Better way to output PHP objects to console? - php

Question

I'm trying to find the best way to debug PHP in the browser.
EDIT:
I know about xdebug, var_dump and print_r; that's not the point. a) I want a way to debug when a site is live (I know, that's bad but sometimes have to for a quick fix), b) dumping PHP variables all over a web page is ugly, and c) I sometimes need to know why AJAX calls aren't working, and it's easier to look at JSON in the console.
Possible other solutions include wrapping your debug output in a hidden div, e.g.
<div id="debug"><?php var_dump($all_the_junk) ?></div><script>document.getElementById('debug').style.display='none'</script>
and then typing something like
document.getElementById('debug')style.display = 'block'
into the console.
But let's be honest, that kind of sucks, and defeats the purpose of being able to see everything in a tidy way.
ENDEDIT;
So, anyways, I have the following code:
function log_as_json($vars)
{
if(is_resource($vars)) {
return;
}
else {
if($vars)
{
$json = #json_encode($vars);
print "<script>console.log($json);</script>";
}
}
}
The problem is that once in a while I get these terrible recursions, but I'm not sure what the best way is to check for types that json_encode supports. Also I suspect there is a better way to do this. Advice would be appreciated. Thanks!

Answer 1

If you are really keen on debugging in the browser, I'd recommend Chrome Logger (formerly known as ChromePHP).

Answer 2

The best way is to use xdebug And an IDE that allows you to set breakpoints

Answer 3

I use print_r or var_dump - it is what they are designed for.
I suppose you could embed these calls in console.log - not sure how they would display though.

Answer 4

I know this is old but here is a way that works for me, just call the function when you need to..
function console_log( $data ){
echo '<script>';
echo 'console.log('. json_encode( $data ) .')';
echo '</script>';
}