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up vote 0 down vote favorite

I have this code, but the first query doesn't run (at phpmyadmin works!), I tried to run the code in 2 servers (maybe the config of php/mysql) but the results are the same.

$habitaciones = "SELECT habitacion.id AS habid, habitacion.nombre AS habnom, tipo.num_cama AS cantidad FROM habitacion, tipo WHERE id_tipo = tipo.id";
$enviar_sql = mysql_query($habitaciones, $enlace);

while($mostar_habs = mysql_fetch_array($enviar_sql)){
    echo "<table><tr>";

    $habid = $mostrar_habs['habid'];
    $habnom = $mostrar_habs['habnom'];
    echo "valor de habid: " .$habid;

    if($mostrar_habs['cantidad'] == 1){
        $i = 0;
        echo "<td>" . $habnom . "</td>";
        $fecha = $fechas[$i];

        $ocupacion1 = "SELECT cama.id AS camaid, cliente.nombre AS nombre, cama.ocupada AS ocupada FROM cliente, evento, cama, habitacion 
        WHERE cliente.id = id_cliente AND id_habitacion = habitacion.id AND cama.id = id_cama AND habitacion.id = " . $habid . " 
        AND checkin = \"" . $fecha . "\"";
        $enviar_ocupacion1=mysql_query($ocupacion1, $enlace);

        for($cliens=1; $mostrar_clien = mysql_fetch_array($enviar_ocupacion1); $cliens+=1){
            echo "<td>" . $mostrar_clien['nombre'] . "</td>";
        }
        $i++;
    }
    else{

        $i = 0;
        echo "<td>" . $habnom . "</td>";
        echo "<tr>";
        $fecha = $fechas[$i];
        $camas = 'SELECT cama.numero AS nombre, cama.id AS camaid FROM cama, habitacion WHERE habitacion.id = id_habitacion AND habitacion.id = '.$habid;
        $enviar_camas = mysql_query($camas, $enlace);
        //echo $camas;

        for($cams=1; $mostrar_camas = mysql_fetch_array($enviar_camas); $cams+=1){
            echo "<td>" . $mostrar_camas['nombre'] . "</td>";
            $fecha = $fechas[$i];
            $ocupacion2 = "SELECT cliente.id AS clienid, cliente.nombre AS nombre FROM cliente, evento, cama WHERE
            cliente.id = id_cliente AND cama.id = id_cama AND id_habitacion = " . $mostrar_habs['camaid'] . " AND checkin = \"" . $fecha . "\"";
            $enviar_ocupacion2 = mysql_query($ocupacion2, $enlace);

            for($cliens = 1; $mostrar_cliens = mysql_fetch_array($enviar_ocupacion); $cliens+=1){
                echo "<td>" . $mostrar_cliens['nombre'] . "</td>";
            }
            $i++;
        }
        echo "</tr>";
    }
    echo "</tr></table>";
}

The problem is in the first mysql_query

$habitaciones = "SELECT habitacion.id AS habid, habitacion.nombre AS habnom, tipo.num_cama AS cantidad FROM habitacion, tipo WHERE id_tipo = tipo.id";
                $enviar_sql = mysql_query($habitaciones, $enlace);

All the code depends on this query, at the browser returns

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

because the querys into if() doesn't have the value of the first query

Any idea? I don't understand why doesn't works

Thanks for all and sorry for my english

share|improve this question
5  
Why do people still use mysql functions without any wrapping/proper error handling? –  ThiefMaster♦ Mar 4 '11 at 23:39
    
a tiny guess, did you try to mention the table name in the where clause? Try WHERE tablename.id_tipo = tipo.id –  sled Mar 4 '11 at 23:55

3 Answers 3

up vote 4 down vote

Try this:

$enviar_sql=mysql_query($habitaciones, $enlace) or trigger_error(mysql_error());

It will show you the error, allowing you to debug it.

share|improve this answer
1  
+1 for not using the die() crap –  ThiefMaster♦ Mar 4 '11 at 23:44
    
The same that @ThiefMaster, mysql_error() doesn't return anything :( –  Javierh Mar 4 '11 at 23:52
    
have you tried to run the query in phpmyadmin or mysql console? If the query is fine, maybe something is wrong with the connection? –  sled Mar 4 '11 at 23:53
    
The query in phpmyadmin runs ok and the connection too, I have another php file with the same connection to the same database and works fine. –  Javierh Mar 5 '11 at 1:32
up vote 3 down vote

Replace $enviar_sql = mysql_query($habitaciones, $enlace); with this:

$enviar_sql = mysql_query($habitaciones, $enlace) or die(mysql_error());

While this is an awful way to handle errors and should not be used except to fix whatever issue there is with your query it's a good way to quickly find out what's going wrong.

share|improve this answer
    
Thanks for the answer but mysql_error() doesn't return anything –  Javierh Mar 4 '11 at 23:49
up vote 0 down vote 
error_reporting(E_ALL);
$habitaciones = 'SELECT h.id AS habid, h.nombre AS habnom, t.num_cama AS cantidad 
FROM habitacion h, tipo t
WHERE h.id_tipo = t.id'; // Less query? jajaa
$enviar_sql = mysql_query($habitaciones, $enlace);

And try with: 

while ($mostar_habs = mysql_fetch_assoc($result)) {...
share|improve this answer
    
Seguro $enlace = mysql_connect('localhost', 'usuario', 'password'); está mal o $enviar_sql = mysql_query($habitaciones); –  Joseadrian Mar 5 '11 at 2:15
    
$enlace que es el mysql_connect funciona perfectamente (otras partes de la misma web funcionan. El mysql_query debería funcionar ya que la consulta en phpmyadmin funciona y muestra los datos correctos –  Javierh Mar 5 '11 at 13:13
    
The only notices that php retuns are because the $mostrar_habs['habid']; and $mostrar_habs['habnom']; are undefined variable :S –  Javierh Mar 5 '11 at 13:18
    
Ya vi el problema... has puesto while(**mostar_habs**) y dentro de la lógica usas **mostrar** :) Jeje –  Joseadrian Mar 5 '11 at 19:24
    
@Joseadrian eres un genio, funciona! Te debo una cerveza! // @Joseadrian, you are a genius, it works! I owe a beer! –  Javierh Mar 5 '11 at 19:45

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