英文:
Go : Unmarshall XML List containing different entities
问题
我已经给出了以下的XML结构：
<properties>
<property name="element1">data1</property>
<property name="element2">data2</property>
<datetime name="start">13 Jan 2015 14:13:15GMT</datetime>
<property name="element3">data3</property>
<property name="element4">data4</property>
<property name="element5">data5</property>
<categorisation name="category">cat1</categorisation>
<property name="element6">data6</property>
<property name="element7">data7</property>
</properties>
我想要将其解组成以下结构：
type Property struct {
Name string
Value string
}
type Properties struct {
Props []Property
DateTime time.Time
Category string
}
但是我不知道如何处理无序的Properties。
英文:
I have been given the following XML structure
<properties>
<property name="element1">data1</property>
<property name="element2">data2</property>
<datetime name="start">13 Jan 2015 14:13:15GMT</datetime>
<property name="element3">data3</property>
<property name="element4">data4</property>
<property name="element5">data5</property>
<categorisation name="category">cat1</categorisation>
<property name="element6">data6</property>
<property name="element7">data7</property>
</properties>
I want to unmarshall it into the following:
type Property string {
Name string
Value string
}
type Properties struct {
Props []Property
DateTime time.Time
Category string
}
but I don't know how to process the out of order Properties
答案1
得分: 2
encoding/xml 包可以帮助你处理这个问题。你只需要正确地为结构体字段添加标签，告诉 xml 包它们在 XML 源中的位置：
type Property struct {
Name string `xml:"name,attr"`
Value string `xml:",cdata"`
}
type Properties struct {
Props []Property `xml:"property"`
DateTime string `xml:"datetime"`
Category string `xml:"categorisation"`
}
请注意，我将 DateTime 字段的类型更改为 string，因为 XML 中的时间戳（13 Jan 2015 14:13:15GMT）不符合标准格式（标准格式应为 ISO 8601，在 Go 中的布局为 2006-01-02T15:04:05Z07:00，对应 time.RFC3339）。
解析的代码如下：
ps := Properties{}
if err := xml.Unmarshal([]byte(src), &ps); err != nil {
panic(err)
}
fmt.Printf("%+v", ps)
输出结果如下（你可以在 Go Playground 上尝试运行）：
{Props:[{Name:element1 Value:data1} {Name:element2 Value:data2} {Name:element3 Value:data3}
{Name:element4 Value:data4} {Name:element5 Value:data5} {Name:element6 Value:data6}
{Name:element7 Value:data7}] DateTime:13 Jan 2015 14:13:15GMT Category:cat1}
你可以看到，Props 字段正确地按照顺序包含了所有 7 个属性，即使在它们之间插入了 <datetime> 和 <categorisation> 标签。
英文:
The encoding/xml package handles that for you. You just have to properly tag your struct fields to tell the xml package how they are found in the XML source:
type Property struct {
Name string `xml:"name,attr"`
Value string `xml:",cdata"`
}
type Properties struct {
Props []Property `xml:"property"`
DateTime string `xml:"datetime"`
Category string `xml:"categorisation"`
}
Note that I changed the type of DateTime field to string because the timestamp in the XML (13 Jan 2015 14:13:15GMT) does not conform to standard (which would be ISO 8601, in Go layout: 2006-01-02T15:04:05Z07:00 which is time.RFC3339).
Code to unmarshal:
ps := Properties{}
if err := xml.Unmarshal([]byte(src), &ps); err != nil {
panic(err)
}
fmt.Printf("%+v", ps)
Output (try it on the Go Playground):
{Props:[{Name:element1 Value:data1} {Name:element2 Value:data2} {Name:element3 Value:data3}
{Name:element4 Value:data4} {Name:element5 Value:data5} {Name:element6 Value:data6}
{Name:element7 Value:data7}] DateTime:13 Jan 2015 14:13:15GMT Category:cat1}
You can see that Props properly contains all 7 properties in correct order even though <datetime> and <categorisation> tags are inserted between them.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论