CREATE TABLE DONORS ( donor_id INT NOT NULL, donor_name varchar(255) NOT NULL, donor_age int, PRIMARY KEY (donor_id ) ); CREATE TABLE DONATIONS( pledge_id INT NOT NULL, pledge_date varchar(255) NOT NULL, amount_pledged int, Is_paid int, donor_id int, PRIMARY KEY (pledge_id ), FOREIGN KEY (donor_id) REFERENCES DONORS(donor_id) ); 1. Populate every relation with three rows.

CREATE TABLE DONORS ( donor_id INT NOT NULL, donor_name varchar(255) NOT NULL, donor_age int, PRIMARY KEY (donor_id ) ); CREATE TABLE DONATIONS( pledge_id INT NOT NULL, pledge_date varchar(255) NOT NULL, amount_pledged int, Is_paid int, donor_id int, PRIMARY KEY (pledge_id ), FOREIGN KEY (donor_id) REFERENCES DONORS(donor_id) ); 1. Populate every relation with three rows.

Oracle 12c: SQL

3rd Edition

ISBN:9781305251038

Author:Joan Casteel

Publisher:Joan Casteel

Chapter3: Table Creation And Management

Section: Chapter Questions

Problem 3HOA: Add two columns to the EMPLOYEES table. One column, named EmpDate, contains the date of employment...

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Add two columns to the EMPLOYEES table. One column, named EmpDate, contains the date of employment for each employee, and its default value should be the system date. The second column, named EndDate, contains employees date of termination.
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create table patients ( patient_id VARCHAR(255),name VARCHAR(255),insurance DECIMAL(15,2),date_admitted DATE,date_checked_out DATE,PRIMARY KEY(patient_id));INSERT INTO patients (ID , name , insurance , date_admitted , date_checked_out) values (2 , 'Ali' , '22.2' , '2/1/2020' , '3/1/2020')SELECT * FROM table patients create table doctor (doctor_id VARCHAR(255),name VARCHAR(255),specialization VARCHAR(255),PRIMARY KEY (doctor_id) ); create table test (test_id VARCHAR(255),test_name VARCHAR(255),test_date DATE,test_time TIME,result VARCHAR(255),PRIMARY KEY(test_id)); create table doctor_patient( patient_id VARCHAR(255), doctor_id VARCHAR(255),PRIMARY KEY(patient_id),FOREIGN KEY(doctor_id) REFERENCES doctor(doctor_id) ); create table test_log (test_log_id varchar(255),patient_id VARCHAR(255),test_id VARCHAR(255),doctor_id VARCHAR(255),comments VARCHAR(255),PRIMARY KEY(test_log_id),FOREIGN KEY(test_id) REFERENCES test(test_id),FOREIGN KEY(patient_id) REFERENCES…
create table patients ( patient_id VARCHAR(255),name VARCHAR(255),insurance DECIMAL(15,2),date_admitted DATE,date_checked_out DATE,PRIMARY KEY(patient_id));INSERT INTO patients (ID , name , insurance , date_admitted , date_checked_out) values (2 , 'Ali' , '22.2' , '2/1/2020' , '3/1/2020')SELECT * FROM table patients create table doctor (doctor_id VARCHAR(255),name VARCHAR(255),specialization VARCHAR(255),PRIMARY KEY (doctor_id) ); create table test (test_id VARCHAR(255),test_name VARCHAR(255),test_date DATE,test_time TIME,result VARCHAR(255),PRIMARY KEY(test_id)); create table doctor_patient( patient_id VARCHAR(255), doctor_id VARCHAR(255),PRIMARY KEY(patient_id),FOREIGN KEY(doctor_id) REFERENCES doctor(doctor_id) ); create table test_log (test_log_id varchar(255),patient_id VARCHAR(255),test_id VARCHAR(255),doctor_id VARCHAR(255),comments VARCHAR(255),PRIMARY KEY(test_log_id),FOREIGN KEY(test_id) REFERENCES test(test_id),FOREIGN KEY(patient_id) REFERENCES…
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Add two columns to the EMPLOYEES table. One column, named EmpDate, contains thedate of employment for each employee, and its default value should be the system date. Thesecond column, named EndDate, contains employees’ date of termination. When I put this formal it says, line 2 has invalid identifier! Can you help me what I need to fix from line 2? alter table EMPLOYEES2add column EmpDate date = getdate(),add column EndDate;
create table students ( student_id serial primary key, citizen_number char(11) unique not null, first_name varchar(100) not null, middle_name varchar(100), last_name varchar(100) not null, birth_date date not null, address varchar(1024) not null ); create table lectures ( lecture_code char(7) primary key, name varchar(100) not null, credits int ); create table grades ( grade_id serial primary key, description char(2) not null, value double precision not null ); insert into grades (description, value) values ('AA', 4.0), ('BA', 3.5), ('BB', 3.0), ('CB', 2.5), ('CC', 2.0), ('DC', 1.5), ('DD', 1.0), ('FD', 0.5), ('FF', 0.0), ('NA', -1), ('P', -1); create table enrolls ( enroll_id serial primary key, student_id int references students(student_id) not null, lecture_code char(7) references lectures(lecture_code) not null, grade_id int references grades(grade_id) ) 2. Create a dummy data. You can use…
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...