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PS C:\> $array

ReadWrite(AllHostsSpecified) : 
ReadOnly(AllHostsSpecified)  : 
ReadWrite(NegateSpecified)   : 
ReadWrite(Negate)            : 
SecFlavor                    : sys
ActualPathname               : 
ReadOnly(AllHosts)           : 
ReadOnly(Name)               : 
ReadWrite(Name)              : 
Anon                         : 
Root                         : {vin1,vin2,vin3,vin4...}
ReadOnly(NegateSpecified)    : 
ReadWrite(AllHosts)          : 
NosuidSpecified              : False
Nosuid                       : 
ReadOnly(Negate)             : 
Pathname                     : /vol/vin_binaries

嗨,伙计们,我有属性根,它是一个动态属性,可以从 (vin1,vin2,vin3,vin4.......) 变化,甚至可以无穷无尽。

现在,当我执行 export-csv 时,我需要如下获取输出对象

ReadWrite(AllHostsSpecified) : 
ReadOnly(AllHostsSpecified)  : 
ReadWrite(NegateSpecified)   : 
ReadWrite(Negate)            : 
SecFlavor                    : sys
ActualPathname               : 
ReadOnly(AllHosts)           : 
ReadOnly(Name)               : 
ReadWrite(Name)              : 
Anon                         : 
Root (Property 1)                         : vin1
Root (Property 1)                         : vin2
Root (Property 1)                         : vin3
Root (Property 1)                         : vin4
.
.
.
.

Root (Property n)                         : vinn

ReadOnly(NegateSpecified)    : 
ReadWrite(AllHosts)          : 
NosuidSpecified              : False
Nosuid                       : 
ReadOnly(Negate)             : 
Pathname                     : /vol/vin_binaries

有没有办法可以做到这一点?就像在 for 循环中迭代 Root 的所有属性???

4

1 回答 1

0

尝试这样的事情:

filter ExpandProperties {
    $obj = $_
    $obj | gm -MemberType *property | % {
        #Find objects with array value
        if( @($obj.($_.Name)).Count -gt 1 ) {
            $count = 1
            $prop = $_.Name
            $obj.($prop) | % {
                #Foreach value in the property
                $obj | Add-Member NoteProperty -Name "$prop (Property $($count))" -Value $_
                $count++
            }
        }
    }
    #Output object
    $obj
}

$o1 = New-Object psobject -Property @{
    Name = @("Name1", "Name2")
    Root = @("vin1","vin2")
}
$o2 = New-Object psobject -Property @{
    Name = "Name2"
    Root = @("vin1","vin2","vin3")
}
$o = $o1, $o2

$o | ExpandProperties


Name              : {Name1, Name2}
Root              : {vin1, vin2}
Name (Property 1) : Name1
Name (Property 2) : Name2
Root (Property 1) : vin1
Root (Property 2) : vin2

Name              : Name2
Root              : {vin1, vin2, vin3}
Root (Property 1) : vin1
Root (Property 2) : vin2
Root (Property 3) : vin3

这也显示了原始数组属性(例如“Root”)。如果要排除它,则需要select -exclude ...在函数内部使用。但是,这会创建一个新的“pscustomobject”(您会丢失原始对象类型),因此我没有将其包含在上面的解决方案中。

于 2013-02-27T06:48:29.940 回答