what is wrong with this piece of code? html + php - php

Question

i've a simple function, which has two parameters, one for the image url, and other for the attributes for the image
function image_found($url,$attributes)
{
if(#getimagesize($url))
{
echo '<img src="'.$url.'" '.$attributes.'/>';
}
else
{
echo '<img src="'.base_url().'/site_images/image_not_found.svg" '.$attributes.'/>';
}
}
now what i'm trying to do is create a clickable image, if the image is found, now this is the html code
echo '<div class="panel-body">';
echo '<div class="col-md-12 col-lg-12 col-sm-12 text-center">';
$url = base_url().'product_images/'.$result->product_image.'.'.$result->image_type;
$attributes = 'height="200px" width="100%"';
echo ''.image_found($url,$attributes).'';
echo '</div>';
echo '</div>';
and this is the output i'm getting
<div class="panel-body">
<div class="col-md-12 col-lg-12 col-sm-12 text-center">
<img src="http://localhost/nsc/product_images/7908076366784972032090.jpg" height="200px" width="100%"/>
</div>
</div>
i don't know what is wrong here, i'm using bootstrap

Answer 1

Just use return statements instead of echo in your function and your problem should be solved ;-)

Answer 2

The better way is verify if your image exists (remove #) and then return (instead of echo):
...
if(file_exists('your/path/to/image'))
return '<img src="'.$url.'" '.$attributes.'/>';
else
return '<img src="'.base_url().'/site_images/image_not_found.svg" '.$attributes.'/>'
...

Answer 3

When you need to return a value from a function, use return statement instead of echo
When echo is used the output immediately gets printed instead of getting returned to the place where the function call is. Here is an illustration.
function printer(){
echo 'second';
}
echo 'first'.' '.printer().' '.'last';
The Output:
secondfirst last
This is the exact same thing happening with your code. The echo in image_found() gets printed as
<img src="http://localhost/nsc/product_images/7908076366784972032090.jpg" height="200px" width="100%"/>
The rest of the echo statement gets printed as
So using a return statement should solve your problem