json - 从 powershell 中的 JSON 对象中删除一个属性

Question

我有以下 JSON，我想从数组地址下的 JSON 对象中删除街道。我正在尝试在 powershell 中执行此操作

{
  "Customer": [
    {
      "id": "123"
    }
  ],
  "Nationality": [
    {
      "name": "US",
      "id": "456"
    }
  ],
  "address": [
    {
      "$type": "Home",
      "name": "Houston",
      "streets": [
        {
          "name": "Union",
          "postalCode": "10",
        }
      ]
    },
    {
      "$type": "Office",
      "name": "Hawai",
      "streets": [
        {
          "name": "Rock",
          "postalCode": "11",
        }
      ]
    }
  ],
  "address": [
    {
      "$type": "Home1",
      "name": "Houston",
      "streets": [
        {
          "name": "Union1",
          "postalCode": "14",
        }
      ]
    },
    {
      "$type": "Office1",
      "name": "Hawaii1",
      "streets": [
        {
          "name": "Rock1",
          "postalCode": "15",
        }
      ]
    }
  ],
}

我想从 JSON 对象中删除街道，这是我的 powershell 脚本，但它不起作用!我正在尝试将 JSON 转换为对象，然后遍历属性以删除它们。

$FileContent = Get-Content -Path "Test.json" -Raw | ConvertFrom-Json
foreach ($content in $FileContent) {
    #Write-Host $content.address
    $content.address = $content.address | Select-Object * -ExcludeProperty streets
}

$FileContent | ConvertTo-Json -Depth 100 | Out-File "Test.json" -Force

Answer 1

当您使用 ConvertFrom-Json 时，它会自动为您将事物转换为对象。一旦它们成为对象，您就可以使用 Select-Object 指定要包含在管道中的属性，您只需设置 $FileContent.address(一个数组对象)等于自身，排除数组中每个对象的街道属性。

$FileContent = Get-Content -Path "Test.json" -Raw | ConvertFrom-Json
$FileContent.address = $FileContent.address | Select-Object * -ExcludeProperty streets
$FileContent | ConvertTo-Json