Hot answers tagged json
14
It seems to me that this is the correct way to extract values from a nested list of rules:
Data
ad = {"accept_rate" -> 75, "account_id" -> 395497, "age" -> 41,
"badge_counts" -> {"bronze" -> 35, "gold" -> 0, "silver" -> 11},
"creation_date" -> 1326833982, "display_name" -> "Verbeia",
"is_employee" -> False, ...
9
It is possible to get rid of all but one ReplaceAll. First, you need to remember that ReplaceAll tests each Rule in order, so to deal with the missing "badge_counts" condition, you can simply add that to the list, as follows
{"display_name", "creation_date", "reputation", "reputation_change_week",
"is_employee", "last_access_date", "user_type",
...
7
The reason is that the JSON format is quite limited. It doesn't support arbitrary expressions. You can e.g. see that by trying the folllowing:
Export["test.json",someSymbol,"JSON"]
You'll get the same error.
If your goal is just to pass the expressions around (i.e. on the other side is another Mathematica instance to interpret them), the simplest ...
6
The short answer is that, as @FJRA noted in the comment, only certain types are supported. Which types? Enter the long answer.
Why the converter behaves as it does
Long answer: JSON supports only certain types, and their nested combinations, as defined e.g. here. Mathematica converter maps JSON objects to lists of rules, arrays to lists, strings to ...
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