Parse error - what am I doing wrong.

Zepher2

I am using the following line in opening up the next page of results from a query where the results are filtered by SelectID.

I have to pass the variable SelectID to the new page but het a parse error with this code:

echo "<a href=\"".$_SERVER['PHP_SELF']."?page=$next\"  ."?SelectID = $SelectID\">"Next>></a>";

The parse erroris generated with the SelectID portion of the code since if I eliminate it I don't get the error message.

Can someone help me correct this code.

zingbats

THis is a stupid way to do it, but I use it and it works fine.

$next_page = $_SERVER['PHP_SELF'];
$next_page .= "?page=$next";
$next_page .= "&SelectID=$SelectID";

Zepher2

I still can't get it to work. The parse error is gone, but it opens to a non existent url.

My code now looks like this:

// Build Next Link
if($page < $total_pages){

$next = ($page + 1);
$next_page = $_SERVER['PHP_SELF'];
$next_page .= "?page=$next";
$next_page .= "&SelectID=$SelectID";

echo $next_page;

echo "<a href='<?=$next_page?>'>Next Page</a>";

}

This returns Page No Found with this url

http://www.aarda.org/<?=/show_survey_results2.php?page=2&SelectID=70?>

aarda.org is the charity I volunteered to do this for.

laserlight

More likely:

// Build Next Link
if ($page < $total_pages){
	$next = $page + 1;
	$next_page = $_SERVER['PHP_SELF'] . "?page=" . $next
		. "&amp;SelectID=" . $SelectID";
	echo $next_page;
	echo '<a href="' . $next_page . '">Next Page</a>';
}

Zepher2

Thank you. It is working now.