Create these two tables exactly as in the 1. figure above using SQL Developer or SQL Command Line 2. Provide Screenshots for the results obtained: 2.1 SELECT b.*, p.* FROM Branch1 b, PropertyForRent1 p WHERE b.bCity = p.pCity; 2.2 SELECT b.*, p.* FROM Branch1 b LEFT JOIN Implement the following Queries and PropertyForRent1 p ON b.bCity = p.pCity; 2.3 SELECT b.*, p.* FROM Branch1 b RIGHT JOIN PropertyForRent1 p ON b.bCity = p.pCity; 2.4 SELECT b.*, p.* FROM Branch1 b FULL JOIN PropertyForRent1 p ON b.bCity = p.pCity;

Create these two tables exactly as in the 1. figure above using SQL Developer or SQL Command Line 2. Provide Screenshots for the results obtained: 2.1 SELECT b., p. FROM Branch1 b, PropertyForRent1 p WHERE b.bCity = p.pCity; 2.2 SELECT b., p. FROM Branch1 b LEFT JOIN Implement the following Queries and PropertyForRent1 p ON b.bCity = p.pCity; 2.3 SELECT b., p. FROM Branch1 b RIGHT JOIN PropertyForRent1 p ON b.bCity = p.pCity; 2.4 SELECT b., p. FROM Branch1 b FULL JOIN PropertyForRent1 p ON b.bCity = p.pCity;

Oracle 12c: SQL

3rd Edition

ISBN:9781305251038

Author:Joan Casteel

Publisher:Joan Casteel

Chapter12: Subqueries And Merge Statements

Section: Chapter Questions

Problem 19MC

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PLz help with the following Please implement the following using Oracle live/SQL Developer: 1- Create the DONORS table with the following columns: donor_id ( PRIMARY KEY), donor_name (Not Null), donor_age. Then Create DONATIONS table with the following columns: pledge_id (PRIMARY KEY), pledge_date, amount_pledged, amount_paid, Is_paid, donor_id that REFERENCES to DONORS table (donor_id). 2- Populate every relation with three rows. 3- Now, your supervisor has asked you to modify the donor_age column in the DONORS table. He wants the column to be configured to accept a value that must be >=20. Which statement should you issue to accomplish this task? 4- Create a Trigger to update Is_paid column to 1 when the amount_paid equal to amount_pledged.
Write the SQL statements for the following User table. User_id Name City Order_date Purchase_amount 101 Alex Paris 2012-10-05 150.5 102 James London 2012-10-05 270.3 103 Paul Paris 2012-10-05 120.8 104 Lawson Rome 2012-10-05 230.5 105 Kate Beijing 2012-10-05 30.2 Write a query to create a user table containing following columns. Whereas user_id as primary key. Write a query to insert multiple records into user table.
Write the SQL statements for the following User table. User_id Name City Order_date Purchase_amount 101 Alex Paris 2012-10-05 150.5 102 James London 2012-10-05 270.3 103 Paul Paris 2012-10-05 120.8 104 Lawson Rome 2012-10-05 230.5 105 Kate Beijing 2012-10-05 30.2 1. Write a query to create a user table containing following columns. Whereas user_id as primary key. 2. Write a query to insert multiple records into user table. 3. Write a query to add two columns, country and currency in user table. 4. Write a query to update user table, set city as shanghai for a user having user_id 105. 5. Write a query to remove all records from user table. 6. Write a query to drop a column purchase _amount from user table. 7. Write a query to display name of user in descending order whose order purchase amount is in between 200 to 270. 8. Write a query to display name of user who belong to either Paris, London or CS Beijing. 9. Write query to display name of an employee who belong to paris. 10. Write a…
So here I have 2 tables and I need help understanding the basics of joins. I have developed some pieces of SQL code, but im not sure what they will output. I need to know what each one will output. INNER JOIN - SELECT StudentDetails.NAME, StudentMarks.MARKS FROM StudentDetails INNER JOIN StudentMarks ON StudentDetails.S_ID = StudentMarks.ID; LEFT JOIN - SELECT Name, Address, Age FROM StudentDetails LEFT JOIN StudentMarks ON S_ID=ID; RIGHT JOIN - SELECT Name, Address, Age FROM StudentDetails RIGHT JOIN StudentMarks ON S_ID=ID; FULL JOIN - SELECT StudentDetails.NAME, StudentMarks.MARKS FROM StudentDetails FULL OUTER JOIN StudentMarks ON StudentDetails.S_ID = StudentMarks.ID;
THIS MODULE IS ABOUT SUBQUERIES, SO YOU MUST USE SUBQUERIES INSTEAD OF TABLE JOINS (a) From which two fields in which table can you calculate the enrollment of a section?(b) From which table can you get both course and section numbers?(c) Which field in table SECTION uniquely identifies a section, section id or section no?(d) Write a SQL statement to show sections and their enrollment.(e) Write a SQL statement to show sections with enrollment greater than five. Display both course andsection numbers
SQL ( ORACLE SQL DEVELOPER ) : THIS IS FIRST , SECOND AND THIRD TABLE for the Task 1 to solve : create table author_45( auth_name char(50), gender char(1), age number(3), nationality char(50), PRIMARY KEY (auth_name) ); create table book_45( isbn number(10), title char(50), ver_no number(2), publisher char(50), PRIMARY KEY (isbn) ); create table writtenBy_45( isbn number(10), auth_name char(50), yr_published number(4), price number(3), no_copy number(4), PRIMARY KEY (isbn,auth_name), foreign key (isbn) references book_45, foreign key (auth_name) references author_45 ); insert into author_45 values ('Benoit','F',42,'French'); insert into author_ 45 values ('Antonini','M',33,'Italian'); insert into author_ 45 values ('McAven','F',31,'Australian'); insert into author_ 45 values ('Williams','M',90,'British'); insert into book_45 values (101,'atabases',1,'Possum'); insert into book_45 values (202,'SQL Primer',2,'Hall'); insert into book_45 values…
In the SQL Oracle Application Write one query to return the employee_id, job_id, hire_date, and department_id of all employees and a second query listing employee_id, job_id, start_date, and department_id from the job_history table and combine the results as one single output. Make sure you suppress duplicates in the output.
Build the Students Table that is shown in the picture using python SQL hint: The SQL command to build the table is create table student( SID int primary key not null, Name varchar(20), GPA char(3), Major varchar(25))
How do I do this question in SQL command? Use the Oracle database employees table and CASE expression to decode the departmentid. Display the department id, last name, salary, and a column called “New Salary” whose value is based on the following conditions: If the department id is 10 then 1.25 * salary If the department id is 90 then 1.5 * salaryIf the department id is 130 then 1.75 * salary Otherwise, display the old salary.
Consider the following tables: Branch1 PropertyForRent1 branchNo bCity propertyNo pCity B003 Glasgow PA14 Aberdeen B004 Bristol PL94 London B002 London PG4 Glasgow 1. Create these two tables exactly as in the figure above using SQL Developer or SQL Command Line 2. Provide Screenshots for the results obtained: 2.1 SELECT b.*, p.* FROM Branch1 b, PropertyForRent1 p WHERE b.bCity = p.pCity; 2.2 SELECT b.*, p.* FROM Branch1 b LEFT JOIN Implement the following Queries and %3D PropertyForRent1 p ON b.bCity = p.pCity; 2.3 SELECT b.*, p.* FROM Branch1 b RIGHT JOIN PropertyForRent1 p ON b.bCity = %3D p.pCity; 2.4 SELECT b.*, FROM Branch1 b FULL JOIN р.* PropertyForRent1 p ON b.bCity = %3D p.pCity;
Q3: How many unique titles do we have in the employee table? Hint: You need to create TWO queries, 1. The first query returned all the unique titles, i.e., Q03_Unique Titles (use the Distinct keyword and use the SQL View to type in the SQL statement directly (Design View will not work). Please refer to SQL_MIS310 slide#105-107 for guidance. You cannot use Distinct and Count together in Access and this is why you have to do it in two related two queries.). The second query is a query against the first query (i.e., Q03_UniqueTitles) and count the number of unique titles (Use the aggregate function Count) and the query should return a number. 2. The second query is named Q03_UniqueTitlles_Count and it has been created for you already. You should use Q03_UniqueTitles (not a table) to create this query. This query should return a result that is a simple number.
My SQL WORKBENCH **Please note that you may actually use as many tables as you need. You are not limited to use only twotables.Example:SELECT *FROM DEPARTMENTS;SELECT *FROM EMPLOYEES;Now join departments and employees tables.SELECT FIRST_NAME, DEPARTMENT_NAMEFROM EMPLOYEES, DEPARTMENTSWHERE EMPLOYEES.DEPARTMENT_ID = DEPARTMENTS.DEPARTMENT_ID;NOTE: If you omit the join condition, the result will be the Cartesian Product of the two tables. A CartesianProduct joins all rows of the first table with all rows of the second table.That is if the first table has n rows and the second table has m rows, the output will have n*m rows.The following query will produce a Cartesian Product:SELECT FIRST_NAME, DEPARTMENT_NAMEFROM EMPLOYEES, DEPARTMENTS;**When table names are long, qualifying column names might be very time consuming. Usually peopleprefer to use table aliases for this purpose.Example: List the name of all employees who work for 'Purchasing' departmentSELECT E.FIRST_NAME "EMPLOYEE…

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