PHP MySQL - Error: No Database selected

Question

I am trying to read and write to a database. Here is the code I have so far:

$mysql = mysqli_connect("example.com", "johndoe", "abc123"); // replace with actual credidentials
$sql = "CREATE DATABASE IF NOT EXISTS dbname";
if (!mysqli_query($mysql, $sql)) {
    echo "Error creating database: " . mysqli_error($mysql);
}
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_close($mysql);
$mysql = mysqli_connect("example.com", "johndoe", "abc123", "dbname"); // replace with actual credidentials
$sql = "CREATE TABLE IF NOT EXISTS Users(ID INT NOT NULL AUTO_INCREMENT, PRIMARY KEY(ID), username CHAR(15), password CHAR(15), email CHAR(50))";
if (!mysqli_query($mysql, $sql)) {
    echo "Error creating table: " . mysqli_error($mysql);
}
$sql = "INSERT INTO Customers(username, password, email) VALUES(" . $username . ", " . $password . ", " . $email . ")";
if (!mysqli_query($mysql, $sql)) {
    echo "Error: " . mysqli_error($mysql);
}
mysqli_close($mysql);

However, when I try to run it, it has an error:

Error: 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 ' , )' at line 1

Could anybody tell me how to fix this?

Answer 1

First check mysqli_select_db if it returns false then create database.

try like this:

 $mysql = mysqli_connect("example.com", "johndoe", "abc123") or die(mysqli_connect_error()); // replace with actual credidentials

if (!mysqli_select_db($mysql,'hardestgame_accounts')) {
 $sql = "CREATE DATABASE IF NOT EXISTS hardestgame_accounts";
 if (!mysqli_query($mysql, $sql)) {
    echo "Error creating database: " . mysqli_error($mysql);
 }
}
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "CREATE TABLE IF NOT EXISTS Users(ID INT NOT NULL AUTO_INCREMENT, PRIMARY KEY(ID), username CHAR(15), password CHAR(15), email CHAR(50))";
if (!mysqli_query($mysql, $sql)) {
    echo "Error creating table: " . mysqli_error($mysql);
}
mysqli_close($mysql);

here is a good answer: Php mysql create database if not exists