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Start your free trialAlfredo Prince
6,175 PointsWhat does it mean by "return the pagination value"
Am I supposed to use a literal return? Where?
<?php
include "pagination.php";
/* add function here */
function pagination ($total_pages, $current_page, $section, $search){
$pagination = "<div class=\"pagination\">";
$pagination .= "Pages: ";
for ($i = 1;$i <= $total_pages;$i++) {
if ($i == $current_page) {
$pagination .= " <span>$i</span>";
} else {
$pagination .= " <a href='catalog.php?";
if (!empty($search)) {
$pagination .= "s=".urlencode(htmlspecialchars($search)) . '&' ;
} else if (!empty($section)) {
$pagination .= "cat=".$section . '&';
}
$pagination .= "pg=$i'>$i</a>";
}
}
$pagination .= "</div>";}
/* setting pagination variable */
/* displaying the pagination */
echo $pagination;
1 Answer
Simon Woodard
6,545 PointsYou've created the function correctly but remember it wont do anything by itself, you also need to execute (call) the function for it to run.
It would seem that in this particular challenge you actually need to 'return $pagination' instead of 'echo $pagination' inside the function. Then after closing the function you can execute it with the following command and set of arguments (or required variables).
echo pagination(Total Pages, Current Page, Section, Search);
Where you have replaced the arguments with the correct values that the course requires.
Alfredo Prince
6,175 PointsAlfredo Prince
6,175 Pointsthis is what i did and it says "Task 1 no longer passing"
Alfredo Prince
6,175 PointsAlfredo Prince
6,175 PointsI tried what you said and it says "task 1 is no longer passing"