Asked  7 Months ago    Answers:  5   Viewed   39 times

How do I write a PHP ternary operator with the elseif portion?

I see basic examples with the if and else portions of the PHP ternary operator like this:

echo (true)  ? "yes" : "no";    //prints yes
echo (false) ? "yes" : "no";    //prints no

How do I get the "elseif" portion like this into the ternary operator?

  if($result->vocation == 1){
    echo "Sorcerer"; 
  }else if($result->vocation == 2){
    echo 'Druid';
  }else if($result->vocation == 3){
    echo 'Paladin';
  }else if($result->vocation == 4){
    echo 'Knight';
  }else if($result->vocation == 5){
    echo 'Master Sorcerer';
  }else if($result->vocation == 6){
    echo 'Elder Druid';
  }else if($result->vocation == 7){
    echo 'Royal Paladin';
    echo 'Elite Knight';



A Ternary is not a good solution for what you want. It will not be readable in your code, and there are much better solutions available.

Why not use an array lookup "map" or "dictionary", like so:

$vocations = array(
    1 => "Sorcerer",
    2 => "Druid",
    3 => "Paladin",

echo $vocations[$result->vocation];

A ternary for this application would end up looking like this:

echo($result->group_id == 1 ? "Player" : ($result->group_id == 2 ? "Gamemaster" : ($result->group_id == 3 ? "God" : "unknown")));

Why is this bad? Because - as a single long line, you would get no valid debugging information if something were to go wrong here, the length makes it difficult to read, plus the nesting of the multiple ternaries just feels odd.

A Standard Ternary is simple, easy to read, and would look like this:

$value = ($condition) ? 'Truthy Value' : 'Falsey Value';


echo ($some_condition) ? 'The condition is true!' : 'The condition is false.';

A ternary is really just a convenient / shorter way to write a simple if else statement. The above sample ternary is the same as:

if ($some_condition) {
    echo 'The condition is true!';
} else {
    echo 'The condition is false!';

However, a ternary for a complex logic quickly becomes unreadable, and is no longer worth the brevity.

echo($result->group_id == 1 ? "Player" : ($result->group_id == 2 ? "Gamemaster" : ($result->group_id == 3 ? "God" : "unknown")));

Even with some attentive formatting to spread it over multiple lines, it's not very clear:

echo($result->group_id == 1 
    ? "Player" 
    : ($result->group_id == 2 
        ? "Gamemaster" 
        : ($result->group_id == 3 
            ? "God" 
            : "unknown")));
Wednesday, March 31, 2021
answered 7 Months ago

When your first argument is null, they're basically the same except that the null coalescing won't output an E_NOTICE when you have an undefined variable. The PHP 7.0 migration docs has this to say:

The null coalescing operator (??) has been added as syntactic sugar for the common case of needing to use a ternary in conjunction with isset(). It returns its first operand if it exists and is not NULL; otherwise it returns its second operand.

Here's some example code to demonstrate this:


$a = null;

print $a ?? 'b'; // b
print "n";

print $a ?: 'b'; // b
print "n";

print $c ?? 'a'; // a
print "n";

print $c ?: 'a'; // Notice: Undefined variable: c in /in/apAIb on line 14
print "n";

$b = array('a' => null);

print $b['a'] ?? 'd'; // d
print "n";

print $b['a'] ?: 'd'; // d
print "n";

print $b['c'] ?? 'e'; // e
print "n";

print $b['c'] ?: 'e'; // Notice: Undefined index: c in /in/apAIb on line 33
print "n";

The lines that have the notice are the ones where I'm using the shorthand ternary operator as opposed to the null coalescing operator. However, even with the notice, PHP will give the same response back.

Execute the code:

Of course, this is always assuming the first argument is null. Once it's no longer null, then you end up with differences in that the ?? operator would always return the first argument while the ?: shorthand would only if the first argument was truthy, and that relies on how PHP would type-cast things to a boolean.


$a = false ?? 'f'; // false
$b = false ?: 'g'; // 'g'

would then have $a be equal to false and $b equal to 'g'.

Wednesday, March 31, 2021
answered 7 Months ago

You created a mocked object, but you didn't bind it into the container. Without binding, when Laravel runs your command, it will just generate a new ExportApiService instance.

public function setUp()

    $mock = $this->createMock(ExportApiService::class);


    // bind the mock in the container, so whenever you ask for
    // a new ExportApiService, you'll get your mocked object.
    $this->app->instance(ExportApiService::class, $mock);

Another option you have, instead of doing this, is to mock your Guzzle requests. This way, all of your code executes as it normally would, but when you make your API call, Guzzle returns a pre-determined response instead of actually making the call. You can find the Guzzle docs on testing here.

After you configure your client with the mocked responses, you would bind that client instance into the container, so that when your ExportApiService is created, Laravel injects the client you've setup with the mocked responses.

Wednesday, March 31, 2021
answered 7 Months ago

You need to add some parenthesis.

$b = 'a';
$c = 'd';
echo ($b == 'a') ? 2 : ($c == 'a' ? 1 : 0);
Saturday, May 29, 2021
answered 5 Months ago

In fact you can, you almost had it with the last one:

(&operator float())[3];

As for the question of whether or not a typedef is ever necessary, I think it is from reading the comments on (which answer is what helped me get the syntax for the above as well).


Apparently, this syntax is incorrect and returning a reference to an array is forbidden as chris discovered for us. I guess you'll just have to settle for a typedef.

Monday, September 27, 2021
answered 3 Weeks ago
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