Is there a concise way to check if a variable is set, and then echo it without repeating the same variable name?
Instead of this:
Use isset function of php like this:
<?php
echo $result = isset($this->variable) ? $this->variable : "variable not set";
?>
i think this will help.
Update:
PHP 7 introduces a new feature: Null coalescing operator
Here is the example from php.net.
<?php
// Fetches the value of $_GET['user'] and returns 'nobody'
// if it does not exist.
$username = $_GET['user'] ?? 'nobody';
// This is equivalent to:
$username = isset($_GET['user']) ? $_GET['user'] : 'nobody';
// Coalescing can be chained: this will return the first
// defined value out of $_GET['user'], $_POST['user'], and
// 'nobody'.
$username = $_GET['user'] ?? $_POST['user'] ?? 'nobody';
?>
For those not using PHP7 yet here is my original answer...
I use a small function to achieve this:
function ifset(&$var, $else = '') {
return isset($var) && $var ? $var : $else;
}
Example:
$a = 'potato';
echo ifset($a); // outputs 'potato'
echo ifset($a, 'carrot'); // outputs 'potato'
echo ifset($b); // outputs nothing
echo ifset($b, 'carrot'); // outputs 'carrot'
Caveat: As Inigo pointed out in a comment below one undesirable side effect of using this function is that it can modify the object / array that you are inspecting. For example:
$fruits = new stdClass;
$fruits->lemon = 'sour';
echo ifset($fruits->peach);
var_dump($fruits);
Will output:
(object) array(
'lemon' => 'sour',
'peach' => NULL,
)
The closest you can get to what you are looking for is using short form of ternary operator (available since PHP5.3)
echo $a ?: "not set"; // will print $a if $a evaluates to `true` or "not set" if not
But this will trigger "Undefined variable" notice. Which you can obviously suppress with @
echo @$a ?: "not set";
Still, not the most elegant/clean solution.
So, the cleanest code you can hope for is
echo isset($a) ? $a: '';