如何在该特定页面上选择导航菜单？

Question

I am building a navigation system on a webpage using PHP, that basically queries data from MySQL database and display as ul, li menu on main menu. But I couldn't figure out how do I make navigation menu selected when user goes to that page. I have one idea but that doesn't make home menu selected by default when user enters that page.

I have also a CSS class written to give that class to the anchor tag if it is that page.

here is my PHP code:

//Query Functions 
function queryMenu() {
    $query = "SELECT * FROM menu_en";
    return $query;  
}

// Render Functions
function renderNav() {  
    $menus = queryMenu();
    $queryMenu = mysql_query($menus);
    while( $navigation = mysql_fetch_array($queryMenu) ){
        if( $navigation['id'] == $_GET['pageId'] ){
            echo '<li>
                    <a href="index.php?pageId="'.$navigation['id'].' class="selected" >'.$navigation['menu_title'].
                 '</a></li>';
        }else{
            echo '<li><a href="index.php?pageId="'.$navigation['id'].'>'.$navigation['menu_title'].'</a></li>';
        }
    }
}

Suggest me a better way to do this, as I want to enhance my skill in PHP. And also please let me know if there is any wrong approach in above code.

Answer 1

If the home page is when $_GET['pageId'] is not set you could check using isset()

if(isset($_GET['pageId'])) {
    $page = $_GET['pageId']
} else {
    $page = $homePageId; // the ID of the home page. 
}

Then you would replace this line:

if( $navigation['id'] == $_GET['pageId'] ){

with this line:

if( $navigation['id'] == $page ){

I think this would work, if i understand you correctly.