Using return in ternary operator

I'm trying t开发者_JS百科o use return in a ternary operator, but receive an error:

Parse error: syntax error, unexpected T_RETURN 

Here's the code:

$e = $this->return_errors();
(!$e) ? '' : return array('false', $e);

Is this possible?

Thanks!

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This is the correct syntax:

return  !$e ? '' : array('false', $e);

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Close. You'd want return condition?a:b

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It doesn't work in most languages because return is a statement (like if, while, etc.), rather than an operator that can be nested in an expression. Following the same logic you wouldn't try to nest an if statement within an expression:

// invalid because 'if' is a statement, cannot be nested, and yields no result
func(if ($a) $b; else $c;); 

// valid because ?: is an operator that yields a result
func($a ? $b : $c); 

It wouldn't work for break and continue as well.

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No it's not possible, and it's also pretty confusing as compared to:

if($e) {
    return array('false', $e);
}

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No. But you can have a ternary expression for the return statement.

return (!$e) ? '' : array('false', $e);

Note: This may not be the desired logic. I'm providing it as an example.

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No, that's not possible. The following, however, is possible:

$e = $this->return_errors();
return ( !$e ) ? '' : array( false, $e );

Hope that helps.

---

Plop,

if you want modify your return with ternary ?

it's totally possible.

In this exemple, i have a function with array in parameters. This exemple function is used to parse users array. On return i have an array with user id and user username. But what happens if I do not have any users?

<?php

public funtion parseUserTable(array $users) {
   $results = [];
   foreach ($users as $user) {
      $results[$users['id']] = $user['username'];
   }
  return $results ?: array('false', $e, "No users on table."); // Ternary operator.
}

Sorry for my bad english, i'm french user haha.

N-D.