PowerShell如何将空XML文件转换为JSON格式？

Question

我正在尝试使用PowerShell将一个XML文件转换为JSON格式。直到这一步都很简单，但是源文件看起来编码有问题，当我应用convertto-json时，JSON结构为空：

XML源代码：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<PnCG3_configuration>
  <domains>
    <domain Name="xxxxx">
      <players>
        <player targetId="00-1c-e6-02-0a-00" targetIdType="mac" label="test" middlewareFamily="gekkota-3" />
        <player />
      </players>
    </domain>
    <domain Name="xxxxxxx2">
      <players>
        <player targetId="00-1c-e6-02-0a-00" targetIdType="mac" label="OfficeChateaugiron-TV01" middlewareFamily="gekkota-3" />
        <player targetId="00-1c-e6-02-20-00" targetIdType="mac" label="test-01" middlewareFamily="gekkota-3" />
        <player targetId="00-1c-e6-02-0b-00" targetIdType="mac" label="test-test-TV01" middlewareFamily="gekkota-3" />
        <player targetId="00-1c-e6-02-20-00" targetIdType="mac" label="test-TV01-test" middlewareFamily="gekkota-3" />
        <player targetId="00-1c-e6-02-20-00" targetIdType="mac" label="test" middlewareFamily="gekkota-3" />
        <player targetId="00-1c-e6-02-09-00" targetIdType="mac" label="test-test-TV01" middlewareFamily="gekkota-3" />
        <player targetId="00-1c-e6-02-04-00" targetIdType="mac" label="test-test-TV01" middlewareFamily="gekkota-3" />
        <player />
      </players>
    </domain>

并输出 Json :

[
    [

    ],
    [
        [
            [
                [
                    [

                    ],
                    [

                    ]
                ]
            ],
            [
                [
                    [

                    ],
                    [

                    ],
                    [

                    ],
                    [

                    ],
                    [

                    ],
                    [

                    ],

我用于转换的代码：

$xmlObject = [XML](Get-Content -Path $o)
$xmlObject | ConvertTo-JSON -depth 100 | Out-File "$o.json"

- Steven Poulain

Answer 1

很遗憾，这并不容易。看一下我之前使用过的这个GitHub Repo，可以完成你所要求的转换。你的代码将变成：

Add-Type -Path .\Newtonsoft.Json.dll
$xmlObject = [XML](Get-Content -Path $o)
[Newtonsoft.Json.JsonConvert]::SerializeXmlNode($xmlObject ) | Out-File "$o.json"

- TobyU

Answer 2

使用这种方法对我来说会产生很多“空格”噪音。

[Newtonsoft.Json.JsonConvert]::SerializeXmlNode($xmlObject) | Out-File "$o.json"

这种方式重新编译XML文件可以去除空格并创建一个整洁的对象。

[Newtonsoft.Json.JsonConvert]::SerializeXmlNode([xml] $xmlObject.OuterXML ) | Out-File "$o.json"

- Rogier Langeveld