if db exist php sql

[JurajSlovakID]

cavte,
po 2 rokoch ucenia sa js a php zacinam konecne vyvijat svoju aplikaciu a zistil som ze vela toho neviem, robit je ine ako pozerat na to...
v skratke chcem spravit chovanie - databaza vytvorena nieje -> vytvorit, ak je vypisat ze je
google mi s tymto vobec nepomohol
dalej preco ked spravim $conndb = mysqli_connect($serverName, $dBUsername, $dBPassword, $dBName);
mi vypise error lebo tu databazu zatial nemam ? preco tu premennu vobec spracovava? nieje to tak ze ju len deklarujem/iniciujem?
dakujem.

Kód: Vybrať všetko

<?php

$serverName = "localhost";
$dBUsername = "root";
$dBPassword = "";
$dBName = "nanochat";

$conn = mysqli_connect($serverName, $dBUsername, $dBPassword);



if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$sqlCreate = "CREATE DATABASE nanochat";

$sqlCheck = "SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = 'nanochat'";
if (mysqli_query($conn, $sqlCheck)) {
    echo "Database exist";
  } else {
    $sqlCreate;
    echo "Database created sucessfully.";
}

mysqli_close($conn);

[JurajSlovakID]

Poradil som si sam
Riesenie je:

Kód: Vybrať všetko

<?php

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "nanochat";

$conn = new mysqli($servername, $username, $password);


if (empty (mysqli_fetch_array(mysqli_query($conn,"SHOW DATABASES LIKE '$dbname'")))) 
{
    echo "DB not exist<br>"; 
    echo "Creating DB";
    mysqli_query($conn, "CREATE DATABASE nanochat");
}
else
{
    echo "DB exist<br>";
}



mysqli_close($conn);