SQL
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
Füge "Peter Pan" in die Tabelle customers einINSERT 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 customersUPDATE 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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