The REPLACE
statement allows us to overwrite a new row when we insert. When using REPLACE, MySQL will delete a row if there is a duplicate, then insert the new row. In this article, we will learn how to use REPLACE in MySQL.
The basic syntax of REPLACE is as follows:
REPLACE INTO [table]([columns])
VALUES([values]);
This syntax will allow us to delete multiple rows based on the join and conditions we supply.
We will be using docker in this article, but feel free to install your database locally instead. Once you have docker installed, create a new file called docker-compose.yml
and add the following.
version: '3'
services:
db:
image: mysql:latest
container_name: db
environment:
MYSQL_ROOT_PASSWORD: root_pass
MYSQL_DATABASE: app_db
MYSQL_USER: db_user
MYSQL_PASSWORD: db_user_pass
ports:
- "6033:3306"
volumes:
- dbdata:/var/lib/mysql
phpmyadmin:
image: phpmyadmin/phpmyadmin
container_name: pma
links:
- db
environment:
PMA_HOST: db
PMA_PORT: 3306
PMA_ARBITRARY: 1
restart: always
ports:
- 8081:80
volumes:
dbdata:
Next, run docker-compose up
.
Now, navigate to http://localhost:8081/
to access phpMyAdmin. Then log in with the username root
and pass root_pass
.
Click the SQL tab and you are ready to go.
In this article, we will need some data to work with. We will be using the sample db provided here: https://dev.mysql.com/doc/employee/en/. However, we will only enter what we need rather than import the whole db.
With the SQL tab open (or your own sql cli going), let's first create our DB and select it.
create DATABASE if not EXISTS sakila;
USE sakila;
CREATE TABLE employees (
emp_no INT NOT NULL,
birth_date DATE NOT NULL,
first_name VARCHAR(14) NOT NULL,
last_name VARCHAR(16) NOT NULL,
gender ENUM ('M','F') NOT NULL,
hire_date DATE NOT NULL,
PRIMARY KEY (emp_no)
);
CREATE TABLE salaries (
emp_no INT NOT NULL,
salary INT NOT NULL,
from_date DATE NOT NULL,
to_date DATE NOT NULL,
FOREIGN KEY (emp_no) REFERENCES employees (emp_no) ON DELETE CASCADE,
PRIMARY KEY (emp_no, from_date)
);
Now, let's enter a few rows
INSERT INTO employees VALUES (10001,'1953-09-02','Georgi','Facello','M','1986-06-26'),
(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21'),
(10003,'1959-12-03','Parto','Bamford','M','1986-08-28'),
(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01'),
(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12'),
(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02'),
(10007,'1957-05-23','Tzvetan','Zielinski','F','1989-02-10'),
(10008,'1958-02-19','Saniya','Kalloufi','M','1994-09-15'),
(10009,'1952-04-19','Sumant','Peac','F','1985-02-18'),
(10010,'1963-06-01','Duangkaew','Piveteau','F','1989-08-24'),
(10011,'1953-11-07','Mary','Sluis','F','1990-01-22'),
(10012,'1960-10-04','Patricio','Bridgland','M','1992-12-18'),
(10013,'1963-06-07','Eberhardt','Terkki','M','1985-10-20'),
(10014,'1956-02-12','Berni','Genin','M','1987-03-11'),
(10015,'1959-08-19','Guoxiang','Nooteboom','M','1987-07-02'),
(10016,'1961-05-02','Kazuhito','Cappelletti','M','1995-01-27'),
(10017,'1958-07-06','Cristinel','Bouloucos','F','1993-08-03'),
(10018,'1954-06-19','Kazuhide','Peha','F','1987-04-03'),
(10019,'1953-01-23','Lillian','Haddadi','M','1999-04-30'),
(10020,'1952-12-24','Mayuko','Warwick','M','1991-01-26');
INSERT INTO salaries VALUES (10001,60117,'1986-06-26','1987-06-26'),
(10002,62102,'1987-06-26','1988-06-25'),
(10003,66074,'1988-06-25','1989-06-25'),
(10004,66596,'1989-06-25','1990-06-25'),
(10005,66961,'1990-06-25','1991-06-25'),
(10006,71046,'1991-06-25','1992-06-24'),
(10007,74333,'1992-06-24','1993-06-24'),
(10008,75286,'1993-06-24','1994-06-24'),
(10009,75994,'1994-06-24','1995-06-24'),
(10010,76884,'1995-06-24','1996-06-23'),
(10011,80013,'1996-06-23','1997-06-23'),
(10012,81025,'1997-06-23','1998-06-23'),
(10013,81097,'1998-06-23','1999-06-23');
Let’s start by inserting some data:
INSERT INTO employees VALUES
(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'),
(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21');
Now, let’s select the rows to see the ids.
SELECT * FROM employees;
emp_no | birth_date | first_name | last_name | gender | hire_date |
---|---|---|---|---|---|
10006 | 1953-04-20 | Anneke | Preusig | F | 1989-06-02 |
10007 | 1957-05-23 | Tzvetan | Zielinski | F | 1989-02-10 |
10008 | 1958-02-19 | Saniya | Kalloufi | M | 1994-09-15 |
10009 | 1952-04-19 | Sumant | Peac | F | 1985-02-18 |
10010 | 1963-06-01 | Duangkaew | Piveteau | F | 1989-08-24 |
10011 | 1953-11-07 | Mary | Sluis | F | 1990-01-22 |
If we want to replace row 2, we can insert using the following.
REPLACE INTO employees(emp_no, birth_date, first_name, last_name, gender, hire_date)
VALUES (10002,'1953-01-23','Lillian','Haddadi','M','1999-04-30');
Now, select the data again.
SELECT * FROM employees;
emp_no | birth_date | first_name | last_name | gender | hire_date |
---|---|---|---|---|---|
10002 | 1953-01-23 | Lillian | Haddadi | M | 1999-04-30 |
10006 | 1953-04-20 | Anneke | Preusig | F | 1989-06-02 |
10007 | 1957-05-23 | Tzvetan | Zielinski | F | 1989-02-10 |
10008 | 1958-02-19 | Saniya | Kalloufi | M | 1994-09-15 |
Notice that there was a conflict on the emp_no key, so the full row was replaced.
We can also use REPLACE to update our data. The REPLACE works like an UPDATE, except we don’t need to use the WHERE clause as REPLACE will search for a matching record first. For example, we can update row 3, like so:
REPLACE INTO employees
SET emp_no = 1003,
birth_date = '1953-01-23',
first_name = 'Alex',
last_name = 'Tam',
gender = 'M',
hire_date = '1999-04-30';
And the result:
SELECT * FROM employees;
emp_no | birth_date | first_name | last_name | gender | hire_date |
---|---|---|---|---|---|
10002 | 1953-01-23 | Lillian | Haddadi | M | 1999-04-30 |
10003 | 1953-01-23 | Alex | Tam | M | 1999-04-30 |