID) StudentID - foreign key references Student(ID) LessonDateTime - partial primary key Write a SELECT statement to create a lesson schedule with the lesson date/time, horse ID, and the student's first and last names. Order the results in ascending order by lesson date/time, then by horse ID. Unassigned lesson times (student ID is NULL) should not appear in the schedule. Hint: Perform a join on the Student and LessonSchedule tables, matching the student IDs.
MySQL CREATE TABLE students ( id INT PRIMARY KEY, first_name VARCHAR(50), last_name VARCHAR(50), age INT, major VARCHAR(50), faculty VARCHAR(50)); CREATE TABLE location ( id INT PRIMARY KEY, name VARCHAR(50), rooms INT); CREATE TABLE faculty ( id INT PRIMARY KEY, name VARCHAR(50), department_id INT); 4. Find the number of employees in each department who get no commission or have salary less than5000.5. Find the maximum salary of employees in each department that the employee was hired 15 yearsbefore now. *hint: user TIMESTAMPDIFF(<unit type>,<Date_value 1>,<Date_value 2>), the unitcan be YEAR, MONTH, DAY, HOUR, etc...6. Find the last name of all employees that were not hired on Tuesday *hint: (UseDATE_FORMAT() function amd the format you need is %W, also use UPPER()7. Find the number of employees in each department who have a manager.8. Find the number of employees for each manager whose employees' minimum salary is greaterthan 5000.9. Find the number of…
Write Mysql code for: The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key RegisteredName Breed Height BirthDate Student with columns: ID - primary key FirstName LastName Street City State Zip Phone EmailAddress LessonSchedule with columns: HorseID - partial primary key, foreign key references Horse(ID) StudentID - foreign key references Student(ID) LessonDateTime - partial primary key Write a SELECT statement to create a lesson schedule for Feb 1, 2020 with the lesson date/time, student's first and last names, and the horse's registered name. Order the results in ascending order by lesson date/time, then by the horse's registered name. Make sure unassigned lesson times (student ID is NULL) appear in the results. Hint: Perform a join on the LessonSchedule, Student, and Horse tables, matching the student IDs and horse IDs.
A unique identifier for a data row that consists of more than one field is commonly called a: primary plus key composite primary key foreign key none of the above
Delete the index named ITEM_INDEX3.
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A new table has been requested to support tracking automated emails sent to customers. Create the table and add data as described below. Tablename: email_log Columns: emailid (numeric), emaildate (datetime), customer# (numeric) Primary key: emailid column, define as an Identity Column Add the following data rows and display resulting rows (if any errors occur, explain why the error is expected) Emaildate = current date, customer# = 1007 Emailid = specify to use the column default value, emaildate = current date, customer# = 1008 Emailid = 25, emaildate = current date, customer# = 1009
Cascading Referential Integrity Constraint defines rules about how referencing tuples are affected when referenced tuples are deleted. In SQL Server Management Studio, while specifying Cascading Referential Integrity Constraint (Delete/Update Rules) you have four options: (1) ______________ _________ (2) ___________ ___________, (3) ______________ _________ (4) ___________ ___________.
DataBaseMid2021-2.doc [Compatibility Mode] - Word Sign in Table Tools File Home Insert Design Layout References Mailings Review View Help Design Layout O Tell me what you want to do 요 Share SECOND OUENIVAR The following Data Base shows some sample data values for an extended form of the Zoo and Visitors. The significance of Visiting row is that the Visitor visits a specified Zoo in a specified date. Zoo : Zno zi Visitors : City Gaza Nusirat Yno Name Address Age V1 Fuad Gaza 55 Z2 V2 Rami Gaza 20 Z3 Gaza V3 Ahmed Nusirat 31 V4 Ahmed Gaza 45 V5 Ali Rafa 20 Visiting: Vno Zno Z1 Date V2 11-11-1998 15-10-1999 V1 Z3 11-01-2003 11-01-2003 06-03-1997 V2 Z2 V2 Z3 V1 Z2 V3 Z1 06-031997 V3 Z3 11-11-2005 V4 Z3 11-11-2005 V2 Z2 12-12-2005 Given the following relational Algebra querya Visitors Þyisitora.Nno = Visiting.VnolIIxna.Znal Visiting) A IIzag [8 City = “Caza' ( Zoo)]1 A- Find the result relation of the above query B What was the request of the relational algebra query above (make a query…
Cust_id(NOT NULL)Txn_date(Not NULL)Txn_type(NOT NULL)Txn_amount(NOT NULL)1 2020-01-21 deposit 821 2020-01-10 deposit 7121 2020-01-01 deposit 1962 2020-01-14 deposit 5632 2020-01-29 deposit 6263 2020-01-13 Withdraw 9953 2020-01-20 Deposit 4853 2020-01-03 Deposit 7063 2020-01-13 Purchase 6014 2020-01-11 Deposit 5205 2020-01-21 Deposit 826 2020-01-10 Deposit 7127 2020-01-01 Withdraw 1967 2020-01-14 Purchase 563 how can I make this table in oracle sql?