GOLANG xml decoding

Question

In Go, I'm trying to decode this XML into the "fileRetriever" struct. I only care about the file names:

 <?xml version="1.0" encoding="ISO-8859-1" ?>
 <FileRetriever>
   <FileList>
       <File name="Name1" />
       <File name="Name2" />
   </FileList>
 </FileRetriever>

I think this code snippet is close, but I can't seem to see where I'm going wrong. It produces no errors, but also a zero-length list of file names:

import (
    "encoding/base64"
    "encoding/xml"
    "fmt"
    "net/http"

    "golang.org/x/net/html/charset"

)
type fileRetriever struct {
    Files []file `xml:"FileRetriever>FileList>File"`
}

type file struct {
    Name string `xml:"name,attr"`
}

func Main(){
    retrieve()
}

func retrieve()(retriever *fileRetriever){
    req := ... //set up http.NewRequest()
    client := &http.Client{}
    rsp, err := client.Do(req)

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

    defer rsp.Body.Close()

    decoder := xml.NewDecoder(rsp.Body)
    decoder.CharsetReader = charset.NewReaderLabel

    retriever = &fileRetriever{}

    err = decoder.Decode(&retriever)

    if err != nil {
        fmt.Println(err)
    }

    return retriever, xTidx
}

mkopriva · Accepted Answer · 2017-04-19 01:23:01Z

4

The root element is automatically decoded into the value you pass to Decode so you don't need to mention it in the Files field tag.

So just change xml:"FileRetriever>FileList>File" to xml:"FileList>File".

answered Apr 19, 2017 at 1:23

mkopriva

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1 Comment

Brent Over a year ago

Perfect. Thanks. That took care of it.

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