I'm trying to replicate this condition:
$sth = $this->getResult();
if($sth !== true){
return $sth;
}
with ternary operator I tried with;
($sth !== true) ? return $sth : ($sth == true) ? null;
But I got:
Expected colon
What you're ultimately trying to achieve is not possible with a ternary operator.
What you try to do is ONLY return in 1 situation, and continue the code in the other. The only way you could do this using a ternary operator is like this:
$result = ($sth !== true) ? true : false;
if ($result) return;
But that kinda defeats the purpose of the ternary operator.
In fact, your code has a couple of problems:
return inside the truthy or falsy part.A ternary operator needs 3 parts
(condition) ? truthy result : falsy result
Note: Since PHP 5.3, it is possible to leave out the middle part of the ternary operator. Expression expr1 ?: expr3 returns expr1 if expr1 evaluates to TRUE, and expr3 otherwise. - source: php.net
No need to check things twice:
Again, let's take our operator:
($sth !== true)
? "I return if '$sth' is not true"
: "I return if '$sth' is true";
There is no need to have a 2nd check. You have both situations covered already. :)
No return values inside the truthy or falsy part
Finally: it is an operator that returns a value. You can't put a return in the truthy or falsy part, you need to put it in front of it:
return ($sth !== true)
? "I return if '$sth' is not true"
: "I return if '$sth' is true";
You have two '?'s in your code, and only one ':'. I believe that's the matter. You must have the same amount of question marks as you have colons.
If you are using it to return from a function I would do it like so:
function testFunc()
{
//arbitrary setting of $sth only to explain the point
$sth = false;
//condition ? true : false
$outcome = ($sth === true) ? true : false;
return $outcome;
}
// false
var_dump(testFunc());
Remember, (condition) ? value if true : value if false;
You can also miss out the [value if true] parameter, like so:
$outcome = ($sth === true) ?: false;
That expression would assign the value of $sth to $outcome if $sth was true.
Your replacement code indeed misses a colon at the very end. It is syntactically incorrect. If $sth !== true you return $sth. If not, you start a new ternary statement (that is not complete) which I cannot properly fathom. It now does not look at all like the original statement.
The ternary operator is nice and useful, but it's not a block of commands. It is a decision on a value. Rule of thumb, if you can assign something as value to a variable you can use it within the ternary operator :
$var = 3; // valid
($condition?3:0); // valid
$var = return 3; //Invalid
($condition?return 3:null) //Invalid
You need to:
return ($sth !== true?$sth:null);