Return the object property that is actually accessible. #3371
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Since there is a check for two types of property, we should also return the one we found.
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Hm, the nightly build has failed. Is that something my code did? It looks more like an issue with the nightly build or some other part of Twig? |
fabpot
requested changes
Dec 26, 2022
| if (isset($object->$item)) { | ||
| return $object->$item; | ||
| } | ||
| return ((array) $object)[(string) $item]; |
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Suggested change
| return ((array) $object)[(string) $item]; | |
| return (isset($object->$item) ? $object->$item : ((array) $object)[(string) $item]; |
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Since there is a check for two types of property, we should also return the one we found.
Without this, cases where a property is accessed that exists in the cast of the object, but not directly, will not be available and will throw notices and end up with NULL.
We noticed this in our project because of using a DateTimeInterface's date property (which I wouldn't recommend, but still). The if check will respond
truebecause if you cast a DateTimeImmutable to an array it has a 'date' key, but trying to access it directly does not actually work (at least not in PHP 7.4. It apparently worked before as the code had been running fine before we upgraded.)My proposed change should work in all situations.