Question

我有一个脚本，它有一个自定义PSObject数组，每个只有2个项目。我需要对每个对象中的第一个值进行排序，然后吐出两个项目，但它们都是@{Name=和@{Line=。这是我的代码片段：

$expinfo = @()
<<<>>>
$dbinfo = New-Object -TypeName PSObject
$dbname = Get-Item $folder | select Name
$dbinfo | Add-Member -NotePropertyName DBName -NotePropertyValue "$dbname"

$dbinfo | Add-Member -NotePropertyName ExportLine -NotePropertyValue "$eline"
$expinfo += $dbinfo
<<<>>>
foreach ($expdb in ($expinfo.GetEnumerator() | Sort-Object DBName))
{
    $dbn = $expdb | Select-Object -ExpandProperty DBName
    $dbe = $expdb | Select-Object -ExpandProperty ExportLine
    $dbn
    $dbe
}

输出如下：

@{Name=AIT1PD}
@{Line=20180312.1700 20180312.1704 All items successfully completed.}
@{Name=APAC1PD}
@{Line=20180313.0100 20180313.0120 All items successfully completed.}

我希望它看起来像这样：

AIT1PD 20180312.1700 20180312.1704 All items successfully completed.
APAC1PD 20180313.0100 20180313.0120 All items successfully completed.

Answer 1

更改您的Add-Member语句，以添加您感兴趣的属性值：

$dbinfo | Add-Member -NotePropertyName DBName -NotePropertyValue $dbname.Name
$dbinfo | Add-Member -NotePropertyName ExportLine -NotePropertyValue $eline.Line

假设您使用的是PowerShell 3.0或更高版本，我建议您使用[pscustomobject]类型加速器：

$dbinfo = [pscustomobject]@{
    DBName     = (Get-Item $folder).Name
    ExportLine = $eline.Line
}

试图获得@ {Name =输出结果

1 个答案: