Parse error: syntax error, unexpected T_STRING don't know why

Question

I was trying to make a form to insert data into a mysql database and I keep getting the same error:

Parse error: syntax error, unexpected T_STRING in /home/u272231605/public_html/add_reg.php3 on line 20

Here's the code where the error is showed:

<html>
<body>

<?php
if (isset($_POST['Nombre Monstruo']) && !empty($_POST['Nombre Monstruo'])&&
    isset($_POST['Ataque']) && !empty($_POST['Ataque'])&&
    isset($_POST['Defensa']) && !empty($_POST['Defensa'])&&
    isset($_POST['Tipo']) && !empty($_POST['Tipo'])&&
    isset($_POST['Nivel']) && !empty($_POST['Nivel'])&&
    isset($_POST['Efecto']) && !empty($_POST['Efecto'])){
    $nombremonstruo = $_POST['Nombre Monstruo'];
    $ataque = $_POST['Ataque'];
    $defensa = $_POST['Defensa'];
    $tipo = $_POST['Tipo'];
    $nivel = $_POST['Nivel'];
    $efecto = $_POST['Efecto'];
    $db = u272231605_data;
    $host = mysql.hostinger.es;
    $user = u272231605_data;
    $password = 4Ym8ix2A;
    $con = mysql_connect($host,$user,$password)or die('No se ha podido conectar con      la base de datos');
    mysql_select_db($db,$con) or die('No se pudo conectar con la base de datos');
    mysql_query("INSERT INTO (Nombre del monstruo,Ataque,Defensa,Tipo,Nivel,Efecto" values('$nombremonstruo','$ataque','$defensa','$tipo','$nivel','$efecto')");
    echo "Guardado correctamente"
    echo "<a href=index.php"> Registrar otro monstruo </a>;
}
    else
    {
        echo "Han de introducirse todos los datos para que el registro de la   carta se complete.";
    } 

?>
</body>
</html>

Thanks in advance for the help provided :)

Might be an idea NOT to post your db details (including password) so publicly — Mark Baker
– Mark Baker, Commented Jul 26, 2013 at 20:18

Bart Friederichs · Accepted Answer · 2013-07-26 20:11:38Z

2

Your problem is here:

$db = u272231605_data;

and on other places. You should put double quotes around them:

$db = "u272231605_data";

answered Jul 26, 2013 at 20:11

Bart Friederichs

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Comments

Marc B · Accepted Answer · 2013-07-26 20:11:54Z

0

You've forgotten quotes in multiple places:

$db = u272231605_data;
$host = mysql.hostinger.es;
$user = u272231605_data;
$password = 4Ym8iJ2A;

Each of those values should be quoted. As they stand now, they're treated by PHP as undefined constants. But the last one (4Ym8iJ2A) is the cause of the error, because constants cannot start with numbers.

They should probably be:

$db = 'u272231605_data';
$host = 'mysql.hostinger.es';
$user = 'u272231605_data';
$password = '4Ym8iJ2A';

answered Jul 26, 2013 at 20:11

Marc B

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Comments

Fallen · Accepted Answer · 2013-07-26 20:12:03Z

0

As well as the missing quotes for variables, there are multiple issues with these 4 lines of codes. Please use a smart editor to highlight syntax and etc. Following lines are updated:

mysql_select_db($db,$con) or die('No se pudo conectar con la base de datos');
mysql_query("INSERT INTO (Nombre del monstruo,Ataque,Defensa,Tipo,Nivel,Efecto) values('$nombremonstruo','$ataque','$defensa','$tipo','$nivel','$efecto')");
echo "Guardado correctamente";
echo "<a href='index.php'> Registrar otro monstruo </a>";

answered Jul 26, 2013 at 20:12

Fallen

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Comments

Adam MacDonald · Accepted Answer · 2013-07-26 20:12:03Z

0

you need to wrap your variables in single quotes:

$db = 'u272231605_data';
$host = 'mysql.hostinger.es';
$user = 'u272231605_data';

answered Jul 26, 2013 at 20:12

Adam MacDonald

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Comments

Dave · Accepted Answer · 2013-07-26 20:12:04Z

0

$password = "4Ym8iJ2A";

The value needs single or double quotes.

answered Jul 26, 2013 at 20:12

Dave

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Comments

Steve · Accepted Answer · 2013-07-26 20:12:34Z

0

You are missing a semi-colon on this line:

echo "Guardado correctamente"

answered Jul 26, 2013 at 20:12

Steve

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