Answer the question
In order to leave comments, you need to log in
Debug information output. How are you doing?
What do you think, is it adequate to display debug info in this way?
<pre class="debug text-center center-block">
<?php
$debug = ["memory" => 'Памяти использовано: '.round(memory_get_usage()/1024/1024,2).' MB',
"time" => 'Страница загружена за: '.round(microtime(true) - $start, 4) .' сек.'
];
foreach($debug as $data) {
print("<span>".$data."</span>");
}
?>
</pre>
Answer the question
In order to leave comments, you need to log in
Didn't find what you were looking for?
Ask your questionAsk a Question
731 491 924 answers to any question