[Nico]

My guess is you made a spelling mistake in line 23. The error you get says:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /opt/lampp/htdocs/Sites/prueba.php on line 23

... but according to my code, you shouldn't even get to this point if the query failed:

PHP Code:

$result = mysql_query("select * from clientes") OR die(mysql_error());

$row = mysql_fetch_array($result); 

... the second line triggers the error, but the first one would have stopped the script on failure.

[tirengarfio]

Sorry... i forget adding to my code "OR die(mysql_error())" ....


I've just added it, and when i open the page, this message is showed: "No database selected".


My steps are:

- Start the apache mysql php server (XAMPP).
- Open the page with the code we have been discusing.

Should i need something more??

[Nico]

Post your new code. I've added database selection and error handing in mine, but I'm not sure what you've done now...

[tirengarfio]

My code is below. NOW the error is: "Error conectando a la base de datos".

PHP Code:

<?php

$nombre="Javier";

$nombre2;


if (!$link = mysql_connect("localhost","",""))
{
    echo "Error conectando a la base de datos.";
    exit();
}

if (!mysql_select_db("ejemplo"))
{
    echo "Error conectando a la base de datos.";
    exit();
}

mysql_query("insert into clientes(nombre) values('$nombre')") OR die(mysql_error());

$result = mysql_query("select * from clientes") OR die(mysql_error());
$row = mysql_fetch_array($result);

echo $row["nombre"];

?>

[Nico]

What does this output?

PHP Code:

<?php

$nombre = "Javier";

mysql_connect("localhost", "", "") OR die(mysql_error());
mysql_select_db("ejemplo") OR die(mysql_error());

mysql_query("insert into clientes(nombre) values('$nombre')") OR die(mysql_error());

$result = mysql_query("select * from clientes") OR die(mysql_error());
$row = mysql_fetch_array($result);

echo $row["nombre"];

?>

[tirengarfio]

Originally Posted by Nico

What does this output?

PHP Code:

<?php



$nombre = "Javier";



mysql_connect("localhost", "", "") OR die(mysql_error());

mysql_select_db("ejemplo") OR die(mysql_error());



mysql_query("insert into clientes(nombre) values('$nombre')") OR die(mysql_error());



$result = mysql_query("select * from clientes") OR die(mysql_error());

$row = mysql_fetch_array($result);



echo $row["nombre"];



?>

It gives this error:

"Access denied for user ''@'localhost' to database 'ejemplo'"

[ianbrind]

Try using "root" as your database username.

[tirengarfio]

Originally Posted by ianbrind

Try using "root" as your database username.

Hi, as you said i have wroten this:

Code:

mysql_connect("localhost", "root", "") OR die(mysql_error());

And now, when i open the file with my browser i just can see this written: first.

Any idea?!?

[tirengarfio]

Hi again,

I have done a "select * from clientes" and i have seen the first element of my table is "nombre => first", so everything is working OK using "root" as username.

Anyway, why must i use "root" as username?

[Nico]

Because root is usually the default user name. You could change it, but if you're working locally, I wouldn't really bother, to be honest.