如何使用 Go 将 JSON 文件解析为结构体
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How Do I Parse a JSON file into a struct with Go
提问by Charles
I'm trying to configure my Go program by creating a JSON file and parsing it into a struct:
我正在尝试通过创建一个 JSON 文件并将其解析为一个结构来配置我的 Go 程序:
var settings struct {
serverMode bool
sourceDir string
targetDir string
}
func main() {
// then config file settings
configFile, err := os.Open("config.json")
if err != nil {
printError("opening config file", err.Error())
}
jsonParser := json.NewDecoder(configFile)
if err = jsonParser.Decode(&settings); err != nil {
printError("parsing config file", err.Error())
}
fmt.Printf("%v %s %s", settings.serverMode, settings.sourceDir, settings.targetDir)
return
}
The config.json file:
config.json 文件:
{
"serverMode": true,
"sourceDir": ".",
"targetDir": "."
}
The Program compiles and runs without any errors, but the print statement outputs:
程序编译和运行没有任何错误,但打印语句输出:
false
(false and two empty strings)
(假和两个空字符串)
I've also tried with json.Unmarshal(..)but had the same result.
我也尝试过,json.Unmarshal(..)但结果相同。
How do I parse the JSON in a way that fills the struct values?
如何以填充结构值的方式解析 JSON?
回答by Daniel
You're not exporting your struct elements. They all begin with a lower case letter.
您没有导出结构元素。它们都以小写字母开头。
var settings struct {
ServerMode bool `json:"serverMode"`
SourceDir string `json:"sourceDir"`
TargetDir string `json:"targetDir"`
}
Make the first letter of your stuct elements upper case to export them. The JSON encoder/decoder wont use struct elements which are not exported.
将 stuct 元素的第一个字母设为大写以导出它们。JSON 编码器/解码器不会使用未导出的结构元素。

