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']

Answer 1

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.

Answer 2

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.

Answer 3

echo @$variable;

Answer 4

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.

Answer 5

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);

Answer 6

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...