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SQL Challenges-1: Find Student Supporter

SQL Challenges-1: Exercise-15 with Solution

From the following table, write a SQL query to find those students who have referred by the teacher whose id not equal to 602. Return the student names.

Input:

Table: students

Structure:

FieldTypeNullKeyDefaultExtra
student_idint(11)YES
student_namevarchar(25)YES
teacher_idint(11)YES

Data:

student_idstudent_nameteacher_id
1001Alex601
1002Jhon
1003Peter
1004Minto604
1005Crage
1006Chang601
1007Philip602

Sample Solution:

SQL Code(MySQL):

CREATE TABLE IF NOT EXISTS students (student_id INT,student_name VARCHAR(25),teacher_id INT);
TRUNCATE TABLE students;

CREATE TABLE IF NOT EXISTS students (student_id INT,student_name VARCHAR(25),teacher_id INT);
INSERT INTO students (student_id, student_name, teacher_id) values ('1001', 'Alex', '601');
INSERT INTO students (student_id, student_name, teacher_id) values ('1002', 'Jhon', NULL);
INSERT INTO students (student_id, student_name, teacher_id) values ('1003', 'Peter', NULL);
INSERT INTO students (student_id, student_name, teacher_id) values ('1004', 'Minto', '604');
INSERT INTO students (student_id, student_name, teacher_id) values ('1005', 'Crage', NULL);
INSERT INTO students (student_id, student_name, teacher_id) values ('1006', 'Chang', '601');
INSERT INTO students (student_id, student_name, teacher_id) values ('1007', 'Philip', '602');
SELECT student_name 
FROM students WHERE teacher_id <> 602 
OR teacher_id IS NULL;

Sample Output:

student_name|
------------|
Alex        |
Jhon        |
Peter       |
Minto       |
Crage       |
Chang       |

SQL Code Editor:


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