php - Why is my ternary expression not working?


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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?


所有的回答
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    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);
    

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    The value of 'false' is true. You need to remove the quotes:

    $canMerge = ($condition1 && $condition2) ? true : false;
    

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    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';
    }
    

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    $canMerge = ($condition1 && $condition2);
    

    then

    if ($canMerge){
        echo "Stuff";
    }