PHP is not creating MySQL database - php

Question

So I'm not sure as to why PHP is not creating a database. Let me show what my code looks like first, and then I will quote what warning/error it's giving me:
<?php
// Connection
$con = mysqli_connect("localhost", "root", "1monica1");
// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// Creating a database
// The database is called alarm_db
$sql = "CREATE DATABASE alarm_db";
if (mysql_query($sql, $con))
{
echo "Database alarm_db created sucessfully";
}
else
{
echo "Error creating database: " . mysqli_error($con);
}
?>
The error and warning I keep getting is this:
Warning: mysql_query(): A link to the server could not be established in B:\DropBox\Dropbox\EbadlyAgha\Ebad\reminders.php on line 51
Error creating database:
Line 51 is where the if (mysql_query($sql, $con)) is. Also, it won't give me the mysqli_error($con). It is just blank.
My MySQL connection is fine, because I never see the "Failed to connect to MySQL" statement.

Answer 1

It should be in mysqli_* while you executing DB creating query.
if (mysqli_query($sql,$con)) {
echo "Database alarm_db created sucessfully";
}
Try to avoid mysql_* statements due to the entire ext/mysql PHP extension, which provides all functions named with the prefix mysql_*, is officially deprecated as of PHP v5.5.0 and will be removed in the future.
There are two other MySQL extensions that you can better Use: MySQLi and PDO_MySQL, either of which can be used instead of ext/mysql.

Answer 2

Your connection is being generated by the mysqli library:
$con = mysqli_connect
but you are trying to query it using the mysql library:
mysql_query($sql,$con)
You need to be consistent about which library you use. Stick to mysqli throughout (since mysql is deprecated).
Note that mysqli_query expects the connection before the SQL:
mysqli_query($con, $sql);

Answer 3

You can try this to create a database using PHP:
<?php
$con = mysql_connect('localhost', 'root', '');
$query = mysql_query('create database alarm_db') or die(mysql_error());
if($query)
{
echo 'Database is created';
}
else
{
echo 'Database is not created';
}
?>