变量中的比较对象输出

Question

我正在构建一个脚本，用于比较使用中的 IPv4 地址的最后一个八位字节与所有可用的八位字节（2 到 254）。

我已经这么远了，我确实通过比较数组得到了结果，但我的最终结果也是一个数组，我似乎不能只得到数字。

我的脚本：

$guestIP = @("192.168.31.200","192.168.31.31","192.168.31.90","192.168.31.25","192.168.31.100")

$AllLastOcted = $("2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31","32","33","34","35","36","37","38","39","40","41","42","43","44","45","46","47","48","49","50","51","52","53","54","55","56","57","58","59","60","61","62","63","64","65","66","67","68","69","70","71","72","73","74","75","76","77","78","79","80","81","82","83","84","85","86","87","88","89","90","91","92","93","94","95","96","97","98","99","100","101","102","103","104","105","106","107","108","109","110","111","112","113","114","115","116","117","118","119","120","121","122","123","124","125","126","127","128","129","130","131","132","133","134","135","136","137","138","139","140","141","142","143","144","145","146","147","148","149","150","151","152","153","154","155","156","157","158","159","160","161","162","163","164","165","166","167","168","169","170","171","172","173","174","175","176","177","178","179","180","181","182","183","184","185","186","187","188","189","190","191","192","193","194","195","196","197","198","199","200","201","202","203","204","205","206","207","208","209","210","211","212","213","214","215","216","217","218","219","220","221","222","223","224","225","226","227","228","229","230","231","232","233","234","235","236","237","238","239","240","241","242","243","244","245","246","247","248","249","250","251","252","253","254")

$guestIP = $guestIP | sort -Property {$_-replace '[\d]'},{$_-replace '[a-zA-Z\p{P}]'-as [int]}
$AllLastOcted = $AllLastOcted | sort -Property {$_-replace '[\d]'},{$_-replace '[a-zA-Z\p{P}]'-as [int]}

$guestIP = $guestIP -replace('192.168.31.','')

Compare-Object -ReferenceObject ($AllLastOcted ) -DifferenceObject ($guestIP) | select -Property InputObject -First 1 | set-Variable AvailableOcted -PassThru

$AvailableOcted

我的目标是，作为结果，我拥有第一个可用的八位字节。喜欢：

write-host "IP that can be used is 192.168.31.$AvailableOcted"
PS > IP that can used is 192.168.31.2

Answer 1

好吧，就这么简单：

$AvailableOcted.InputObject

会 return 只有值。

所以它看起来像这样：

write-host ("IP that can be used is 192.168.31." + $AvailableOcted.InputObject)

Answer 2

你可以把它简化很多。

您可以使用 范围运算符 创建数组，而不是定义从 2 到 254 的所有数字。您也不需要 [int] 强制转换。您可以使用 Where-Object cmdlet，而不是使用 Compare-Object cmdlet 来过滤八位字节：

$guestIP = @("192.168.31.200","192.168.31.31","192.168.31.90","192.168.31.25","192.168.31.100")

$AllLastOcted = 2 .. 254
$usedOcted = $guestIP -replace '.*\.'
$nextAvailableOcted = $AllLastOcted | Where { $_ -NotIn $usedOcted } | select -first 1

write-host "IP that can be used is 192.168.31.$nextAvailableOcted"

输出：

IP that can be used is 192.168.31.2

变量中的比较对象输出

Compare-object output in variable

powershell

scripting

compareobject