Error: You have an error in your SQL syntax; check the manual that corresponds t

[Dann]

having problems inserting data into the mysql db.
the values from the form are actually being inserted into the bd but the test statement after that is showing the following:
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1

here's the php code

PHP Code: [Select]

<?php
include("../../lib/conectar.inc");

if(isset($_POST['submit'])){

$grava=mysql_query("INSERT INTO filmes (id,nome,ano,elenco,comentario) VALUES ('','$_POST[nome]','$_POST[ano]','$_POST[elenco]','$_POST[comentario]')",$conn);

if(!(mysql_query($grava,$conn))){

die("Error: ".mysql_error());

}else{

echo("$_POST[nome] inserido com sucesso.");

}

mysql_close($conn);

}
?>

here's the html form

Code: [Select]

<div class="center">
<form action="insert.php" method="post">
<p>Name:<input type="text" name="nome" /></p>
<p>Year:<input type="text" name="ano" /></p>
<p>Cast:<input type="text"  name="elenco" /></p>
<p>Comment:<textarea name="comentario"></textarea></p>
<p><input type="submit" name="submit" value="Send" /></p>
</form>
</div>
any help would be appreciated.
thanks

[Smokey PHP]

Change this line:

PHP Code: [Select]

if(!(mysql_query($grava,$conn))){

to this:

PHP Code: [Select]

if($grava === false) {

The function mysql_query is setting $grava to a boolean value, depicting the success of the query in the function. Hence your if statement needs to check the value of $grava to know whether or not the query was successful.