Active menu class with php function

Question

I'm not getting it done, I need to put a class on the output of the function.

Can you help, when the menu item is active there has to be put a class to the link.

This is the code:

function loadMenu()
      {
     $result = mysql_query(" SELECT * FROM  cms_page WHERE  site_id =2  AND  page_inmenu =1 AND  page_active =1");
    if ($DEBUG)
         echo "<pre>$result</pre>";
    while($row = mysql_fetch_array($result))
    {   $last_parent = $row['page_name']; 
        echo "<a href={$row['link_name']}>{$row['page_name']}</a>";

         }  
     }

Where's your Mysql connection? Also, i think your href is missing quotes for the link_name. — Charles Forest, May 30 '13 at 18:58

score 2 · Answer 1 · answered May 30 '13 at 18:59

2

function loadMenu()
{
    $result = mysql_query(" SELECT * FROM  cms_page WHERE  site_id =2  AND  page_inmenu =1 AND  page_active =1");
    if ($DEBUG)
        echo "<pre>$result</pre>";
    while($row = mysql_fetch_array($result))
    {
        $last_parent = $row['page_name']; 
        $class = $row['active'] === true ? ' class="active"' : '';
        echo "<a href=\"{$row['link_name']}\"{$class}>{$row['page_name']}</a>";

    }   
}

Also your

answered May 30 '13 at 18:59

beiller

3,075
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17

Hi Beiller, thanks, but this one is not working for me. It ads nothing to the active link. – user2001054 May 30 '13 at 19:03
$row['active'] === true -- that needs to be swapped with however you determine if a link is 'active' – beiller May 30 '13 at 19:04
thats the issue where i'm trying to getting a sollution for... How can i determine when a link is active – user2001054 May 30 '13 at 19:05
Its impossible for me to answer that without knowing more. You could say something like $thispagename == $row['page_name'] – beiller May 30 '13 at 19:07

score 0 · Answer 2 · edited May 23 '17 at 11:43

I'm not 100% sure about what's your issue.

So, if you aren't displaying anything, probably a MySQL issue.

1- Reference for Mysql (Look at OZ_'s answer) : How to put mysql inside a php function?

2- As mentionned by Beiller, you'll need a way to see if the link is active.

Solution A : Might need some adjustment according $row['link_name'] value.

    $class = ($row['link_name'] == $_SERVER['REQUEST_URI'].)?' class="active" ' : '';

Solution B : Simply stick to CSS - http://www.echoecho.com/csslinks.htm

    .mylink:active{ background-color:pink; }

3- Href needs quotes on around the URL. also, I'm not a big fan of {} inside strings.

    echo "<a href=\"". $row['link_name'] ."\"". $class .">". $row['page_name'] ."</a>";

score 0 · Answer 3 · edited May 30 '13 at 19:19

Thanks beiller for the help..

This is the function now:

function loadMenu(){
     $result = mysql_query(" SELECT * FROM  cms_page WHERE  site_id =2  AND  page_inmenu =1 AND  page_active =1");
    if ($DEBUG)
         echo "<pre>$result</pre>";
    while($row = mysql_fetch_array($result))
    {   $last_parent = $row['page_name']; 
        echo "<a href={$row['link_name']}>{$row['page_name']}</a>";
}}

This is the call:

<div id="header_menu">
           <? loadMenu (); ?>
</div>

And mysql has the following tables:

page_id
page_name
link_name
site_id
order_id
page_menu
page_inmenu
page_leftmenu
page_text
page_active

Active menu class with php function

3 Answers3