xml - 从 zip 中解码特定的 XML 文件而不提取

Question

我有一个 zip 文件，里面有几个 xml 文件，使用 zip和 encoding/xml来自 Go 存档的包。我想做的是将 only a.xml 解码为一个类型 - 即不遍历里面的所有文件:

test.zip
├ a.xml
├ b.xml
└ ...

a.xml 的结构如下:

<?xml version="1.0" encoding="UTF-8"?>
<root>
    <app>
        <code>0001</code>
        <name>Some Test App</name>
    </app>
    <app>
        <code>0002</code>
        <name>Another Test App</name>
    </app>
</root>

如何选择和解码其名称在注释掉的行中作为参数提供的文件，例如:

package marshalutils

import (
    "archive/zip"
    "log"
    "fmt"
    "encoding/xml"
)

type ApplicationRoot struct {
    XMLName xml.Name `xml:"root"`
    Applications []Application `xml:"app"`
}

type Application struct {
    Code string `xml:"code"`
    Name string `xml:"name"`
}

func UnmarshalApps(zipPath string, fileName string) {
    // Open a zip archive for reading.
    reader, err := zip.OpenReader(zipFilePath)
    if err != nil {
        log.Fatal(`ERROR:`, err)
    }

    defer reader.Close()

    /* 
     * U N M A R S H A L   T H E   G I V E N   F I L E ...
     * ... I N T O   T H E   T Y P E S   A B O V E
     */
}

Answer 1

好吧，这是我通过添加到示例函数的返回类型声明找到的答案:

func UnmarshalApps(zipPath string, fileName string) ApplicationRoot {
    // Open a zip archive for reading.
    reader, err := zip.OpenReader(zipFilePath)
    if err != nil {
        log.Fatal(`ERROR:`, err)
    }

    defer reader.Close()

    /* 
     * START OF ANSWER
     */
    var appRoot ApplicationRoot
    for _, file := range reader.File {
        // check if the file matches the name for application portfolio xml
        if file.Name == fileName {
            rc, err := file.Open()
            if err != nil {
                log.Fatal(`ERROR:`, err)
            }

            // Prepare buffer
            buf := new(bytes.Buffer)
            buf.ReadFrom(rc)

            // Unmarshal bytes
            xml.Unmarshal(buf.Bytes(), &appRoot)
            rc.Close()
        }
    }   
     /* 
     * END OF ANSWER
     */     
}