Why is my ternary expression not working?
-
08-06-2019 - |
Question
I am trying to set a flag to show or hide a page element, but it always displays even when the expression is false.
$canMerge = ($condition1 && $condition2) ? 'true' : 'false';
...
<?php if ($canMerge) { ?>Stuff<?php } ?>
What's up?
Solution
This is broken because 'false' as a string will evaluate to true as a boolean.
However, this is an unneeded ternary expression, because the resulting values are simple true and false. This would be equivalent:
$canMerge = ($condition1 && $condition2);
OTHER TIPS
The value of 'false' is true. You need to remove the quotes:
$canMerge = ($condition1 && $condition2) ? true : false;
Seems to me a reasonable question especially because of the discrepancy in the way PHP works.
For instance, the following code will output 'its false'
$a = '0';
if($a)
{
echo 'its true';
}
else
{
echo 'its false';
}
$canMerge = ($condition1 && $condition2);
then
if ($canMerge){
echo "Stuff";
}
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