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Start your free trialLewis Cowles
74,902 PointsYour usage of {'varname'} to access json vars
So I must've skipped over how you access the JSON in this because one of my Juniors just showed me, and I was not impressed... Sure PHP can allow you to access invalid object names using object syntax, but surely array syntax would be simpler if you know the developer did not follow, what are standard variable declaration rules. I would say its better to deal with JSON using array rather than use this syntax, but I dont see accessing as array covered... Is there a reason for this?
1 Answer
miguelcastro2
Courses Plus Student 6,573 PointsBy default, json_decode() will return an stdClass object from a JSON string. Since with JSON we are dealing with objects, it makes sense to return an object within PHP as well. If you want any array just add an extra boolean flag to the json_decode function and it will return an associative array:
json_decode($json_str, TRUE);
Lewis Cowles
74,902 PointsLewis Cowles
74,902 PointsHi Miguel, looks like we agree? (so a bit confusing), but I was advocating using 'true' and therefore returning an array. (note 'true' lowercase is what is being advocated by php-fig), I just happen to use it but it does make more sense as even modern C compilers use the lowercase true nowdays.
Thanks for the backup though