Go XML Unmarshal示例不編譯

Question

go docs中的Xml示例已損壞。有誰知道如何使它工作？當我編譯它，結果是：

xmlexample.go:34: cannot use "name" (type string) as type xml.Name in field value 
xmlexample.go:34: cannot use nil as type string in field value 
xmlexample.go:34: too few values in struct initializer 

下面是相關代碼：

package main 

import (
     "bytes" 
     "xml" 
) 

type Email struct { 
     Where string "attr"; 
     Addr string; 
} 

type Result struct { 
     XMLName xml.Name "result"; 
     Name string; 
     Phone string; 
     Email []Email; 
} 

var buf = bytes.NewBufferString (` 
<result> 
     <email where="home"> 
       <addr>gre@example.com</addr> 
     </email> 
     <email where='work'> 
       <addr>gre@work.com</addr> 
     </email> 
     <name>Grace R. Emlin</name> 
     <address>123 Main Street</address> 
</result>`) 


func main() { 
     var result = Result{ "name", "phone", nil } 
     xml.Unmarshal (buf , &result) 
     println (result.Name) 
} 

Answer 1

的type Result被定義爲：

type Result struct { 
    XMLName xml.Name "result" 
    Name string 
    Phone string 
    Email []Email 
} 

的type xml.Name，嵌入type Result定義爲：

// A Name represents an XML name (Local) annotated 
// with a name space identifier (Space). 
// In tokens returned by Parser.Token, the Space identifier 
// is given as a canonical URL, not the short prefix used 
// in the document being parsed. 
type Name struct { 
    Space, Local string 
} 

因此，使用複合lite RALS，使用類似的東西一個：

var result = Result{xml.Name{}, "name", "phone", nil} 

var result = Result{xml.Name{"space", "local"}, "name", "phone", nil} 

var result = Result{Name: "name", Phone: "phone", Email: nil} 

Answer 2

線

var result = Result{ "name", "phone", nil } 

需求，成爲

var result = Result{ Name: "name", Phone: "phone", Email: nil } 

然後應該按預期工作。我提交了一個補丁來修復文檔，並且很快發佈了一個版本，所以沒有人應該再次遇到這個問題。

Answer 3

，如果你與其他參數一起提供xml.Name {}，像這樣它也適用：

var result = Result{ xml.Name{"", "result"}, "name", "phone", nil } 

Answer 4

這裏

var result Result 

作品。