Trouble with simple PHP-MySQL form

Question

I want to make a 2 field input form to my MySQL database. I connect to the database with no problem and even can post if I want, but the form is giving an error. (I'm already connected to the database at this point and I can test that is working)

This is the form:

<form action="insert.php" method="post">
Title: <input type="text" name="title" />
Privacy: <select type="text" name="privacy" />
  <option value="public">Publico</option>
  <option value="private">Privado</option>
</select>
<input type="submit" />
</form>

This is the insert.php file:

<?  
mysql_select_db("copoetry", $con);
$sql="INSERT INTO Poems (Title, Privacy)
VALUES
('$_POST[title]','$_POST[privacy]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";
mysql_close($con)
?>

When I press submit I get this error:

Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in /home/content/02/6945202/html/copoetry/insert.php on line 2

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/content/02/6945202/html/copoetry/insert.php on line 7
Error:

What am I doing wrong? Thanks

In addition to the DB connection missing the query string isn't correct: do echo $sql; to see why. Don't forget to sanitize your input too. — JJJ
– JJJ, Commented Jul 27, 2011 at 9:11
$con is defined on my header, putting the PHP code directly in the file writes a blank row in the database so the connection works. However when put on a different file, I get this error. — lisovaccaro
– lisovaccaro, Commented Jul 27, 2011 at 9:20

Dunhamzzz · Accepted Answer · 2011-07-27 09:45:15Z

2

Your $con variable should hold a connection to the database with mysql_connect();, you appear to removed this line at some point.

EG:

$con = mysql_connect('host', 'user', 'pass') or die(mysql_error());

Once you have done this successfully all your mysql_* calls will use that connection, so you could get rid of the $con variable anyway.

ALSO Don't forget to escape your inputs so that are safe, inserting a $_POST, $_GET or $_REQUEST variable straight into mysql is very unsafe. Make at the very least you run mysql_real_escape_string(); on each and every input you get from a form or cookie.

EG

// Create a shortcut function somewhere early in your script
function mes($input) { 
    return mysql_real_escape_string($input);
}

// SQL Example
$sql = "INSERT INTO Poems (Title, Privacy) VALUES ('".mes($_POST['title'])."','".mes($_POST['privacy'])."')";

Note the string concatenation (using fullstop) to separate the strings/variables.

edited Jul 27, 2011 at 9:45

answered Jul 27, 2011 at 9:09

Dunhamzzz

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3 Comments

lisovaccaro Over a year ago

I had defined $con in the header, I didn't realize I had to define it again in the insert file. Thanks

lisovaccaro Over a year ago

Also thanks for the info, could you tell me where should I add mysql_real_escape_string(); ? I have no idea in which part of the code it should be. Thanks

Dunhamzzz Over a year ago

@Liso22 updates my answer. You shouldn't need to redefine $con, my guess is it's just not available to this script for whatever reason.

Alex Ackerman · Accepted Answer · 2011-07-27 09:10:01Z

1

You're missing mysql_connect(). Try adding this. Also it's very unsafe to run a query with whatever comes from $_GET/$_POST/$_REQUEST.

answered Jul 27, 2011 at 9:10

Alex Ackerman

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Comments

clement · Accepted Answer · 2011-07-27 09:12:17Z

0

There's no connexion to your DB ;-)

answered Jul 27, 2011 at 9:12

clement

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Comments

Rukmi Patel · Accepted Answer · 2011-07-27 09:12:24Z

0

I think you have forgotten to specify mysql_connect() command.

answered Jul 27, 2011 at 9:12

Rukmi Patel

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