Question

我试图运行此命令来获取A和CNAME记录，但它似乎无法正常工作，我的语法错误。它只检索CNAME类型的DNS记录。

实际命令是： Get-DnsServerResourceRecord -RRType＆＃34; CNAME＆＃34;

$results = Get-DnsServerZone | % {
$zone = $_.zonename
Get-DnsServerResourceRecord $zone -filter {(RRType -like "CNAME") -or (RRType -like "A")}   | select @{n='ZoneName';e={$zone}}, HostName, RecordType, @{n='RecordData';e={if ($_.RecordData.IPv4Address.IPAddressToString) {$_.RecordData.IPv4Address.IPAddressToString} else {$_.RecordData.NameServer.ToUpper()}}}
}

$results | Export-Csv -NoTypeInformation c:\temp\DNSRecords10.csv -Append

Answer 1

您应该使用Where-Object进行过滤，因为cmdlet没有自己的参数：

Get-DnsServerResourceRecord -ZoneName $zone |
    Where-Object { $_.RRType -in 'CNAME', 'A' } |
    Select-Object -Property @(
        @{ Name       = 'ZoneName'
           Expression = { $zone }
         }
        'HostName'
        'RecordType'
        @{ Name       = 'RecordData'
           Expression = {
                if ($_.RecordData.IPv4Address.IPAddressToString) {
                    $_.RecordData.IPv4Address.IPAddressToString
                } else {
                    $_.RecordData.NameServer.ToUpper()
                }
            }
         }
    )

作为脚注，如果您不使用通配符，则-like与使用-eq相同。

文档：

Answer 2

the solution proposed by TheIncorrigible1的替代方法是简单地发出两个查询：

@(
  Get-DnsServerResourceRecord $zone -RRType CNAME
  Get-DnsServerResourceRecord $zone -RRType A
) | select @{n='ZoneName';e={$zone}}, HostName, RecordType, @{n='RecordData';e={if ($_.RecordData.IPv4Address.IPAddressToString) {$_.RecordData.IPv4Address.IPAddressToString} else {$_.RecordData.NameServer.ToUpper()}}}

PowerShell - OR运算符

2 个答案: