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Create an image of a colorful database with SQL code snippets and a pen, representing learning and mastery in SQL skills. Include graphical elements that suggest quizzes and challenges.

SQL Mastery Quiz

Test your SQL skills with our engaging quiz designed to challenge your knowledge of SQL commands and database manipulation. Whether you're a beginner or a seasoned developer, this quiz will help you assess your understanding of SQL concepts.

Key Features:

6 challenging questions
Instant feedback on your answers
Track your progress and improve your skills

6 Questions2 MinutesCreated by CodingNinja102

Füge "Peter Pan" in die Tabelle customers ein

INSERT INTO TABLE customers VALUES name='Peter',lastname='Pan''

INSERT name='Peter',lastname='Pan' INTO TABLE customers

INSERT INTO customers(name,lastname) VALUES ('Peter','Pan')

INSERT INTO customers VALUES ('Peter','Pan)

Ändere den Nachnamen von "Müller" in "Obermüller" , in der Tabelle customers

UPDATE lastname = 'Obermüller' WHERE lastname = 'Müller'

UPDATE customers SET lastname = 'Obermüller' WHERE lastname = 'Obermüller'

UPDATE customers lastname = 'Obermüller' WHERE lastname = 'Müller'

UPDATE customers SET (lastname) VALUES ('Obermüller') WHERE lastname = 'Müller'

Selektiere alle Einträge der Tabelle customers, die ein "m" im name Feld enthalten

SELECT * FROM customers WHERE name = 'm'

SELECT * FROM customers WHERE name LIKE 'm'

SELECT * FROM customers WHERE name ='%m%'

SELECT * FROM customers WHERE name LIKE '%m%'

Füge ein weiteres Feld namens "Adresse" in die Tabelle customers ein

TABLE customers ADD Adresse varchar(50)

ALTER TABLE customers ADD Adresse varchar(50)

ALTER TABLE customers ADD Adresse TINYINT

ADD COLUMN Adresse varchar(50) TABLE customers

Vereine die Tabellen customer und adress

customer

id	name	lastname	adressId
1	Thomas	Müller	1

adress

id	name
1	Hauptstraße 6

SELECT customers.*,adress* FROM customer INNER JOIN adress ON customer.adressId = adress.id

SELECT * FROM customers FROM adress COMBINE adress AND TABLE ON adress.id = customers.adressId

SELECT * customers.*,adress* FROM customers WHERE adress.id = customers.adressId

SELECT adress INTO customers JOIN ON adress.id = customers.adressId

Zähle , wieviel Personen in der Tabelle customers mit dem lastname Müller existieren

COUNT FROM customers WHERE lastname LIKE 'Müller'

SELECT COUNT(id) FROM customers WHERE lastname = 'Müller'

SELECT * FROM customers COUNT('Müller')

SELECT SUM(id) FROM customers WHERE id = 'Müller'

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