what does this structure mean:
$var = isset($var_1) ? $var_1 : $var_2;
I came across it and of course with other values, not $va, $var_1 and $var_2.
thanks.
This is the ternary operator, and means the same as:
if (isset($var_1)) {
$var = $var_1;
}
else {
$var = $var_2;
}
The ternary operator provides a short-hand method of creating simple if/else statements.
It has some syntax errors, correctly:
$var = isset($var_1) ? $var_1 : $var_2;
This means:
if (isset($var_1))
{
$var = $var_1;
}
else
{
$var = $var_2;
}
That means:
if(isset($var_1))
$var = $var_1;
else
$var = $var_2;
It is short syntax for that.
just for your information from the php manual i copy pasted good things to know about ternary comparision operators
The expression (expr1) ? (expr2) : (expr3) evaluates to expr2 if expr1 evaluates to TRUE, and expr3 if expr1 evaluates to FALSE.
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.
Note: Please note that the ternary operator is a statement, and that it doesn't evaluate to a variable, but to the result of a statement. This is important to know if you want to return a variable by reference. The statement return $var == 42 ? $a : $b; in a return-by-reference function will therefore not work and a warning is issued in later PHP versions.
Note:
It is recommended that you avoid "stacking" ternary expressions. PHP's behaviour when using more than one ternary operator within a single statement is non-obvious:
Example #3 Non-obvious Ternary Behaviour
<?php
// on first glance, the following appears to output 'true'
echo (true?'true':false?'t':'f');
// however, the actual output of the above is 't'
// this is because ternary expressions are evaluated from left to right
// the following is a more obvious version of the same code as above
echo ((true ? 'true' : false) ? 't' : 'f');
// here, you can see that the first expression is evaluated to 'true', which
// in turn evaluates to (bool)true, thus returning the true branch of the
// second ternary expression.
?>
Related
This question already has answers here:
Stacking Multiple Ternary Operators in PHP
(11 answers)
Closed 4 years ago.
I have a method that I'm checking if param is null, but if I use a ternary operator to make sure the false result isn't a string, I don't get the same expected result... I am a full stack .NET dev by day, but do some PHP free lance and this just stumped me...
$param = null;
// $active evaluates to true
$active = is_null($param) ? true : false;
// $active evaluates to false
$active = is_null($param) ? true : is_string($param)
? (strtolower($param) === 'true')
: true;
I have used nested ternary operators in C# and JavaScript what feels like countless times, but I don't know if I have ever tried in PHP... does PHP attempt to evaluate all nested ternary operations prior to expressing a result or is there something I'm missing here since from my understanding in this case, the ternary operator should be short circuited and evaluated to true in both circumstances.
The ternary operator is left associative unlike most other languages such as C#. The code:
$active = is_null($param)
? true
: is_string($param)
? (strtolower($param) === 'true')
: true;
is evaluated as follows:
$active = ((is_null($param) ? true : is_string($param))
? (strtolower($param) === 'true') : true);
You must explicitly add parenthesis to make sure ?: works the way it does in familiar languages:
$active = is_null($param)
? true
: (is_string($param)
? (strtolower($param) === 'true')
: true);
You need to wrap your second ternary condition with parenthesis (),
<?php
$param = null;
// $active evaluates to true
$active = is_null($param) ? true : false;
echo "Simple ternary result = $active".PHP_EOL;
// $active evaluates to true
$active = is_null($param) ? true : (is_string($param)? (strtolower($param) === 'true'): true);
echo "Nested ternary result = $active";
?>
Note:
It is recommended that you avoid "stacking" ternary expressions. PHP's
behaviour when using more than one ternary operator within a single
statement is non-obvious:
See Example #4 here at http://php.net/manual/en/language.operators.comparison.php
Example #4 Non-obvious Ternary Behaviour
<?php
// on first glance, the following appears to output 'true'
echo (true?'true':false?'t':'f');
// however, the actual output of the above is 't'
// this is because ternary expressions are evaluated from left to right
// the following is a more obvious version of the same code as above
echo ((true ? 'true' : false) ? 't' : 'f');
// here, you can see that the first expression is evaluated to 'true', which
// in turn evaluates to (bool)true, thus returning the true branch of the
// second ternary expression.
?>
DEMO: https://3v4l.org/gW8pk
This is a well-known problem with PHP. I doubt it will ever be fixed. Use parentheses, or if..else or switch statements, to get the behaviour you want.
(In technical terms, the ternary operator in PHP is "left associative", while that in every other language with this operator is "right associative". The latter is the more logical behaviour for this operator.)
Is there a way to prevent using if to check if an element exists before append ing itto a string, so it does not cause an error?
e.g.
$output = "";
if($data['element'])
{
$output .= $data['element'];
}
in Javascript I can use:
var output = "";
output += data['element'] || '';
You could always use the Ternary Operator ?:
$output .= (isset($data['element'])) ? $data['element'] : '';
From the PHP Docs
The expression (expr1) ? (expr2) : (expr3) evaluates to expr2 if expr1 evaluates to TRUE, and expr3 if expr1 evaluates to FALSE.
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.
use isset to check it.
$output = "";
if(isset($data['element']))
{
$output .= $data['element'];
}
Use the PHP Ternary Operator like this:
(isset($data['element'])) ? $output=$data['element'] : $output="";
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 ($result) return;
But that kinda defeats the purpose of the ternary operator.
In fact, your code has a couple of problems:
Ternary operator always needs 3 parts (*)
There is no need to check a condition twice, as that's the entire point of the operator
The operator returns a value, so you can't put a return inside the truthy or falsy part.
A ternary operator needs 3 parts
(condition) ? truthy result : falsy result
Note: Since PHP 5.3, it is possible to leave out the middle part of the ternary operator. Expression expr1 ?: expr3 returns expr1 if expr1 evaluates to TRUE, and expr3 otherwise. - source: php.net
No need to check things twice:
Again, let's take our operator:
($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.
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);
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.
Given the following expression:
$att['menutext'] = isset($attrib_in['i_menu_text']) ? : $this->getID();
If it evaluates to true, will $att['menutext'] be set to true or $this->getID()?
According to this reference:
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.
Yes, in version 5.3+ the middle expression is optional and returns true.
$a = (true ? : 1); // $a evaluates to true.
$a = (false ? : 1); // $a evaluates to 1.
It's just the same as the following
$att['menutext'] = isset($attrib_in['i_menu_text']) ? true : $this->getID();
never tested before, but its quite easy to test:
<?php var_dump(TRUE ? : 'F'); ?>
and its says: bool(true)
This won't execute, it's invalid syntax for PHP < 5.3.
Parse error: syntax error, unexpected ':' on line X
If you want the value to be set to true, then use true:
$att['menutext'] = isset($attrib_in['i_menu_text']) ? true : $this->getID();
Or it may be more likely that you want:
$att['menutext'] = isset($attrib_in['i_menu_text']) ? $attrib_in['i_menu_text'] : $this->getID();
Simple question. Here is this code.
$r = rand(0,1);
$c = ($r==0)? rand(65,90) : rand(97,122);
$inputpass .= chr($c);
I understand what it does in the end result, but I'd like a better explanation on how it works, so I can use it myself. Sorry if this is a bad question.
If you're unsure of what I'm asking about, its the (function?) used here:
$c = ($r==0)? rand(65,90) : rand(97,122);
That's called a ternary operator. It's effectively the equivalent of
if ($r == 0) {
$c = rand(65, 90);
} else {
$c = rand(97, 122);
}
But it's obviously a bit more compact. Check out the docs for more info.
That simply means:
if($r==0){
$c = rand(65,90);
else{
$c = rand(97,122);
}
If the statement is true the first operation after that ? is executed, else the operation after : is executed.
Its called a ternary operator.
Its the Ternary Operator
<?php
// Example usage for: Ternary Operator
$action = (empty($_POST['action'])) ? 'default' : $_POST['action'];
// The above is identical to this if/else statement
if (empty($_POST['action'])) {
$action = 'default';
} else {
$action = $_POST['action'];
}
?>
The expression (expr1) ? (expr2) : (expr3) evaluates to expr2 if expr1 evaluates to TRUE, and expr3 if expr1 evaluates to FALSE.
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.
It's called the ternary operator. It's similar to an if/else construct, but the difference is that an expression using the ternary operator yields a value.
That is, you can't set a variable to the result of an if/else construct:
// this doesn't work:
$c = if ($r == 0) {
rand(65, 90);
} else {
rand(97, 122);
}
You can read more about it here: http://php.net/ternary#language.operators.comparison.ternary
The ternary operator is frequently misused. There's little benefit to using it in the example you showed. Some programmers love to use compact syntactic sugar even when it's not needed. Or even when it's more clear to write out the full if/else construct.
The ternary operator can also obscure test coverage if you use a tool that measures lines of code covered by tests.