Including php into new page

[wazda21]

Hiya,

I'm trying to include the output file (output.php) into a new webpage, but all the webbrowser shows is a blanc page, with any content. I don't get it...?

<?php include("output.php"); ?>

Any help would be appreciated!

here's the output.php code:

<?
$username="wsb";
$password="macuser";
$database="wsb";

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM berichten ORDER BY titel DESC";
$result=mysql_query($query);

$num=mysql_num_rows($result);
$ret = mysql_query($query) or die(mysql_error());
mysql_close();

$i=0;
while ($i < $num) {

$taal=mysql_result($result,$i,"taal");
$titel=mysql_result($result,$i,"titel");
$auteur=mysql_result($result,$i,"auteur");
$doc=mysql_result($result,$i,"doc");
$datum=mysql_result($result,$i,"datum");
$website=mysql_result($result,$i,"website");
$nieuwsbrief=mysql_result($result,$i,"nieuwsbrief");
$totdatum=mysql_result($result,$i,"totdatum");
$inhoud=mysql_result($result,$i,"inhoud");


echo
"<br><h2><font face=Arial color=#3366FF>$titel</font></h2>
<font face=Arial size=2 color=#999999>$inhoud</font></br>
<br><i><font face=Arial size=1 color=#222222>$auteur</font></i></br>";

$i++;
}

?>

<? echo $variablename; ?>

[Kudose]

@mysql_select_db($database) or die( "Unable to select database"); would suppress a fatal error if the database could not be selected, generating nothing but a blank page.

Post up the code that includes said file.

[shamly]

Remove the @ sign from @mysql_select_db($database). As Kudose said, the @ sign will suppress any errors that may pop up if the database could not be selected.

If this does not work check your php.ini file to see if error reporting has been enabled and if display errors has been set to 1.

[sharad_impinge]

Try this,

$con= mysql_connect(localhost,$username,$password);

@mysql_select_db($database, $con);

Now it should work

[bradgrafelman]

sharad.impinge;10940788 wrote:

Try this,

$con= mysql_connect(localhost,$username,$password);

@mysql_select_db($database, $con);

Now it should work

There are two problems with that code, one of which was discussed just before you replied.