mysql-error-1064

Tried mysql> GRANT ALL PRIVILEGES ON *.* TO 'root'@'%' IDENTIFIED BY 'root' WITH GRANT OPTION; Getting ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'IDENTIFIED BY 'root' WITH GRANT OPTION' at line 1. Note: The same is working when tried in previous versions. Also tried mysql> GRANT ALL PRIVILEGES ON *.* TO 'root'@'%' WITH GRANT OPTION; Getting ERROR 1410 (42000): You are not allowed to create a user with GRANT MySQL (8.0.11.0) username/password is root/root. Mike Lischke Starting with

This question already has an answer here: Syntax error due to using a reserved word as a table or column name in MySQL 1 answer I've looked through all the other StackOverflow (and google) posts with the same problem, but none seemed to address my problem. I am using PDO and PHP. My code: $vals = array( ':from' => $email, ':to' => $recipient, ':name' => $name, ':subject' => $subject, ':message' = >$message ); print_r($vals); try { $pdo = new PDOConfig(); $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $sql = "SELECT * FROM messages WHERE `message` LIKE :message"; $q = $pdo-

SELECT airline, airports.icao_code, continent, country, province, city, website FROM airlines FULL OUTER JOIN airports ON airlines.iaco_code = airports.iaco_code FULL OUTER JOIN cities ON airports.city_id = cities.city_id FULL OUTER JOIN provinces ON cities.province_id = provinces.province_id FULL OUTER JOIN countries ON cities.country_id = countries.country_id FULL OUTER JOIN continents ON countries.continent_id = continents.continent_id It says that You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'outer

This question already has an answer here: Syntax error due to using a reserved word as a table or column name in MySQL 1 answer I am trying to create 2 tables in the same MySQL database with a PHP-script: table 'user' with primary key 'user_id' and table 'order' with primary key 'order_id' and foreign key 'user_id' from the 'user' table (1 to many relationship). Table user creates successfully without problems: $sql="CREATE TABLE user( user_id INT UNSIGNED NOT NULL AUTO_INCREMENT, type ENUM('member','admin') NOT NULL, username VARCHAR(30) NOT NULL, email VARCHAR(80) NOT NULL, pass VARBINARY(32