[RESOLVED] Help.

[tranxek23]

I have made a script where two usernames are entered into a form and then these are put in the database however it is not working. I get no error message or anything, the names just don't get put in the database.

here is the code for the form....

<form name="form1" method="post" action="addfriendprocesssub.php">
  <p>
    Username One<input type="text" name="usernameone">
</p>
  <p>
    Username Two<input type="text" name="usernametwo">
</p>
  <p>
    <input type="submit">
  </p>
</form>

and here is the PHP bit:

$hostname='******************';
$username='*********';
$password='***********';
$dbname='******';
$usertable='addfriend';
$username1=$_POST['usernameone'];
$username2=$_POST['usernametwo'];
mysql_connect($hostname,$username, $password) OR DIE ('Unable to connect to database!');
mysql_select_db($dbname);
$query = "INSERT INTO '$usertable' ('username1','username2', 'id') VALUES ('$username1', '$username2', '')" or die('cannot enter data into table');
mysql_query($query);
mysql_close();

Any Help Would Be Appreciated!

[rincewind456]

Your "or die" code is in the wrong place, try this

$hostname='h50mysql7.secureserver.net'; 
$username='resizetest'; 
$password='***********'; 
$dbname='resizetest'; 
$usertable='addfriend'; 
$username1=$_POST['usernameone']; 
$username2=$_POST['usernametwo']; 
mysql_connect($hostname,$username, $password) OR DIE ('Unable to connect to database!'); 
mysql_select_db($dbname); 
$query = "INSERT INTO '$usertable' ('username1','username2', 'id') VALUES ('$username1', '$username2', '')";
mysql_query($query) or die(mysql_error('cannot enter data into table'));
mysql_close();

It's also a good idea to use [man]mysql_error/man or [man]mysql_errno/man or both to aid with debugging.

[tranxek23]

thanks for your help. I repositioned the 'or die' bit and now instead of nothing happening i get this error:

Warning: mysql_error(): supplied argument is not a valid MySQL-Link resource in /home/content/s/o/c/socomkiwiseal/html/addfriendprocesssub.php on line 2

[paulnaj]

Hi there,

Did this before I saw rincewind456's post so a bit of cross-over ...

Firstly, comment out all your connection details as you did for password.

Your code has lot's of potential glitches, so here's a cleaned up version in my formatting style.

I'm not able to debug it (I would need access to your db) so check to see if haven't mis-spelled the field names etc. before trying it out.

Things to note:

1) Use back-ticks to enclose fieldnames and tablenames in SQL - not single-quotes.
2) My advice is NOT to embed variables in strings (cut in and out)
3) Do as much error-checking as you can

Anyway

<?php
$hostname = '***********';
$username = '***********';
$password = '***********';
$dbname = '***********';

$usertable = 'addfriend';

///////////////////////////////////////////////////////////////
// Do some checking to see if data is as expected
if(!isset($_POST['usernameone']) || $_POST['usernameone'] == ''){
	die('[usernameone] not sent.');
} else {
	$username1 = $_POST['usernameone'];
}

if(!isset($_POST['usernametwo']) || $_POST['usernametwo'] == ''){
	die('[usernametwo] not sent.');
} else {
	$username2 = $_POST['usernametwo'];
}
///////////////////////////////////////////////////////////////

mysql_connect($hostname,$username, $password) 
	or die('Unable to connect to database!');

mysql_select_db($dbname)
	or die('Cannot select db');

$query = "
INSERT INTO `".$usertable."` (
	`username1`
, 	`username2`
, 	`id`
) VALUES (
	'".$username1."'
, 	'".$username2."'
, 	''
)";

if(mysql_query($query)){
	echo 'Query executed successfully';
} else {
	echo 'Query failed';
}

mysql_close(); 
?>

Paul

[tranxek23]

its working great...thanks.