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PHP Integrating PHP with Databases Limiting Records in SQL Pagination Function

Victoria Loisi
Victoria Loisi
7,070 Points
Posted by Victoria Loisi
Victoria Loisi
7,070 Points

Why $pagination won't work?

I need to do this: Move all the code that sets the $pagination variable inside the pagination function. Return the $pagination variable. Change the echo on the final line to call the function instead of the variable. Make sure you pass the four variables: \$total_pages, \$current_page, \$section, \$search.

So, I've done this:

index.php 
<?php
include "pagination.php";

/* add function here */
function pagination($total_pages, $current_page, $section, $search){
  /* setting pagination variable */
$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>";

  return $pagination;
}

/* displaying the pagination */
pagination($total_pages, $current_page, $section, $search);

Now, when I try to submit the code, it says that the previous challenge is no longer passing.

This was the previous challenge: A function does not know about variables outside the scope of the function. To use a function for displaying the pagination, we need to pass our function the required variables. Create a function named pagination that accepts four arguments, $total_pages, $current_page, $section and $search.

index.php 
<?php
include "pagination.php";

/* add function here */
function pagination($total_pages, $current_page, $section, $search){
}

/* setting pagination variable */
$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>";

/* displaying the pagination */
echo $pagination;

Why the error keeps appearing?

2 Answers

Kallil Belmonte
Kallil Belmonte
35,561 Points
Kallil Belmonte
Kallil Belmonte
35,561 Points

Hi Victoria, try this instead: 

<?php
include "pagination.php";

/* add function here */
function pagination($total_pages, $current_page, $section, $search) {

/* setting pagination variable */
$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>";

return $pagination;

} 

/* displaying the pagination */
echo pagination($total_pages, $current_page, $section, $search);
Victoria Loisi
Victoria Loisi
7,070 Points
Victoria Loisi
Victoria Loisi
7,070 Points

Thank you!! I didn't realized that I needed to actually echo the pagination function. :)