PowerShell -打印JSON输出与排序数组的对象?

Question

如何打印JSON输出排序的对象数组?我的$result对象必须保持原样，即"Good"或";糟糕的;无所谓,我试着对数组中的对象进行排序按计数排序属性。

我代码:

$result = [PSCustomObject]@{
Good = @() 
Bad  = @()
}
$food = [PSCustomObject]@{
name  = "Banana"
count = 2
}
if ($food.count -gt 3) { $result.Good += $food }
else { $result.Bad += $food }
$sortGood = $result.Good | Sort-Object count
$sortBad = $result.Bad | Sort-Object count
Write-Output ($result | ConvertTo-Json)

我的JSON输出是:

{
"Good": [
{
"name": "Apple"
"count": 10
},
{
"name": "Lime"
"count": 5
},
{
"name": "Peach"
"count": 7
}
],
"Bad": [
{
"name": "Banana"
"count": 2
},
{
"name": "Kiwi"
"count": 1
},
{
"name": "Orange"
"count": 3
}
] 
}

我如何打印一个JSON看起来像这样?(按"计数"分类的水果;属性(升序排列)

{
"Good": [
{
"name": "Lime"
"count": 5
},
{
"name": "Peach"
"count": 7
},
{
"name": "Apple"
"count": 10
},
],
"Bad": [
{
"name": "Kiwi"
"count": 1
},
{
"name": "Banana"
"count": 2
},
{
"name": "Orange"
"count": 3
}
] 
}

[问题修复]编辑解决方案:

$result.Good = $result.Good | Sort-Object count
$result.Bad  = $result.Bad | Sort-Object count
Write-Output ($result | ConvertTo-Json)

Answer 1

Sort-Object不对对象进行"排序"。它返回对象的排序副本。这

$sortGood = $result.Good | Sort-Object count

将导致$sortGood被正确排序，并且$result.Good与原来完全相同。

$json = @"
{
"Good": [
{"name": "Apple", "count": 10},
{"name": "Lime", "count": 5},
{"name": "Peach", "count": 7}
],
"Bad": [
{"name": "Kiwi", "count": 1},
{"name": "Orange", "count": 4}
] 
}
"@
$data = ConvertFrom-Json $json
$food = @{
name  = "Banana"
count = 2
}
if ($food.count -gt 3) {
$data.Good += $food
} else {
$data.Bad += $food
}
$data.Good = $data.Good | Sort-Object count
$data.Bad = $data.Bad | Sort-Object count
$result = $data | ConvertTo-Json -Depth 10
$result

为

{
"Good":  [
{
"name":  "Lime",
"count":  5
},
{
"name":  "Peach",
"count":  7
},
{
"name":  "Apple",
"count":  10
}
],
"Bad":  [
{
"name":  "Kiwi",
"count":  1
},
{
"count":  2,
"name":  "Banana"
},
{
"name":  "Orange",
"count":  4
}
]
}

请注意，我总是重新分配$data.Good和$data.Bad的值:

• 使用$data.Good += $food创建一个新的数组(!)，$food在末尾，然后将其分配给$data.Good。($data.Good = $data.Good + $food的简写)
• 使用$data.Good = $data.Good | Sort-Object count以不同的顺序创建一个新的数组，然后将其分配给$data.Good。

Answer 2

嘿，我猜你忘记在Sort-Object后添加-Property了即

$sortGood = $result.Good | Sort-Object -Property count

试一试，让我知道!

Answer 3

我会这样做:

ConvertTo-Json @{
Good = $result.Good | sort Count
Bad = $result.Bad | sort Count
}