PHP: printing undefined variables without warning

Question

I'm just wondering if there is a quick way to echo undefined variables without getting a warning? (I can change error reporting level but I don't want to.) The smallest I have so far is:

isset($variable)?$variable:''

I dislike this for a few reasons:

It's a bit "wordy" and complex
$variable is repeated
The echoing of a blank string always kind of annoys me.
My variable names will probably be longer, eg $arrayvar['parameter']

Really not a good idea. If that's seems to be "It's a bit "wordy" and complex", then you'll have some surprise in the future... There is no way this going to be you app bottleneck. Start using the good habit of checking var. Make you own function to speed up the process if you like. — Bite code
– Bite code, Commented Dec 19, 2008 at 10:32

Eran Galperin · Accepted Answer · 2008-12-18 01:38:48Z

15

You can run it with the error suppression operator @.

echo @$variable;

However, it's best not to ignore unset variables. Unset variables could indicate a logical error on the script, and it's best to ensure all variables are set before use.

edited Dec 18, 2008 at 1:38

answered Dec 18, 2008 at 1:31

Eran Galperin

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3 Comments

Matthew Scharley Over a year ago

The one exception I usually find myself making to this rule is unset array indexes. These can crop up all over the place. It's also better to put it on the variable itself as @$variable, so that it only supresses the unset error.

Eran Galperin Over a year ago

Yeah thanks for that. I almost never use this operator for the reasons I mentioned :)

Robert Swisher Over a year ago

Fully agree with the suggestion that this is not best practice.

Tom Haigh · Accepted Answer · 2008-12-19 09:15:56Z

9

you could use the ifsetor() example taken from here:

function ifsetor(&$variable, $default = null) {
    if (isset($variable)) {
        $tmp = $variable;
    } else {
        $tmp = $default;
    }
    return $tmp;
}

for example:

echo ifsetor($variable);
echo ifsetor($variable, 'default');

This does not generate a notice because the variable is passed by reference.

edited Dec 19, 2008 at 9:15

answered Dec 18, 2008 at 14:31

Tom Haigh

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2 Comments

Ólafur Waage Over a year ago

iirc this will still produce the error, since php is sending an undefined variable into a function, even though the isset is within it.

Tom Haigh Over a year ago

It won't, because it is passed by reference.

Alexei Tenitski · Accepted Answer · 2008-12-18 01:34:39Z

4

echo @$variable;

answered Dec 18, 2008 at 1:34

Alexei Tenitski

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Comments

Community · Accepted Answer · 2017-05-23 11:59:18Z

1

This is a long-standing issue with PHP, they intend to fix it with isset_or() (or a similar function) in PHP 6, hopefully that feature will make it into PHP 5.3 as well. For now, you must use the isset()/ternary example in your question, or else use the @ prefix to silence the error. IMHO, this is the only circumstance that justifies using @ in PHP.

I wouldn't worry about speed issues using echo with an empty string, it is probably more expensive to wrap it in an if clause than to just echo empty string.

edited May 23, 2017 at 11:59

CommunityBot

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answered Dec 19, 2008 at 0:52

too much php

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1 Comment

nickf Over a year ago

I find @ to be useful when doing database connections or file operations where a failure would show errors. eg: "if ($fh = @fopen('myfile', 'w'))" - it lets you gracefully handle any problems.

AlperXX · Accepted Answer · 2021-03-03 10:29:39Z

1

You can use ?? selector for PHP 7 and later.

echo $variable??'not defined';

answered Mar 3, 2021 at 10:29

AlperXX

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Comments

Tomasz · Accepted Answer · 2012-07-18 08:29:42Z

0

undefined variables are very common, i suggest you to initialize variable with null at first

$var = null;

or disable error reporting for notices:

error_reporting(E_ALL^E_NOTICE);

answered Jul 18, 2012 at 8:29

Tomasz

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Comments

smartcoder · Accepted Answer · 2008-12-19 09:49:39Z

-1

Suppress errors using the @-operator forces the interpreter to change error level, executing the function and then change back error level. This decreases your scripts runtime.

Build a function like this will eliminate at least 3 of your reasons:

function echoVar($var, $ret=NULL) {
    return isset($var)?$var:$ret;
}

echoVar($arrayvar['parameter']);

But why echoing undefined variables? This sounds like not really well coded...

edited Dec 19, 2008 at 9:49

answered Dec 19, 2008 at 0:16

smartcoder

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1 Comment

Tom Haigh Over a year ago

I think echoVar needs to accept $var by reference, in order to not show an error.

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