考虑JSON格式如下:
"Stuffs": [
{
"Name": "Darts",
"Type": "Fun Stuff"
},
{
"Name": "Clean Toilet",
"Type": "Boring Stuff"
}
]
在PowerShell中3,我们可以得到的东西的清单:
$JSON = Get-Content $jsonConfigFile | Out-String | ConvertFrom-Json
假设我们不知道名单的确切内容,包括对象的排序,我们如何能检索与名称字段的特定值的对象(S)?
蛮力,我们可以遍历列表:
foreach( $Stuff in $JSON.Stuffs ) {
但我希望存在一个更直接的机制(在C#相似到Lync或Lambda表达式)。
$json = @"
{
"Stuffs":
[
{
"Name": "Darts",
"Type": "Fun Stuff"
},
{
"Name": "Clean Toilet",
"Type": "Boring Stuff"
}
]
}
"@
$x = $json | ConvertFrom-Json
$x.Stuffs[0] # access to Darts
$x.Stuffs[1] # access to Clean Toilet
$darts = $x.Stuffs | where { $_.Name -eq "Darts" } #Darts
我刚才问了同样的问题在这里: https://stackoverflow.com/a/23062370/3532136它有一个很好的解决方案。 我希望它能帮助^^。 在简历中,您可以使用此:
在我的情况JSON文件被称为jsonfile.json
:
{
"CARD_MODEL_TITLE": "OWNER'S MANUAL",
"CARD_MODEL_SUBTITLE": "Configure your download",
"CARD_MODEL_SELECT": "Select Model",
"CARD_LANG_TITLE": "Select Language",
"CARD_LANG_DEVICE_LANG": "Your device",
"CARD_YEAR_TITLE": "Select Model Year",
"CARD_YEAR_LATEST": "(Latest)",
"STEPS_MODEL": "Model",
"STEPS_LANGUAGE": "Language",
"STEPS_YEAR": "Model Year",
"BUTTON_BACK": "Back",
"BUTTON_NEXT": "Next",
"BUTTON_CLOSE": "Close"
}
码:
$json = (Get-Content "jsonfile.json" -Raw) | ConvertFrom-Json
$json.psobject.properties.name
输出:
CARD_MODEL_TITLE
CARD_MODEL_SUBTITLE
CARD_MODEL_SELECT
CARD_LANG_TITLE
CARD_LANG_DEVICE_LANG
CARD_YEAR_TITLE
CARD_YEAR_LATEST
STEPS_MODEL
STEPS_LANGUAGE
STEPS_YEAR
BUTTON_BACK
BUTTON_NEXT
BUTTON_CLOSE
由于mjolinor 。
大卫·布拉班特的回答使我对我所需要的,但增加了:
x.Stuffs | where { $_.Name -eq "Darts" } | Select -ExpandProperty Type
这是我的JSON数据:
[
{
"name":"Test",
"value":"TestValue"
},
{
"name":"Test",
"value":"TestValue"
}
]
PowerShell脚本:
$data = Get-Content "Path to json file" | Out-String | ConvertFrom-Json
foreach ($line in $data) {
$line.name
}