首页 > 后端开发 > Golang > 如何最好地将 JSON 数据规范化为 Go 中的 API 结构体

如何最好地将 JSON 数据规范化为 Go 中的 API 结构体

王林
发布: 2024-02-13 18:06:21
转载
982 人浏览过

如何最好地将 JSON 数据规范化为 Go 中的 API 结构体

php小编西瓜在这里带来了一篇关于如何将JSON数据规范化为Go中的API结构体的精简指南。在现代Web应用程序中,与JSON数据打交道是常见的任务。Go语言作为一门强大的后端语言,提供了一种简洁而灵活的方式来处理JSON数据。本文将介绍如何使用Go语言中的结构体来规范化JSON数据,从而更好地处理和操作它们。无论你是初学者还是有经验的开发人员,本文都将为你提供有用的技巧和实用的示例。让我们开始吧!

问题内容

我对 go 还很陌生,正在尝试确定是否有更简洁的方法来完成从前端 (js) 到我的 api 的 json 数据的规范化。为了确保在从结构 (model.expense) 创建变量时使用正确的类型,我将有效负载转储到映射中,然后规范化并保存回结构。如果有人可以教我更好的方法来处理这个问题,我将不胜感激!提前致谢!

模型.费用结构:

type expense struct {
    id        primitive.objectid   `json:"_id,omitempty" bson:"_id,omitempty"`
    name      string               `json:"name"`
    frequency int                  `json:"frequency"`
    startdate *time.time           `json:"startdate"`
    enddate   *time.time           `json:"enddate,omitempty"`
    cost      primitive.decimal128 `json:"cost"`
    paid      []string             `json:"paid,omitempty"`
}
登录后复制

有问题的控制器:

func InsertOneExpense(w http.ResponseWriter, r *http.Request) {
    w.Header().Set("Content-Type", "application/json")
    w.Header().Set("Allow-Control-Allow-Methods", "POST")

    var expense map[string]interface{}
    json.NewDecoder(r.Body).Decode(&expense)

    var expenseName string
    if name, ok := expense["name"]; ok {
        expenseName = fmt.Sprintf("%v", name)
    } else {
        json.NewEncoder(w).Encode("missing required name")
    }

    var expenseFrequency int
    if frequency, ok := expense["frequency"]; ok {
        expenseFrequency = int(frequency.(float64))
    } else {
        expenseFrequency = 1
    }

    // Handle startDate normalization
    var expenseStartDate *time.Time
    if startDate, ok := expense["startDate"]; ok {
        startDateString := fmt.Sprintf("%v", startDate)
        startDateParsed, err := time.Parse("2006-01-02 15:04:05", startDateString)

        if err != nil {
            log.Fatal(err)
        }

        expenseStartDate = &startDateParsed
    } else {
        json.NewEncoder(w).Encode("missing required startDate")
    }

    // Handle endDate normalization
    var expenseEndDate *time.Time
    if endDate, ok := expense["endDate"]; ok {
        endDateString := fmt.Sprintf("%v", endDate)
        endDateParsed, err := time.Parse("2006-01-02 15:04:05", endDateString)

        if err != nil {
            log.Fatal(err)
        }

        expenseEndDate = &endDateParsed
    } else {
        expenseEndDate = nil
    }

    // Handle cost normaliztion
    var expenseCost primitive.Decimal128
    if cost, ok := expense["cost"]; ok {
        costString := fmt.Sprintf("%v", cost)
        costPrimitive, err := primitive.ParseDecimal128(costString)

        if err != nil {
            log.Fatal(err)
        }

        expenseCost = costPrimitive
    } else {
        json.NewEncoder(w).Encode("missing required cost")
        return
    }

    normalizedExpense := model.Expense{
        Name:      expenseName,
        Frequency: expenseFrequency,
        StartDate: expenseStartDate,
        EndDate:   expenseEndDate,
        Cost:      expenseCost,
    }

    // Do more things with the struct var...
}
登录后复制

解决方法

您可以定义 json.unmarshaljson 接口,然后根据需要手动验证数据。尝试这样的事情:

package main

import (
    "encoding/json"
    "fmt"
    "strconv"
)

type CoolStruct struct {
    MoneyOwed string `json:"money_owed"`
}

// UnmarshalJSON the json package will delegate deserialization to our code if we implement the json.UnmarshalJSON interface
func (c *CoolStruct) UnmarshalJSON(data []byte) error {
    // get the body as a map[string]*[]byte
    raw := map[string]*json.RawMessage{}
    if err := json.Unmarshal(data, &raw); err != nil {
        return fmt.Errorf("unable to unmarshal raw meessage map: %w", err)
    }

    // if we don't know the variable type sent we can unmarshal to an interface
    var tempHolder interface{}
    err := json.Unmarshal(*raw["money_owed"], &tempHolder)
    if err != nil {
        return fmt.Errorf("unable to unmarshal custom value from raw message map: %w", err)
    }

    // the unmarshalled interface has an underlying type use go's typing
    // system to determine type conversions / normalizations required
    switch tempHolder.(type) {
    case int64:
        // once we determine the type of the we just assign the value
        // to the receiver's field
        c.MoneyOwed = strconv.FormatInt(tempHolder.(int64), 10)
    // we could list all individually or as a group; driven by requirements
    case int, int32, float32, float64:
        c.MoneyOwed = fmt.Sprint(tempHolder)
    case string:
        c.MoneyOwed = tempHolder.(string)
    default:
        fmt.Printf("missing type case: %T\n", tempHolder)
    }
    // success; struct is now populated
    return nil
}

func main() {
    myJson := []byte(`{"money_owed": 123.12}`)
    cool := CoolStruct{}
    // outside of your struct you marshal/unmarshal as normal
    if err := json.Unmarshal(myJson, &cool); err != nil {
        panic(err)
    }
    fmt.Printf("%+v\n", cool)
}
登录后复制

输出:{moneyowed:123.12}
游乐场链接://m.sbmmt.com/link/87ca4eb840b6f78e3b6d6b418c0fef40

以上是如何最好地将 JSON 数据规范化为 Go 中的 API 结构体的详细内容。更多信息请关注PHP中文网其他相关文章!

相关标签:
来源:stackoverflow.com
本站声明
本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系admin@php.cn
最新问题
热门教程
更多>
最新下载
更多>
网站特效
网站源码
网站素材
前端模板