Error message displaying in address bar

maxwell_

I am trying to echo 2 url variables from a page called softwarelist.php to another page
called fulltitle.php

<a href="fulltitle.php?software_id=<?php echo $row_rsTitles['id']; ?>&id=<?php echo $row_rsTitles['id']; ?>">Read / Make Comments About This Software</a>

when I do the following error message displays in the browser address bar

software_id= Notice: Undefined index: id in C:\server\www\server\public_html\tutor\public\software\softwarelist.php on line 97 &id= Notice: Undefined index: id in C:\server\www\server\public_html\tutor\public\software\softwarelist.php on line 97

bradgrafelman

Where is $row_rsTitles defined? Have you tried using something like [man]print_r/man or [man]var_dump/man on it to make sure it contains the values you think it does? (Hint: If not, do that now and let us know what the result is. 😉)

maxwell_

Looks like the problem was caused by the mysql query.

$query_rsTitles = "SELECT id, title, company, description, resources, location, url, image, keyword, class_id, level_id, copies FROM titles ORDER BY title ASC";

I tried removing class_id, level_id, which appears to have fixed the problem.

Code now working

if(!isset($_POST['softwareLevel'])){
if (!isset($_GET['class_id'])) {
//show all software titles
$query_rsTitles = "SELECT id, title, company, `description`, resources, location, url, image, keyword, copies FROM titles ORDER BY title ASC";

}else{
//show software titles filtered by Literacy of Numeracy (using URL GET variable)
$query_rsTitles = "SELECT id, title, company, `description`, resources, location, url, image, keyword, copies FROM titles WHERE titles.class_id = ". GetSQLValueString($_GET['class_id'], "int") ." ORDER BY id ASC";
}
}else{
//show software titles filtered by Level (using Form POST variable)
$query_rsTitles = "SELECT id, title, company, `description`, resources, location, url, image, keyword, copies FROM titles WHERE titles.level_id = ". GetSQLValueString($_POST['softwareLevel'], "int") ." ORDER BY id ASC";
}

johanafm

Even if you fixed your previous problem, you should not simply assume that things will work. Always make sure that they did, and if not, deal with the problem

$result = mysqli_query($link, $query);

# did it work?
if ($result)
{
	# output the link with the data that you now know you have
}
else
{
	# output no link, a link without the query string or use default values
}