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Hey guys, I need your help with this code, thanks in advance.


pickamaterina
Go to solution Solved by Ch0cu3r,

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<?php
$user="root";
$password="";
$database="base";
$host="localhost";
$table="com";

mysql_connect($host, $user, $password) or die("error");
mysql_select_db($database) or die("error");

$name = $_POST['name'];
$lastname = $_POST['lastname'];
$email = $_POST['email'];
$coment= $_POST['coment'];
$error = '';

 if(empty($name) || empty($lastname) || empty($email) || empty($coment))
{
      $error .= 'error. ';
}
if(!preg_match("/^[a-zA-A]+$/i", $name))
{
     $error .= 'Error ';
}

if(!preg_match("/^[a-zA-A]+$/i", $lastname))
{
     $error .= 'Error ';
}



if(!preg_match("/^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,3})$/i", $email))
{
     $error .= 'errorl. ';
}

if(!preg_match("/^[a-zA-z0-9.,?!]+$/i", $comment))
{
     $error .= 'Error ';
}

if($error == '')
{
 
$result = "INSERT INTO com (name, lastname, email, coment) VALUES('$name', '$lastname', '$email','$coment')";
    echo "<script type='text/javascript'>alert('Thanks.'); window.close();</script>";}
else{
      echo"<script type='text/javascript'>alert ('$error'); window.close(); </script>";}
 
Errors seems to be working its just it won't insert data in table in mysql database....

Thank you so much. :-*

Edited by pickamaterina
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Because you're not doing anything with the query

$result = "INSERT INTO com (name, lastname, email, coment) VALUES('$name', '$lastname', '$email','$coment')";

You have just defined a query to $result. PHP doesn't see it as a query, it just sees it as a string.

 

What do you expect that line to do?

Edited by Ch0cu3r
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Well you need to check if mysql_query is returning any errors

$sql = "INSERT INTO com (name, lastname, email, coment) VALUES('$name', '$lastname', '$email','$coment')";

if(mysql_query($sql))
{
    echo "<script type='text/javascript'>alert('Thanks.'); window.close();</script>";
}
else
{
	echo 'Error: ' . mysql_error();
	// I have commented out the line below.
    //echo"<script type='text/javascript'>alert ('$error'); window.close(); </script>";
}
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  • Solution

When you perform queries on database you need to make sure they execute ok. During development you need to to output as much debug information as possible when something your expect to happen doesn't happen, if that makes sense.

 

You said earlier when you use mysql_query it didn't work. So I recommended you add echo mysql_error to see if the query returns an error. You can remove the debug code when everything is working.

 

So does your query return any errors when your run your script?

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