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我使用以下方法提取了 SQL 实例的一些详细信息：

$SQLInstance1 = Get-DbaInstanceProperty -SqlInstance SQL | Select-Object ComputerName, Name, Value  | Where-Object Name -In ("FullyQualifiedNetName", "Edition", "PhysicalMemory", "Processors", "VersionString", "Collation", "IsClustered", "ResourceLastUpdateDateTime")

将这些值分配给数组输出中的变量的最佳方法是什么，如下所示：

计算机名称 名称 值

SQL Edition Enterprise Edition（64 位） SQL PhysicalMemory 1571485
SQL Processors 40
SQL VersionString 11.0.7493.4
SQL Collat​​ion Latin1_General_CI_AS
SQL IsClustered True
SQL ResourceLastUpdateDateTime 24/12/2019 9:58:55 PM
SQL FullyQualifiedNetName PTHSQL13.FMG.local

所以我想从上面分配$ComputerName = ComputerName等$Edition = Edition，然后我可以用它来更新 SQL 表$ComputerName

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2 回答 2

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好的。首先，您需要将结果转换为字典，然后使用该属性字典创建一个新对象。

例如：

$dict = @{ }

(Get-DbaInstanceProperty -SqlInstance localhost | Select-Object ComputerName, Name, Value | Where-Object Name -In ("FullyQualifiedNetName", "Edition", "PhysicalMemory", "Processors", "VersionString", "Collation", "IsClustered", "ResourceLastUpdateDateTime")).GetEnumerator() | % { $dict.Add($_.Name, $_.Value) }

New-Object -TypeName PSObject -Property $dict

$result = New-Object -TypeName PSObject -Property $dict

$result.FullyQualifiedNetName

$result.Processors

上面的代码应该打印出FullyQualifiedNetName值和Processors计数。

于 2021-02-20T03:51:59.720 回答
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点符号是你的朋友。由于您存储的是一个序列化的实际对象，因此您可以使用.propertiesname.

[array]SQLInstance1 = Get-DbaInstanceProperty -SqlInstance SQL | Select-Object ComputerName, Name, Value  | Where-Object Name -In ("FullyQualifiedNetName", "Edition", "PhysicalMemory", "Processors", "VersionString", "Collation", "IsClustered", "ResourceLastUpdateDateTime")

($SQLInstance1.ComputerName.Count + $SQLInstance1.Name.Count + $SQLInstance1.Value.Count); $i++){
[pscustomobject]@{
            ComputerName = $($SQLInstance1.ComputerName[$i])
            Name = $($SQLInstance1.Name[$i])
            Value = $($SQLInstance1.Value[$i])

                }
        }
于 2021-02-20T04:32:28.710 回答