如何在Go中解析JSON时指定默认值 [英] How to specify default values when parsing JSON in Go
问题描述
我想在Go中解析一个JSON对象,但想为未给出的字段指定默认值。例如,我有struct类型:
I want to parse a JSON object in Go, but want to specify default values for fields that are not given. For example, I have the struct type:
type Test struct {
A string
B string
C string
}
A,B和C的默认值,分别是a,b和c。这意味着当我解析json的时候:
The default values for A, B, and C, are "a", "b", and "c" respectively. This means that when I parse the json:
{"A": "1", "C": 3}
我想得到结构:
I want to get the struct:
Test{A: "1", B: "b", C: "3"}
这可能使用内置程序包 encoding / json
?否则,是否有任何Go库具有此功能?
Is this possible using the built-in package encoding/json
? Otherwise, is there any Go library that has this functionality?
推荐答案
这可以使用encoding / json:调用 json.Unmarshal
,你不需要给它一个空的结构,你可以给它一个默认值。
This is possible using encoding/json: when calling json.Unmarshal
, you do not need to give it an empty struct, you can give it one with default values.
你的例子:
For your example:
var example []byte = []byte(`{"A": "1", "C": "3"}`)
out := Test{
A: "default a",
B: "default b",
// default for C will be "", the empty value for a string
}
err := json.Unmarshal(example, &out) // <--
if err != nil {
panic(err)
}
fmt.Printf("%+v", out)
{A: 1 B:default b C:3}
。
正如您所见, json.Unmarshal(例如,& ; out)
将JSON解组为 out
,覆盖指定的值ied JSON,但保持其他领域不变。
As you can see, json.Unmarshal(example, &out)
unmarshals the JSON into out
, overwriting the values specified in the JSON, but leaving the other fields unchanged.
这篇关于如何在Go中解析JSON时指定默认值的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!