How can I re-formating PHP code, returning duplicate attribute class?

Question

Dears, I would like ask you for help with formating PHP code. Because W3C validator, return me this as error, "Error: Duplicate attribute class." I have now next PHP code:

$html .= '<a href="'.$href.'"';
if($this->hasChilds($item, $nazev)) $html .= ' class="qmparent"';
if ((empty($view) && $item->view == 'main') || (empty($display) && $view == $item->view) || (!empty($display) && $display == $item->display)) $html .= ' class="active"';
$html .= '>';

This giving me html
code: <a href="link.html" class="qmparent" class="active">LINK</a> And I would like, can will be generate, a different html
code: <a href="link.html" class="qmparent active" >LINK</a>
Can I ask you for help, how to rewrite? I don? know if it is enough for correct sense, if not I will add much of code. Thanx!

question from:https://stackoverflow.com/questions/65832596/how-can-i-re-formating-php-code-returning-duplicate-attribute-class

Answer 1

The error is HTML error because you have duplicated class. In HTML it is ok to have empty class attribute. So you can append classes then add them to the HTML class attribute. See the code:

$classes = '';
if($this->hasChilds($item, $nazev)) $classes .= ' qmparent';
if ((empty($view) && $item->view == 'main') || (empty($display) && $view == $item->view) || (!empty($display) && $display == $item->display)) $classes .= ' active';
$html .= '<a href="'.$href.'"class ="'.$classes.'"/>';

To have better structure, it is common to separate the view from model. So you can create a data array that contains or your HTML elements with final classes, type and id. Then pass this array to view to construct the view correctly.