xml - 从 XML 解析日期不起作用

Question

如果我用简单的值执行 time.Parse() - 那么一切都很好，但解析 XML 却不行。

type customDate struct {
    time.Time
}


func (c *customDate) UnmarshalXml(d *xml.Decoder, start xml.StartElement) error {
    var v string
    if err := d.DecodeElement(&v, &start); err != nil{
        return err
    }

    loc, _ := time.LoadLocation("Europe/Moscow")
    prs, err := time.ParseInLocation("02.01.2006", v, loc)
    if err != nil {
        return err
    }

    *c = customDate{prs}
    return nil
}

example on playground

Answer 1

date 是一个 XML 属性，而不是一个元素。因此，您必须实现 xml.UnmarshalerAttr接口(interface)而不是 xml.Unmarshaler :

package main

import (
    "encoding/xml"
    "fmt"
    "time"
)

type xmlSource struct {
    XMLName xml.Name `xml:"BicDBList"`
    Base    string   `xml:"Base,attr"`
    Items   []item   `xml:"item"`
}

// Item represent structure of node "item"
type item struct {
    File string     `xml:"file,attr"`
    Date customDate `xml:"date,attr"`
}

type customDate struct {
    time.Time
}

func (c *customDate) UnmarshalXMLAttr(attr xml.Attr) error {
    loc, err := time.LoadLocation("Europe/Moscow")
    if err != nil {
        return err
    }
    prs, err := time.ParseInLocation("02.01.2006", attr.Value, loc)
    if err != nil {
        return err
    }

    c.Time = prs
    return nil
}

var data = []byte(`<BicDBList Base="/mcirabis/BIK/">
    <item file="bik_db_09122016.zip" date="09.12.2016"/>
    <item file="bik_db_08122016.zip" date="08.12.2016"/>
    <item file="bik_db_07122016.zip" date="07.12.2016"/>
    <item file="bik_db_06122016.zip" date="06.12.2016"/>
</BicDBList>`)

func main() {
    var sample xmlSource

    err := xml.Unmarshal(data, &sample)

    if err != nil {
        println(err.Error())
    }
    fmt.Printf("%#v\n", sample)
}

https://play.golang.org/p/U56qfEOe-A