Creating table using MySQLi Object-oriented Procedure

mysqli mysql table creation 5 years, 3 months ago

<?php $servername = "localhost"; $username = "username"; $password = "password"; $dbname = "newDB"; // checking connection $conn = new mysqli($servername, $username, $password, $dbname); // Check connection if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } // sql code to create table $sql = "CREATE TABLE employees( id INT(2) PRIMARY KEY, firstname VARCHAR(30) NOT NULL, lastname VARCHAR(30) NOT NULL, email VARCHAR(50) )"; if ($conn->query($sql) === TRUE) { echo "Table employees created successfully"; } else { echo "Error creating table: " . $conn->error; } $conn->close(); ?>

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                            <?php
$servername = "localhost";
$username = "username";

mysqli table creation mysql


                            CREATE TABLE `user` (
    `username` VARCHAR(25) NOT NULL,
    `email` VARCHAR(150) NOT NULL,


                            # pip install mysql-connector

from mysql import connector

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                            Drop table Table_name;


                            SELECT 
IFNULL(
  (SELECT NAME


                            SELECT BIN(5);

SELECT BIN(12);


                            RENAME USER 'RAJ' @ 'localhost' TO 'AMIT

RENAME USER 'PRIYA' @ 'localhost' TO 'DI


                            ALTER TABLE TABLE_NAME RENAME TO NEW_TAB

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                            ALTER TABLE TABLE_NAME RENAME COLUMN TO

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                            SELECT DEFAULT (COLUMN_NAME) FROM TABLE_

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                            DROP USER 'user' @ 'host';


                            SET PASSWORD FOR 'user' @ 'localhost' =

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                            ALTER USER user@localhost IDENTIFIED BY

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                            UPDATE USER SET AUTHENTICATION_STRING = 
WHERE USER = 'USER_NAME' AND
HOST = 'LOCALHOST';

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                            CREATE USER 'user_name' @ 'localhost' ID

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                            CREATE USER
'user_name' @ 'localhost' IDENTIFIED BY 
'user_name' @ 'localhost' IDENTIFIED BY

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                            SELECT COUNT(SALARY)
FROM EMPLOYEES
WHERE SALARY >= 40000;


                            SELECT AVG(SALARY) FROM EMPLOYEES;

SELECT MAX(column_name)

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                            SELECT DATABASE();

SELECT CURRENT_USER();

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                            SELECT MAX (column_name) 
FROM table_name 
WHERE column_name NOT IN (SELECT Max (co

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