MySQL – How to Check if JSON Feature Exists and/or Version

You probably want to use this syntax:

CREATE TABLE IF NOT EXISTS `table1` ( -- create table );
-- returns only a warning if the table already exists

Unless you want to check that the table name does not exist instance-wide, in which the way you "program with SQL" should be a bit different, for several reasons:

SET @check_table_name := false;

SELECT true INTO @check_table_name 
FROM information_schema.TABLES 
WHERE TABLE_NAME = 'table1'
LIMIT 1;   

Perhaps doing a COUNT is better since it must be a number always

DELIMITER $$
DROP PROCEDURE IF EXISTS `USER_EXISTS` $$
CREATE DEFINER=`root`@`localhost` PROCEDURE `USER_EXISTS`
(IN `GIVEN_USERNAME` VARCHAR(64) CHARSET utf8mb4)
BEGIN
    SET @User_exists = 0;
    SELECT COUNT(1) INTO @found
    FROM `dbname`.`tablename`
    WHERE `username` = GIVEN_USERNAME;
    IF @found > 0 THEN
        SET @User_exists = 1;
    END IF;
    SELECT @User_exists;
END $$
DELIMITER ;

You could also write it as a function

DELIMITER $$
DROP FUNCTION IF EXISTS `USER_EXISTS` $$
CREATE DEFINER=`root`@`localhost` FUNCTION `USER_EXISTS`
(`GIVEN_USERNAME` VARCHAR(64) CHARSET utf8mb4) RETURNS INT
BEGIN
    SET @User_exists = 0;
    SELECT COUNT(1) INTO @found
    FROM `dbname`.`tablename`
    WHERE `username` = GIVEN_USERNAME;
    IF @found > 0 THEN
        SET @User_exists = 1;
    END IF;
    RETURN @User_exists;
END $$
DELIMITER ;

Then, you could call the function to get a number

SET @x = USER_EXISTS('Synnyster');
SELECT USER_EXISTS('Synnyster');

Give it a Try !!!