在Golang中，执行Marshal和Unmarshal时JSON字段名称的大小写是否重要？

Question

Are Marshal and Unmarshal functions strictly case sensitive?

The following structure defines a ticket:

    type TicketInfo struct {

         TicketKey                  string `json:"ticketKey"`       
         Ticketextnum               string `json:"ticketextnum"`        
         TicketDate                 string `json:"ticketDate"`      
         TicketDesc                 string `json:"ticketDesc"`
    }       

This stringified form of this json will be passed as a single parameter in the payload of an API call by the source systems. What will happen if the source system sends, for example, "TicketKey" instead of "ticketKey" for the first field. Will json.Marshal correctly received the ticketKey field?

Answer 1

This is easily answered by reading the docs, or with a simple test. But specifically, see these notes (emphasis added) for Marshal:

Struct values encode as JSON objects. Each exported struct field becomes a member of the object, using the field name as the object key, unless the field is omitted for one of the reasons given below.

The encoding of each struct field can be customized by the format string stored under the "json" key in the struct field's tag. The format string gives the name of the field, possibly followed by a comma-separated list of options. The name may be empty in order to specify options without overriding the default field name.

and Unmarshal:

To unmarshal JSON into a struct, Unmarshal matches incoming object keys to the keys used by Marshal (either the struct field name or its tag), preferring an exact match but also accepting a case-insensitive match. By default, object keys which don't have a corresponding struct field are ignored (see Decoder.DisallowUnknownFields for an alternative).

So TL;DR; for marshaling, case is observed exactly, and for unmarshaling, exact case is preferred with a fallback to a case-insensitive match.