Summary: in this tutorial, you will learn how to use the PostgreSQL jsonb_agg() function to aggregate values into a JSON array.
Introduction to the PostgreSQL jsonb_agg() function
The jsonb_agg() function is an aggregate function that allows you to aggregate values into a JSON array.
The jsonb_agg() function can be particularly useful when you want to create a JSON array from data of multiple rows.
Here’s the syntax of the jsonb_agg() function:
jsonb_agg(expression)In this syntax:
expression: is any valid expression that evaluates to a JSON value.
The jsonb_agg() function returns a JSON array that consists of data from multiple rows.
PostgreSQL jsonb_agg() function example
Let’s explore some examples of using the jsonb_agg() function.
1) Basic jsonb_agg() function example
First, create a new table called products:
CREATE TABLE products (
id SERIAL PRIMARY KEY,
name VARCHAR(100) NOT NULL,
price DECIMAL(10, 2) NOT NULL
);Second, insert some rows into the products table:
INSERT INTO products (name, price)
VALUES
('Laptop', 1200.00),
('Smartphone', 800.00),
('Headphones', 100.00);Third, use the jsonb_agg() function to aggregate product information into a JSON array:
SELECT
jsonb_agg(
jsonb_build_object('name', name, 'price', price)
) AS products
FROM
products;Output:
products
--------------------------------------------------------------------------------------------------------------------------
[{"name": "Laptop", "price": 1200.00}, {"name": "Smartphone", "price": 800.00}, {"name": "Headphones", "price": 100.00}]
(1 row)2) Using jsonb_agg() function with GROUP BY clause
First, create new tables called departments and employees:
CREATE TABLE departments(
id SERIAL PRIMARY KEY,
department_name VARCHAR(255) NOT NULL
);
CREATE TABLE employees(
id SERIAL PRIMARY KEY,
employee_name VARCHAR(255) NOT NULL,
department_id INT NOT NULL,
FOREIGN KEY (department_id)
REFERENCES departments(id) ON DELETE CASCADE
);Second, insert rows into departments and employees tables:
INSERT INTO departments (department_name)
VALUES
('Engineering'),
('Sales')
RETURNING *;
INSERT INTO employees (employee_name, department_id)
VALUES
('John Doe', 1),
('Jane Smith', 1),
('Alice Johnson', 1),
('Bob Brown', 2)
RETURNING *;The departments table:
id | department_name
----+-----------------
1 | Engineering
2 | Sales
(2 rows)The employees table:
id | employee_name | department_id
----+---------------+---------------
1 | John Doe | 1
2 | Jane Smith | 1
3 | Alice Johnson | 1
4 | Bob Brown | 2
(4 rows)Third, use the jsonb_agg() function to retrieve departments and a list of employees for each department in the form of a JSON array:
SELECT
department_name,
jsonb_agg(employee_name) AS employees
FROM
employees e
INNER JOIN departments d ON d.id = e.department_id
GROUP BY
department_name;Output:
department_name | employees
-----------------+---------------------------------------------
Engineering | ["John Doe", "Jane Smith", "Alice Johnson"]
Sales | ["Bob Brown"]
(2 rows)3) Using jsonb_agg() function with NULLs
First, drop the departments and employees tables:
DROP TABLE employees;
DROP TABLE departments;Second, recreate the departments and employees tables:
CREATE TABLE departments(
id SERIAL PRIMARY KEY,
department_name VARCHAR(255) NOT NULL
);
CREATE TABLE employees(
id SERIAL PRIMARY KEY,
employee_name VARCHAR(255) NOT NULL,
department_id INT NOT NULL,
FOREIGN KEY (department_id)
REFERENCES departments(id) ON DELETE CASCADE
);Third, insert rows into the departments and employees tables:
INSERT INTO departments (department_name)
VALUES
('Engineering'),
('Sales'),
('IT')
RETURNING *;
INSERT INTO employees (employee_name, department_id)
VALUES
('John Doe', 1),
('Jane Smith', 1),
('Alice Johnson', 1),
('Bob Brown', 2)
RETURNING *;Output:
The departments table:
id | department_name
----+-----------------
1 | Engineering
2 | Sales
3 | IT
(3 rows)The employees table:
id | employee_name | department_id
----+---------------+---------------
1 | John Doe | 1
2 | Jane Smith | 1
3 | Alice Johnson | 1
4 | Bob Brown | 2
(4 rows)Third, use the jsonb_agg() function to retrieve departments and a list of employees for each department in the form of a JSON array:
SELECT
department_name,
jsonb_agg (employee_name) AS employees
FROM
departments d
LEFT JOIN employees e ON d.id = e.department_id
GROUP BY
department_name;Output:
department_name | employees
-----------------+---------------------------------------------
Engineering | ["John Doe", "Jane Smith", "Alice Johnson"]
Sales | ["Bob Brown"]
IT | [null]
(3 rows)In this example, the IT department has no employees therefore jsonb_agg() function returns an array that contains a null value.
To skip the null and make the JSON array an empty array, you can use the jsonb_agg_strict() function:
SELECT
department_name,
jsonb_agg_strict (employee_name) AS employees
FROM
departments d
LEFT JOIN employees e ON d.id = e.department_id
GROUP BY
department_name;Output:
department_name | employees
-----------------+---------------------------------------------
Engineering | ["John Doe", "Jane Smith", "Alice Johnson"]
Sales | ["Bob Brown"]
IT | []
(3 rows)The jsonb_agg_strict() function works like the jsonb_agg() except that it skips the null values.
Summary
- Use the
jsonb_agg()function to aggregate values into a JSON array.