Ask for help with php program

[Any Who]

Next php code is being included in a table on a link page. For each category there is a script like this one. But it gives an error for the last php tag. can anybody tell me what's wrong?

<?
$verbinding=mysql_connect("localhost", "phpgebruiker", "php");
mysql_select_db("webleren");
$query = "SELECT sitenaam, url FROM linkpagina WHERE categorie=='science fiction';
$resultaat = mysql_query($query);
while ($T_string) {
$T_string = mysql_fetch_assoc($resultaat);
print $T_string($sitenaam);
$T_string = mysql_fetch_assoc($resultaat);
}?>


<?
mysql_free_result($resultaat);
mysql_close($verbinding);
?>

[imported__sam_]

Originally Posted by Any Who

$query = "SELECT sitenaam, url FROM linkpagina WHERE categorie=='science fiction';

Missing " here

Originally Posted by Any Who

print $T_string($sitenaam);

That should be:
print $T_string['sitenaam'];

And I hope you are aware that this loop prints only every second row and you will have to work on the formating of the ourput.

best regards

- Sam

[imported__sam_]

Originally Posted by Any Who

$query = "SELECT sitenaam, url FROM linkpagina WHERE categorie=='science fiction';

Missing " here

Originally Posted by Any Who

print $T_string($sitenaam);

That should be:
print $T_string['sitenaam'];

And I hope you are aware that this loop prints only every second row and you will have to work on the formating of the output.

best regards

- Sam

[icb01co2]

Looks a bit strange to me, i havent tested this but maybe:



<?
$verbinding=mysql_connect("localhost", "phpgebruiker", "php");
mysql_select_db("webleren");


$query = "SELECT sitenaam, url FROM linkpagina WHERE categorie='science fiction'";

$result = mysql_query($query, $verbinding);

if ($myrow = mysql_fetch_array($result)) {
do
{
$Sitenaam = $myrow['sitenaam'];
echo $Sitenaam;
}while($myrow = mysql_fetch_array($result));
}

?>


Not sure though as im a little unsure what you were trying to do.