2012-06-20 19:29:27 8 Comments
Is there a technical reason why unexported fields are not included by encoding/json? If not and it is an arbitrary decision could there be an additional back door option (say '+') to include even though unexported?
Requiring client code to export to get this functionality feels unfortunate, especially if lower case is providing encapsulation or the decision to marshal structures comes much later than design of them.
How are people dealing with this? Just export everything?
Also, doesn't exporting field names make it difficult to follow suggested idioms. I think if a struct X has field Y, you can not have an accessor method Y(). If you want to provide interface access to Y you have to come up with a new name for the getter and no matter what you'll get something un-idiomatic according to http://golang.org/doc/effective_go.html#Getters
Related Questions
Sponsored Content
33 Answered Questions
[SOLVED] What is the correct JSON content type?
- 2009-01-25 15:25:19
- Oli
- 2374430 View
- 9517 Score
- 33 Answer
- Tags: json http-headers content-type
46 Answered Questions
53 Answered Questions
[SOLVED] How can I pretty-print JSON in a shell script?
- 2008-12-09 08:20:45
- AnC
- 1026179 View
- 2656 Score
- 53 Answer
- Tags: json unix command-line format pretty-print
17 Answered Questions
[SOLVED] How to POST JSON data with Curl from Terminal/Commandline to Test Spring REST?
- 2011-08-24 08:51:11
- kamaci
- 2284019 View
- 2317 Score
- 17 Answer
- Tags: json rest spring-mvc curl http-headers
23 Answered Questions
[SOLVED] How can I pretty-print JSON using JavaScript?
- 2011-01-26 22:33:53
- Mark
- 767422 View
- 1874 Score
- 23 Answer
- Tags: javascript json pretty-print
9 Answered Questions
6 Answered Questions
[SOLVED] Why does Google prepend while(1); to their JSON responses?
- 2010-04-19 18:00:09
- Jess
- 440963 View
- 3677 Score
- 6 Answer
- Tags: javascript json ajax security
39 Answered Questions
24 Answered Questions
[SOLVED] Safely turning a JSON string into an object
- 2008-09-05 00:12:01
- Matt Sheppard
- 1162446 View
- 1213 Score
- 24 Answer
- Tags: javascript json
16 Answered Questions
[SOLVED] Parse JSON in JavaScript?
- 2011-02-08 16:34:03
- user605334
- 1808520 View
- 1571 Score
- 16 Answer
- Tags: javascript json parsing
2 comments
@jorelli 2012-06-20 23:31:05
Stephen's answer is complete. As an aside, if all you really want is lowercase keys in your json, you can manually specify the key name as follows:
In that way, marshaling a Whatever produces the key "some_field" for the field SomeField (instead of having "SomeField" in your json).
If you're dead-set on keeping unexported fields, you can also implement the json.Marshaler interface by defining a method with the signature
MarshalJSON() ([]byte, error)
. One way to do this is to use a struct literal that simply has exported versions of the unexported fields, like this:That can be a bit cumbersome, so you can also use a
map[string]interface{}
if you prefer:However it should be noted that marshaling
interface{}
has some caveats and can do things like marshaluint64
to a float, causing a loss of precision. (all code untested)@Tyler Eaves 2013-05-14 14:21:28
Technically, ALL numbers in Javascript are floats.
@jorelli 2013-05-14 19:19:05
JSON is not JavaScript. The JSON spec merely states which characters are acceptable, not which numeric ranges are valid. A number like 876234958273645982736459827346598237465923847561203947812435968234659827346 is still valid in JSON, even if it can't be understood by JavaScript. For a real world example, the Twitter API represents tweet IDs as 64bit unsigned integers, which is not valid in JavaScript, but is valid JSON.
@quetzalcoatl 2014-07-13 20:28:02
Currently, all true, but still the name is JSON, which historically stands for "JavaScript Object Notation". One of those tiny suprising bits that make you want to rename it to NJSON (NewJSON or rather NotJavaScript) :)
@Stephen Weinberg 2012-06-20 23:10:57
There is a technical reason. The json library does not have the power to view fields using reflect unless they are exported. A package can only view the unexported fields of types within its own package
In order to deal with your problem, what you can do is make an unexported type with exported fields. Json will unmarshal into an unexported type if passed to it without a problem but it would not show up in the API docs. You can then make an exported type that embeds the unexported type. This exported type would then need methods to implement the
json.Marshaler
andjson.Unmarshaler
interfaces.Note: all code is untested and may not even compile.
@Derek 2013-03-28 20:05:47
For the record, I had to Unmarshall using "json.Unmarshal(b, &d.jsonData)". Did I do something wrong, or is that expected?
@Stephen Weinberg 2013-03-29 08:00:40
@Derek, thanks, I updated my answer. As I said the code was untested. I also apparently also forgot a return statement in my
UnmarshalJSON()
method. I fixed this as well.@davidjosepha 2017-02-28 03:17:15
I'm a bit late to the show, but... while the above works, Field1 and Field2 are exported. You can read and write Field1 and Field2 of JsonData (with a capital J) outside of that package. So, while this is cool in theory, it doesn't actually do anything different than exporting both the type and the fields.
@darkwing 2018-05-15 18:42:11
@davidjosepha you're correct as the answer was originally shown although someone would have to be looking into your source code to know that
Field1
andField2
are present since godoc wouldn't list the unexported structjsonData
. However, this issue can be fixed by not using an Anonymous field in the JsonData struct, and having a named field which is unexported. This then requires thatField1
andField2
are accessed through the named, unexported field. I edited the answer to use this methodology.