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Need Help on Outputting SQL

Posted: Wed Nov 29, 2006 6:00 pm
by admaster
I have an sql database with a table in it. This table has two rows. ONe is username and another is password. I want to have a php page that dynamically displays all of the information under the username category.

I tried to create a code, but it does not work. The code I tried is:

Code: Select all

<?php
$link = mysql_connect('localhost', 'root', 'my_password_was_here');

  $sql = "SELECT username FROM beta";
  $query = mysql_query($sql, $var);

    while($row = mysql_fetch_row($query)) {
  
    echo $row[0].'<br />';

  }

?>
However, it gives these errors:

Code: Select all

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\Program Files\xampp\htdocs\php\db_test2.php on line 5

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\Program Files\xampp\htdocs\php\db_test2.php on line 7
Thanks. :P

Posted: Wed Nov 29, 2006 6:03 pm
by Flamie
$var doesnt exist :) I think you want to use $link instead ;)

Posted: Wed Nov 29, 2006 6:47 pm
by admaster
Now this:

Code: Select all

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\Program Files\xampp\htdocs\php\db_test2.php on line 7

Posted: Wed Nov 29, 2006 7:00 pm
by volka
missing mysql_select_db

Code: Select all

<?php
$link = mysql_connect('localhost', 'root', 'my_password_was_here') or die(mysql_error());
mysql_select_db('name_of_the_database', $link) or die(mysql_error());

$query = "SELECT username FROM beta";
$result = mysql_query($query, $link) or die(mysql_error().': '.$query);


while( $row=mysql_fetch_array($result) ) {
	echo $row['username'], "<br />\n";
}
?>

Posted: Wed Nov 29, 2006 9:57 pm
by admaster
Thanks volka. It works now. :D