# Yangi ma'lumotlar bazasi yaratish
sqlite3 mydatabase.db
# Mavjud ma'lumotlar bazasini ochish
sqlite3 existing_database.db
# Xotira bazasini ochish (vaqtinchalik)
sqlite3 :memory:
SQLite CLI ishga tushganda siz quyidagicha prompt ko'rasiz:
SQLite version 3.39.0 2022-06-25 14:57:56
Enter ".help" for usage hints.
sqlite>
CREATE TABLE users (
id INTEGER PRIMARY KEY,
name TEXT NOT NULL,
email TEXT UNIQUE,
age INTEGER,
created_at DATETIME DEFAULT CURRENT_TIMESTAMP
);
INSERT INTO users (name, email, age) VALUES
('Ali Valiyev', '[email protected]', 25),
('Malika Karimova', '[email protected]', 30),
('Bobur Toshmatov', '[email protected]', 28);
# Barcha ma'lumotlar
SELECT * FROM users;
# Belgili ustunlar
SELECT name, email FROM users;
# Shart bilan
SELECT * FROM users WHERE age > 25;
# Sarlavhalarni yoqish
.headers on
# Jadval formatida ko'rsatish
.mode table
# CSV formatida
.mode csv
# JSON formatida
.mode json
# Natijani faylga saqlash
.output results.txt
SELECT * FROM users;
.output stdout
# Ekranga qaytarish
.output stdout
# Ma'lumotlar bazalari
.databases
# Jadvallar
.tables
# Jadval strukturasi
.schema users
# Hozirgi sozlamalar
.show
# Ma'lumotni yangilash
UPDATE users SET age = 26 WHERE name = 'Ali Valiyev';
# Ma'lumotni o'chirish
DELETE FROM users WHERE age < 25;
# Barcha ma'lumotlarni o'chirish
DELETE FROM users;
# SQLite ni ishga tushirish
sqlite3 company.db
# Sozlamalarni o'rnatish
.headers on
.mode table
# Jadvallar yaratish
CREATE TABLE employees (
id INTEGER PRIMARY KEY,
name TEXT NOT NULL,
position TEXT,
salary REAL,
department_id INTEGER
);
CREATE TABLE departments (
id INTEGER PRIMARY KEY,
name TEXT NOT NULL,
location TEXT
);
# Ma'lumotlar qo'shish
INSERT INTO departments (name, location) VALUES
('IT', 'Toshkent'),
('Marketing', 'Samarqand'),
('Sales', 'Buxoro');
INSERT INTO employees (name, position, salary, department_id) VALUES
('Ahmad Karimov', 'Developer', 5000, 1),
('Malika Toshmatova', 'Designer', 4500, 1),
('Bobur Valiyev', 'Manager', 6000, 2);
# Ma'lumotlarni ko'rish
SELECT * FROM employees;
SELECT * FROM departments;
# JOIN so'rov
SELECT e.name, e.position, d.name as department
FROM employees e
JOIN departments d ON e.department_id = d.id;
# Chiqish
.quit