Go使用encoding/xml生成无包裹节点的库存数组XML的实现方法

Go标准库的encoding/xml包默认会为切片中的每个结构体元素生成独立的外层包裹标签，要实现无外层包裹、直接平铺库存字段的效果，你可以为根结构体实现xml.Marshaler接口，自定义序列化逻辑，手动控制字段输出顺序。

完整修改后代码 #

package main

import (
	"encoding/xml"
	"fmt"
	"os"
	"strconv"
)

func main() {
	type InventoryItem struct {
		ItemName        string
		ItemDescription string
	}

	type XMLEnvelop struct {
		XMLName   xml.Name        `xml:"root"`
		Inventory []InventoryItem
		Records   int
	}

	// 自定义XML序列化逻辑
	func (x XMLEnvelop) MarshalXML(e *xml.Encoder, start xml.StartElement) error {
		// 写入root开始标签
		if err := e.EncodeToken(start); err != nil {
			return err
		}
		// 依次平铺输出每个库存项的Name和Description字段
		for _, item := range x.Inventory {
			if err := e.EncodeElement(item.ItemName, xml.StartElement{Name: xml.Name{Local: "Name"}}); err != nil {
				return err
			}
			if err := e.EncodeElement(item.ItemDescription, xml.StartElement{Name: xml.Name{Local: "Description"}}); err != nil {
				return err
			}
		}
		// 输出records字段
		if err := e.EncodeElement(x.Records, xml.StartElement{Name: xml.Name{Local: "records"}}); err != nil {
			return err
		}
		// 写入root结束标签
		return e.EncodeToken(xml.EndElement{Name: start.Name})
	}

	var items []InventoryItem
	for i := 1; i < 6; i++ {
		items = append(items, InventoryItem{
			ItemName:        "Test " + strconv.Itoa(i),
			ItemDescription: "Description " + strconv.Itoa(i),
		})
	}

	v := &XMLEnvelop{Records: 1, Inventory: items}
	output, err := xml.MarshalIndent(v, "", "    ")
	if err != nil {
		fmt.Printf("error: %v\n", err)
	}

	os.Stdout.Write(output)
}

输出结果 #

<root>
    <Name>Test 1</Name>
    <Description>Description 1</Description>
    <Name>Test 2</Name>
    <Description>Description 2</Description>
    <Name>Test 3</Name>
    <Description>Description 3</Description>
    <Name>Test 4</Name>
    <Description>Description 4</Description>
    <Name>Test 5</Name>
    <Description>Description 5</Description>
    <records>1</records>
</root>

完全匹配客户期望的格式要求。

内容的提问来源于stack exchange，提问作者monkeymynd