Question

Possible Duplicate:
putting print_r results in variable

I’m throwing an exception and trying to include a variable in the exception, like so:

throw new Exception('Oh no, an exception! ' . $variable);

(Where $variable is an array)

The problem is, this only puts the following in my log file:

On no, an exception! Array

Unfortunately I’m not an expert at PHP, I’m guessing this could mean one of two things:

1) $variable is an empty array

2) $varialbe is an array with data in it, but outputting it as such in an exception does not output all of its contents

Please let me know if 1) is the case here (I hope it isn’t though)

However, if 2) is the case, how can I get more information about $variable? Is it possible to do print_r or var_dump inside the exception like follows:

throw new Exception('Oh no, an exception! ' . print_r($variable));

Or will that cause problems?

Answer 1

Or will that cause problems?

Yes, it will. The print_r will be evaluated before the string is fully constructed, and the output will look something like this, if you try and print out the Exception message:

Array
(
    [x] => 5
    [y] => 65
)
Oh no, an exception! 1

To fix this, you need to make sure you set print_r‘s $return parameter to true so the value is returned rather than echoed:

throw new Exception('Oh no, an exception! ' . print_r($variable, true));