Check if variable is set and then echo it without repeating?

问题

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:

<?php
    if(!empty($this->variable)) {
        echo '<a href="', $this->variable, '">Link</a>';
    }
?>

I'm thinking about something in the lines of this C-style pseudocode:

<?php
    echo if(!empty($this->variable, '<a href="', %s, '">Link</a>'));
?>

PHP has sprintf, but it doesn't quite do what I was hoping for. If course I could make a method/function out of it, but surely there must be a way to do it "natively"?

Update: Ternary operations would also repeat the $this->variable part, if I understood it?

echo (!empty($this->variable) ? '<a href="',$this->variable,'">Link</a> : "nothing");

回答1:

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: '';

回答2:

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,
)

回答3:

Use isset function of php like this:

<?php

  echo $result = isset($this->variable) ? $this->variable : "variable not set";

 ?>

i think this will help.

来源：https://stackoverflow.com/questions/15970220/check-if-variable-is-set-and-then-echo-it-without-repeating

标签

php

variables

isset