warning if variable does not exist

Question

If I try to use a variable which doesnt even exist in PHP, I dont get any errors or notices. In the example below the return value is "null". I'd like to see some form of notification if that happens so that I can find bugs in my code more easily.

echo $doesNotExist -> something; //returns null

The error reporting in my php.ini is the following

error_reporting = E_ALL

What do I have to do to see notifications when accessing varibles that don't exist?

To clarifiy this:

I know that I can check if a variable exists by using isset(). That's not what I want though. I want to get a notification in case I accidentaly try to use a variable that does not exists. For example if I misspell the name of a variable.

display_errors = On is in my php.ini

only need to google for this http://php.net/manual/en/function.isset.php — Rajeev Ranjan, Sep 20 '13 at 12:14
Interesting, blanket downvoting. The OP is not asking for isset here, or how to check if a variable exists, they're asking how to see their reported errors. — cmbuckley, Sep 20 '13 at 12:17
And if the display_errors directive gets it, then it's a dupe: http://stackoverflow.com/questions/845021/how-to-get-useful-error-messages-in-php — cmbuckley, Sep 20 '13 at 12:21
@marius2k12 For the future, please precisely describe questions. You gave all -1 and then you explain what's going on - not nice. — Adam, Sep 20 '13 at 12:28

score 0 · Answer 1 · answered Sep 20 '13 at 12:13

0

As well as error_reporting, you may need to turn on display_errors, which should be off in production environments.

Production environments can use log_errors instead, which outputs the errors to the file named in the error_log directive.

answered Sep 20 '13 at 12:13

cmbuckley

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Hum. What's the output of `phpinfo()`? It should tell you which INI files are being parsed as well as the values of the config options. – cmbuckley Sep 20 '13 at 13:03
error_reporting 32767 / display_errors On / log_errors On / error_log C:\xampp\php\logs\php_error_log – marius2k12 Sep 20 '13 at 14:03
Do you get the errors in the log? The only other thing that I can think of is that output buffering may be enabled, and causing the error not to be shown? – cmbuckley Sep 20 '13 at 20:40
I once had a problem with php not logging errors all the time and what I did to fix it was deleted the log file and let php create a new one. Perhaps you could try that. – starshine531 Oct 02 '13 at 11:49

score -1 · Answer 2 · answered Sep 20 '13 at 12:22

-1

I generally use thw following;

if(isset($var) && !empty($var)) {
    //Do something with $var
} else {
    //Define $var
}

As said in a previous answer you also need to turn error reporting on in the php.ini

answered Sep 20 '13 at 12:22

Phyore

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Aris · Answer 3 · 2013-09-21T05:44:51.277

-1

what you do is perfectly valid in PHP. It just creates and instantiates that variable at runtime, and doesn't throw an error like other languages.

even isset would not work too well. Say you set an object variable with a wrong name. If you do isset on that name, it will be true. If you do isset on the correct name, it will be false.

Example:

class test{
    public $name1;
}

$test = new test;
$test->name2 = 'wrong';

var_dump($test);

prints: object(test)#1 (2) { ["name1"]=> NULL ["name2"]=> string(5) "wrong" }

edited Sep 21 '13 at 05:44

answered Sep 20 '13 at 13:53

Aris

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Simply not true — PHP does not create anything when accessing an undefined variable, and it *does* give an error. – cmbuckley Sep 20 '13 at 22:21
@cbuckley you are wrong. you can try the example code I added. – Aris Sep 21 '13 at 05:45
The OP was not talking about the member variable not existing, and was not talking about setting a member variable either. They were talking about the object variable not existing. [Like this](http://codepad.org/pfiwCiD8). – cmbuckley Sep 21 '13 at 12:21

warning if variable does not exist

3 Answers3