Problem fixed but caused new problem

[Stan007]

Guys thanks for helping me solve the problem i had but now i have another problem and i am lost. i have included the code below for you to have overview.

On the home page the url shows how it should be: http://localhost/board/italian.php?country=Italian
BUT 
When you mouse over the NEXT link to go to next page you get this : http://localhost/board/?page=1&country=Italian  and then an error 404 when you click on the link
page = 1  no matter if you are mousing over link for page 5 or 2

This is the code:

<table style="width:100%; margin-left:auto; margin-right:auto">
        <?php    
            
            $items_per_page = 5;//how many records/rows to display per page
                        
            $page_url      = "http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];
                                    
            $cleaned_current_page_country = htmlspecialchars($_GET["country"]);
                        
            $page_number = (isset($_GET['page']))? filter_var($_GET['page']) : 1;
            //$page_number = filter_var($_GET['page']);
            //$page_number = $_GET['page'];
                
            $selected_restaurant_type = $_GET['country'];
                                                            
            // find number of rows in the db
                $query = "SELECT * FROM menu_update WHERE special_of_theday = 1 AND restaurant_type = '".$selected_restaurant_type."' ";
                $query_result = mysqli_query($connection, $query); 
                $row_count = mysqli_fetch_array($query_result);
                $total_rows = mysqli_num_rows($query_result);
                                                
            $expected_number_of_pages = ceil($total_rows/$items_per_page); //expected number of pages 
                                                
            $page_position = (($page_number - 1) * $items_per_page); //get starting position to fetch record
                        
            //fetch a group of records using sql LIMIT clause
                $query = "SELECT * FROM menu_update WHERE special_of_theDay = 1 AND restaurant_type = '".$selected_restaurant_type."' LIMIT " .$page_position. ',' .$items_per_page;
                $query_result = mysqli_query($connection, $query);
        ?>
                        
            <table width="100%" border=0>
                <tr bgcolor = "#cccccc">
                    <td>Meal</td>
                    <td>Meal name</td>
                    <td>Meal Price</td>
                    <td>Order Now</td>
                    <td>Full Menu</td>
                    <td>Our Gallery</td>
                    <td>Find Us</td>
                    <td>Our Chef</td>
                    <td>Our Manager</td>
                </tr>
        <?php
                while($row_count = mysqli_fetch_array($query_result)){
                    echo "<tr>";
                        echo"
                            <td style ='width:200px; text-align:center; height:120px; background:#eaedf2'><a href='meal_fullPic.php'><img src= ".$row_count['img']." /></a></td>
                            <td style ='width:119px; text-align:center; height:120px; background:#eaedf2'> ".$row_count['meal_name']."</td>
                            <td style ='width:100px; text-align:center; height:120px; background:#eaedf2'>".$row_count['price']."</td>
                            <td style ='width:100px; text-align:center; height:120px; background:#eaedf2'><a href='order_processor.php'><img src= ".$row_count['order_now']." /></a></td>
                            <td style ='width:100px; text-align:center; height:120px; background:#eaedf2'><a href='full_menu.php'><img src= ".$row_count['full_menu_img']." /></a></td>
                                    
                            <td style ='width:170px; text-align:center; height:120px; background:#eaedf2'><a href='restaurant_gallery.php'><img src= ".$row_count['restaurant_pic']." /></a></td>
                            <td style ='width:100px; text-align:center; height:120px; background:#eaedf2'><a href='restaurant_location.php'><img src= ".$row_count['location_pic']." /></a></td>
                            <td style ='width:119px; height:120px; background:#eaedf2'> ".$row_count['chef_name']."</td>
                            <td style ='width:119px; height:120px; background:#eaedf2'> ".$row_count['admin']."</td>
                        "; 
                    echo "</tr>";
                }
                mysqli_close($connection);
        ?>    
            </table>
        <div class="board_pagination">
        <?php
            //display the links to the pages
            for($current_page = 1; $current_page <= $expected_number_of_pages; $current_page++){
                 if ($page_number != $expected_number_of_pages){
                    $data = array(
                        'page' => $page_number,
                        'country' => $cleaned_current_page_country,
                    );
                    $the_query = http_build_query($data);
                    $page_url .= '?' . $the_query;
                    echo '<li><a href=" '.$the_query.' ">'.$current_page.'</a></li> ';
                }
            }
        ?>
        </div>
    </table>
 

[Barand]

3 minutes ago, Stan007 said:

When you mouse over the NEXT link to go to next page you get this : http://localhost/board/?page=1&country=Italian  and then an error 404 when you click on the link

does it work if you put the "Italian.php" back in the middle of that url?

[Stan007]

Hello Barand,

let me check that now

Thanks so much for replying

 

[Stan007]

Hello Barand,

yes it does work when i put italian and page=2 in the url and takes me rightly to page 2

this is the url that works but i had to put back italian.php in the url manually: http://localhost/board/italian.php?page=2&country=Italian 

[Barand]

What does $page_url contain?

Try

var_dump($page_url)

on line 7 after you set its value;

[Barand]

I have just noticed you don't use the $page_url in those pagination links. It looks like you intended to but then just used the query string

$the_query = http_build_query($data);
$page_url .= '?' . $the_query;
echo '<li><a href=" '.$the_query.' ">'.$current_page.'</a></li> ';

As you go to the same page that should be sufficient but I think you will need the preceding "?"

Try

$the_query = http_build_query($data);
$the_query = '?' . $the_query;
echo "<li><a href=\"$the_query\">$current_page</a></li>";

 

[Stan007]

Hi Barand,

Thanks so much, the italian.php is now automatically added to the url like this: http://localhost/menuboard/italian.php?page=1&country=Italian  But even when you click on link for page 2, the page always = 1 

So you dont actually go to any subsequent page corresponding to the link you clicked. You remain at the index page.

[Barand]

In your pagination loop you set all the links to the same page number!

Should be

$data = array(
    'page' => $current_page,                               <-- CHANGED
    'country' => $cleaned_current_page_country
);

 

Edited January 11, 2020 by Barand

[Stan007]

Hello Barand,

The problem is solved, this code that you  suggested fixed it:  'page' => $current_page,                                                            <-- CHANGED

You really helped me thanks a lot, you have a new fan. Will be making donations pretty soon.

Thanks again

Stanley

[Barand]

Your misleading choice of variables names no doubt contributed to the problem.

There are other things could be improved too

• Use prepared queries instead of putting variables directly into the sql query strings. (I would also recommend PDO in preference to mysqli)
• use css classes instead of repetitive inline style definitions
• you are using deprecated html markup code (eg <tr bgcolor = "#cccccc">)
• you use a lot of unnecessary concatenation when building strings - a common source of errors.

[Stan007]

Hello Barand,

One last request please, i just noticed that when i click link and arrive at second page, i dont have any back link. Can you please help me with the pagination section of the code.

Thanks a million

[Barand]

It would appear from the code that you would always have links

1  2  3  4 ... N

except when you are on the last page. You show the links on this condition...

if ($page_number != $expected_number_of_pages)

Do you, perchance, only have two pages at present? And do you mean $page_number or do you mean $current_page (confusing names)

[Barand]

The more information a query retrieves the slower it will run. In the code below all you want is a single count yet you retrieve every column in every record.

                // find number of rows in the db
                $query = "SELECT * FROM menu_update WHERE special_of_theday = 1 AND restaurant_type = '".$selected_restaurant_type."' ";
                $query_result = mysqli_query($connection, $query); 
                $row_count = mysqli_fetch_array($query_result);
                $total_rows = mysqli_num_rows($query_result);

Far more efficient to

                $query = "SELECT COUNT(*) as total 
                          FROM menu_update 
                          WHERE special_of_theday = 1 
                                AND restaurant_type = '$selected_restaurant_type' ";
                $query_result = mysqli_query($connection, $query); 
                $row = mysqli_fetch_array($query_result);
                $total_rows = $row['total'];