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phpvarisset

What is the PHP shorthand for: print var if var exist

We've all encountered it before, needing to print a variable in an input field but not knowing for sure whether the var is set, like this. Basically this is to avoid an e_warning.

<input value='<?php if(isset($var)){print($var);}; ?>'>

How can I write this shorter? I'm okay introducing a new function like this:

<input value='<?php printvar('myvar'); ?>'>

But I don't succeed in writing the printvar() function.

Solution

  • For PHP >= 7.0:

    As of PHP 7 you can use the null-coalesce operator:

    $user = $_GET['user'] ?? 'guest';

    Or in your usage:

    <?= $myVar ?? '' ?>

    For PHP >= 5.x:

    My recommendation would be to create a issetor function:

    function issetor(&$var, $default = null) {
    return isset($var) ? $var : $default;
}

    This takes a variable as argument and returns it, if it exists, or a default value, if it doesn't. Now you can do:

    echo issetor($myVar);

    But also use it in other cases:

    $user = issetor($_GET['user'], 'guest');