如何将以下json变量分解为数组中的各个项?
[
{
"server":{
"name":"myUbuntuServer1",
"imageRef":"3afe97b2-26dc-49c5-a2cc-a2fc8d80c001",
"flavorRef":"6"
}
},
{
"server":{
"name":"myUbuntuServer2",
"imageRef":"3afe97b2-26dc-49c5-a2cc-a2fc8d80c001",
"flavorRef":"6"
}
},
{
"server":{
"name":"myUbuntuServer3",
"imageRef":"3afe97b2-26dc-49c5-a2cc-a2fc8d80c001",
"flavorRef":"6"
}
}
]
例如,上述内容将转换为数组,包含以下内容:
数组项目0
{
"server":{
"name":"myUbuntuServer1",
"imageRef":"3afe97b2-26dc-49c5-a2cc-a2fc8d80c001",
"flavorRef":"6"
}
}
数组项目1
{
"server":{
"name":"myUbuntuServer2",
"imageRef":"3afe97b2-26dc-49c5-a2cc-a2fc8d80c001",
"flavorRef":"6"
}
}
数组项目2
{
"server":{
"name":"myUbuntuServer3",
"imageRef":"3afe97b2-26dc-49c5-a2cc-a2fc8d80c001",
"flavorRef":"6"
}
}
我想在Powershell 2.0中完成此操作并单独访问每个。到目前为止,这是我设法完成的任务:
$jsonarr = @()
$arr = (Get-Content C:\json.json| Out-String).replace("[","") -split "(})," -replace "]",""
$jsonarr += $arr[0..1] -join ""
$jsonarr += $arr[2..3] -join ""
$jsonarr += $arr[4]
然而,这是非常不灵活的,并且只要我有另一台服务器的详细信息到JSON文件就会停止工作。
答案 0 :(得分:1)
对于PowerShell v2,您可以使用Convert between PowerShell and JSON
PS
PowerShell v3 +,使用上面的工具应该是相同的:
$json = '[
{
"server":{
"name":"myUbuntuServer1",
"imageRef":"3afe97b2-26dc-49c5-a2cc-a2fc8d80c001",
"flavorRef":"6"
}
},
{
"server":{
"name":"myUbuntuServer2",
"imageRef":"3afe97b2-26dc-49c5-a2cc-a2fc8d80c001",
"flavorRef":"6"
}
},
{
"server":{
"name":"myUbuntuServer3",
"imageRef":"3afe97b2-26dc-49c5-a2cc-a2fc8d80c001",
"flavorRef":"6"
}
}
]'
$servers = ConvertFrom-Json $json
$servers.server.imageRef
返回
3afe97b2-26dc-49c5-a2cc-a2fc8d80c001
3afe97b2-26dc-49c5-a2cc-a2fc8d80c001
3afe97b2-26dc-49c5-a2cc-a2fc8d80c001
另外,不要忘记“Get-Member”
PPS
PS C:\Users\joshua\Desktop> $servers.server| where name -EQ myUbuntuServer2
name imageRef flavorRef
---- -------- ---------
myUbuntuServer2 3afe97b2-26dc-49c5-a2cc-a2fc8d80c001 6
PS C:\Users\joshua\Desktop> $servers.server| where name -EQ myUbuntuServer2 | select -Property flavorRef
flavorRef
---------
6
PPPS
也是当然
$servers.server[0]
你应该能够按姓名索引,但我正在制作一些愚蠢的错误atm