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                        Node.js - PostgreSQL LIKE operator
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                                In this article, we would like to show you how to use Postgres LIKE operator in Node.js.
 
Note: at the end of this article you can find database preparation SQL queries.
const { Client } = require('pg');
const client = new Client({
    host: '127.0.0.1',
    user: 'my_username',
    database: 'my_database',
    password: 'my_password',
    port: 5432,
});
const findRows = async (pattern) => {
	const query = `
            SELECT * FROM "users"
            WHERE  "name"  LIKE $1;
    `;
    await client.connect();   // creates connection
    try {
        const { rows } =  await client.query(query, [pattern]);  // sends query
        return rows;
    } finally {
        await client.end();   // closes connection
    }
};
findRows('A%') // Ailisa, Andrew
    .then(result => console.table(result))
    .catch(error => console.error(error.stack));
Result:
┌─────────┬────┬──────────┬────────────┬───────────────┬───────────┐
│ (index) │ id │   name   │  surname   │ department_id │  salary   │
├─────────┼────┼──────────┼────────────┼───────────────┼───────────┤
│    0    │ 4  │ 'Ailisa' │  'Gomez'   │     null      │ '6500.00' │
│    1    │ 8  │ 'Andrew' │ 'Thompson' │     null      │ '2100.00' │
└─────────┴────┴──────────┴────────────┴───────────────┴───────────┘
Database preparation
create_tables.sql file:
CREATE TABLE "users" (
	"id" SERIAL,
	"name" VARCHAR(50) NOT NULL,
	"surname" VARCHAR(50) NOT NULL,
	"department_id" INTEGER,
    "salary" DECIMAL(15,2) NOT NULL,
	PRIMARY KEY ("id")
);
insert_data.sql file:
INSERT INTO "users"
	( "name", "surname", "department_id", "salary")
VALUES
	('John', 'Stewart', 1, '3512.00'),
	('Chris', 'Brown', 2, '1344.00'),
	('Kate', 'Lewis', 3, '6574.00'),
	('Ailisa', 'Gomez', NULL, '6500.00'),
	('Gwendolyn', 'James', 2, '4200.00'),
	('Simon', 'Cousersllins', 4, '3320.00'),
	('Taylor', 'Martin', 2, '1500.00'),
	('Andrew', 'Thompson', NULL, '2100.00');
Native SQL query (used in the above example):
SELECT * FROM "users"
WHERE  "name"  LIKE 'A%';
                                    
                                    
                                