how to use nested if else stament in shortcode [duplicate]

Question

i am trying to make nested if else shortcode but its not working. any one please help as soon as possible.

function fn_services_text_box($attr){

$name           = $attr['service_name'];
$icon           = $attr['icon_path'];
$text           = $attr['content_text'];
$link           = $attr['url_path'];

$output         = if($icon == ''){
                    $output        .='<div class="service-list-title"><h3>'.$name.'</h3></div>
                    <div class="service-list-txt">'.$text.'</div>                                    
                    <div class="service-list-readmore"><a href="'.$link.'">Know More</a></div>';
                }
                else if($link == ''){
                    $output        .='<div class="polygon-shape">
                        <div class="icon">
                            <img class="img" src="'.$icon.'" />
                        </div>
                    </div>
                    <div class="service-list-title"><h3>'.$name.'</h3></div>
                    <div class="service-list-txt">'.$text.'</div>  '; 
                }
                else{
                    $output        .='<div class="polygon-shape">
                        <div class="icon">
                            <img class="img" src="'.$icon.'" />
                        </div>
                    </div>
                    <div class="service-list-title"><h3>'.$name.'</h3></div>
                    <div class="service-list-txt">'.$text.'</div>                                    
                    <div class="service-list-readmore"><a href="'.$link.'">Know More</a></div>';
                }

return $output;

}

i am usig this short code in visual composer by adding it to shortcode mapper.

What does $attr contain? Doat your function return anything? — ClydeFrog
– ClydeFrog, Commented Aug 26, 2019 at 12:02

Vaibhavi S. · Accepted Answer · 2019-08-26 11:29:39Z

2

Please replace this line

$output = if($icon == ''){

to

$output = "";
if($icon == ''){

answered Aug 26, 2019 at 11:29

Vaibhavi S.

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2 Comments

Poonam Katpara Over a year ago

it worked like magic thanks alot . can u please explain why you did this $output == "";

Vaibhavi S. Over a year ago

Most Welcome @PoonamKatpara Let me explain, We can not directly declare variable $output with if statement. so we need to first declare variable empty like $outout = "" then if else statment we can append data like you already did.

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