Question

I’m trying to connect to a database that I created using this code:

<?php
function Conectarse()
{
    $host="localhost";
    $user="root";
    $password="root";
    $bda="toobebe_catalogo";

    if (!($link=mysql_connect($host,$user,$password)))
    {
        echo "Error conectando a la base de datos.<br>";
        exit();
    }
    else
    {
        echo "Éxito conectando con la base de datos.<br>";
    }

    if (!mysql_select_db($bda,$link))
    {
        echo "Error seleccionando la base de datos.<br>";
        exit();
    }
    else
    {
        echo "Éxito al encontrar la base de datos.<br>";
    }
    return $link;
}

$conn=Conectarse();

$sql="SELECT * FROM toobebe-octubre";
$db_fila = mysql_query($sql,$conn);
$ok=1;

while (($row = mysql_fetch_array($db_fila)) && $ok) 
{
    $valor=mysql_query($sql,$conn);

if(!$valor)
{
    $ok=0;
}   
}

?>

But it fires this mistake when I execute it:

 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in ...

I’ve been searching, and to know:
– Database exists

• Permissions are correct

• Table exists

• Table is not null

Any idea on why this mistake is happening?

EDIT:
Added image with the mistake:

Answer 1

The error isn’t with connecting to the Database, but rather with your query. You have a hyphen in your table name, so you should try and enclose it as follows:

$sql = "SELECT * FROM `toobebe-octubre`";
$db_fila = mysql_query($sql, $conn);
$ok = 1;

while(($row = mysql_fetch_array($db_fila)) && $ok) 
{
    $valor=mysql_query($sql, $conn);
    if(!$valor)
    {
       $ok=0;
    }   
}

Just a couple of tips, using mysql_* is severely deprecated now. You should really be using mysqli_* at a very minimum, or PDO.

Also, SELECT * is generally considered a bad practice, because I really doubt you do need everything from the table.