This is for SQL. Show a union that uses more than two select statements. The following select statement shows the number of lunches that each employee will attend, but it does not account for Carol Rose or Paula Jacobs because they are not attending any lunches. Modify this statement to show that these two people will not attend any lunches.select a.first_name, a.last_name, count(b.lunch_id) as number_of_lunchesfrom l_employees a inner join l_lunches b on a.employee_id = b.employee_idgroup by a.first_name, a.last_name; Table Data Indexes ModelQuery Count Rows Insert RowEDITNNNNNNNNNN201202203204205206207208209210EMPLOYEE_IDConstraints GrantsLoad DataSUSANJIMFIRST_NAMEMARTHAELLENHENRYCAROLDANFREDPAULANANCYStatisticsKERNUI DefaultsBROWNLAST_NAMEWOODSOWENSROSEPERKINSSMITHCAMPBELLJACOBSHOFFMANTriggersEXESALSHPSALSALACTSHPSHPMKTSALL_EMPLOYEESDependencies SQLDEPT_CODERESTHIRE_DATE06/01/199808/16/199902/02/200907/01/200803/01/200612/01/200804/01/2008Sample Queries03/17/199902/16/2007302525152525251525CREDIT_LIMIT348487227591683052862259175233572974PHONE NUMBER201201202202203203201203^+MANAGER_ID Query Count RowsRRRRRRRRRREDITInsert Row1234678912131516172021Load DataLUNCH_IDܡܚܐܚܡ ܫܝܚܕ ܒܝܚܝ ܫ ܕ ܕ ܚܕ ܒܚܐDIVER11/16/201111/16/201111/16/201111/16/201111/16/201111/16/201111/25/201111/25/201111/25/201111/25/201111/25/201112/05/201112/05/201112/05/201112/05/2011**bb****LUNCH_DATEL LUNCHESexpenserierea201207203204202210201208204207205201210205203sempre queremEMPLOYEE ID10/13/201110/13/201110/13/201110/13/201110/13/201110/13/201110/14/201110/14/201110/14/201110/18/201110/21/201110/21/201110/21/201110/24/201110/24/2011DATE_ENTERED+v

SQL which refers to the one it is used to communicate with a database. According to ANSI (American…

This is for SQL. Show a union that uses more than two select statements. The following select statement shows the number of lunches that each employee will attend, but it does not account for Carol Rose or Paula Jacobs because they are not attending any lunches. Modify this statement to show that these two people will not attend any lunches. select a.first_name, a.last_name, count(b.lunch_id) as number_of_lunches from l_employees a inner join l_lunches b on a.employee_id = b.employee_id group by a.first_name, a.last_name;

This is for SQL. Show a union that uses more than two select statements. The following select statement shows the number of lunches that each employee will attend, but it does not account for Carol Rose or Paula Jacobs because they are not attending any lunches. Modify this statement to show that these two people will not attend any lunches. select a.first_name, a.last_name, count(b.lunch_id) as number_of_lunches from l_employees a inner join l_lunches b on a.employee_id = b.employee_id group by a.first_name, a.last_name;

Np Ms Office 365/Excel 2016 I Ntermed

1st Edition

ISBN:9781337508841

Author:Carey

Publisher:Carey

Chapter8: Working With Advanced Functions

Section: Chapter Questions

Problem 2.7CP

See similar textbooks

Similar questions

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
Attempt to view the structure of the HOMEWORK13 view.
There's a "users" table that has more than 1M rows of "users" information. The image below are the first 3 rows.Implement a SQL query that retrieves the "email" and "friend_count" of the "user_id" with the most friends.
Textbook: Digital Systems Chapter 8 Question/Problem 44 I'm not sure if I'm running the correct Query for this problem. The question is: "Create a trigger named trg_char_hours that automatically updates the AIRCRAFT table when a new CHARTER row is added. Use the CHARTER table’s CHAR_HOURS_FLOWN to update the AIRCRAFT table’s AC_TTAF, AC_TTEL, and AC_TTER values. (Hint: use temp values, as the INSERT event does not accept the OLD keyword.)" I ran this query without any errors: CREATE TRIGGER trg_char_hours AFTER INSERT ON CHARTER FOR EACH ROW UPDATE AIRCRAFT SET AC_TTAF = AC_TTAF + NEW.CHAR_HOURS_FLOWN, AC_TTEL = AC_TTEL + NEW.CHAR_HOURS_FLOWN, AC_TTER = AC_TTER + NEW.CHAR_HOURS_FLOWN WHERE AIRCRAFT.AC_NUMBER = NEW.AC_NUMBER; But the answer is still incorrect because the last row doesn't have the expected values. I'm not sure why the test query would delete any row with a CHAR_TRIP ABOVE 10018. I attached a screen of the output I'm getting: I saw in another answer this:…
Textbook: Digital Systems Chapter 8 Question/Problem 44 I'm not sure if I'm running the correct Query for this problem. The question is: "Create a trigger named trg_char_hours that automatically updates the AIRCRAFT table when a new CHARTER row is added. Use the CHARTER table’s CHAR_HOURS_FLOWN to update the AIRCRAFT table’s AC_TTAF, AC_TTEL, and AC_TTER values. (Hint: use temp values, as the INSERT event does not accept the OLD keyword.)" I ran this query without any errors: CREATE TRIGGER trg_char_hours AFTER INSERT ON CHARTER FOR EACH ROW UPDATE AIRCRAFT SET AC_TTAF = AC_TTAF + NEW.CHAR_HOURS_FLOWN, AC_TTEL = AC_TTEL + NEW.CHAR_HOURS_FLOWN, AC_TTER = AC_TTER + NEW.CHAR_HOURS_FLOWN WHERE AIRCRAFT.AC_NUMBER = NEW.AC_NUMBER; But the answer is still incorrect because the last row doesn't have the expected values. I'm not sure why the test query would delete any row with a CHAR_TRIP ABOVE 10018. I attached a screen of the output I'm getting:
SQL : Write a statement that allows the same classmate to modify (insert, update, and delete) the data of the newEmpXX table
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.
Refer to the film and inventory tables of the Sakila database. The tables in this lab have the same columns and data types but fewer rows. Write a query that lists the titles of films with the fewest rows in the inventory table. This query requires a subquery that computes the minimum of counts by film_id: SELECT MIN(count_film_id) FROM ( SELECT COUNT(film_id) AS count_film_id FROM inventory GROUP BY film_id ) AS temp_table; This subquery is provided in the template.
Using the WideWorldDistributors database and the Sales schema create a function that takes InvoiceID as an input parameter and returns all the rows from the table InvoiceLines for the InvoiceID value passed into the function. If the result is a NULL return a zero
1- Display contact First name and contact Last name for customers after sorting the customersby their contact Last name in ascending order.2- selects the order line items (orderNumber, orderlinenumber) from the “orderdetails” table.Also, display the subtotal (quantityOrdered * priceEach) for each line item and sorts the resultset based on the subtotal descending .3- Display unique last names from employees table ordered by last name.4- Display customerNumber, customerName and creditLimit for getting the top five customerswho have the highest creditLimit.5- Find customer Name, customer country, sales representative employee number fromcustomers who do not have a sales representative.6- For each order status, display status and count of orders have this status.7- For Each order number in order details table, Display order number, sum of quantity ordered(itemsCount) and sum of (priceEach * quantityOrdered) as total. With condition “total”greater than 1000.8- Display product code, product…
MySql Workbench 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); 1. List last name of all students whose first name is longer than 4 letters in ascending order accordingto the last name. Duplicated rows should be removed from the output.2. Count the total number of rooms in Location.3. Find the number of students in each major.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...
Please help me to transcribe SQL codes to python codes create table ab asselect *from aliWHERE itemprice IS NOT NULL; SELECT COUNT(MerchandiseCode), UserIDFROM abGROUP BY useridORDER BY COUNT(MerchandiseCode) DESC; select sum(totalsales), useridfrom abgroup by useridorder by sum(totalsales) desc; SELECT COUNT(transactiondate), UserIDFROM abGROUP BY useridORDER BY COUNT(transactiondate) DESC;