Asked  7 Months ago    Answers:  5   Viewed   50 times

Based on the examples from this page, I have the working and non-working code samples below.

Working code using if statement:

if (!empty($address['street2'])) echo $address['street2'].'<br />';

Non-working code using ternary operator:

$test = (empty($address['street2'])) ? 'Yes <br />' : 'No <br />';

// Also tested this
(empty($address['street2'])) ? 'Yes <br />' : 'No <br />';

After Brian's tip, I found that echoing $test outputs the expected result. The following works like a charm!

echo (empty($storeData['street2'])) ? 'Yes <br />' : 'No <br />';




(condition) ? /* value to return if condition is true */ 
            : /* value to return if condition is false */ ;

syntax is not a "shorthand if" operator (the ? is called the conditional operator) because you cannot execute code in the same manner as if you did:

if (condition) {
    /* condition is true, do something like echo */
else {
    /* condition is false, do something else */

In your example, you are executing the echo statement when the $address is not empty. You can't do this the same way with the conditional operator. What you can do however, is echo the result of the conditional operator:

echo empty($address['street2']) ? "Street2 is empty!" : $address['street2'];

and this will display "Street is empty!" if it is empty, otherwise it will display the street2 address.

Tuesday, June 1, 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 9 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 7 Months ago

It is the ternary operator, and it works like in C (the parenthesis are not required). It's an expression that works like:

if_this_is_a_true_value ? then_the_result_is_this : else_it_is_this

However, in Ruby, if is also an expression so: if a then b else c end === a ? b : c, except for precedence issues. Both are expressions.


puts (if 1 then 2 else 3 end) # => 2

puts 1 ? 2 : 3                # => 2

x = if 1 then 2 else 3 end
puts x                        # => 2

Note that in the first case parenthesis are required (otherwise Ruby is confused because it thinks it is puts if 1 with some extra junk after it), but they are not required in the last case as said issue does not arise.

You can use the "long-if" form for readability on multiple lines:

question = if question.size > 20 then
  question.slice(0, 20) + "..."
Tuesday, June 1, 2021
answered 7 Months ago

The simple answer is providing operator bool() const, but you might want to look into the safe bool idiom, where instead of converting to bool (which might in turn be implicitly converted to other integral types) you convert to a different type (pointer to a member function of a private type) that will not accept those conversions.

Thursday, July 29, 2021
answered 4 Months ago
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