PostgreSQL jsonb_agg() Function

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.