如果你真的想选择这两条线，那么我想这是最短的方法：

(Get-Content -Path "E:\myfile.txt")[1,3]

或

Get-Content -Path "E:\myfile.txt" | Select-Object -Index 1,3

但是，如果您希望仅从文件中选择偶数行，则可以执行以下操作：

# return only the even lines (for odd lines, do for ($i = 0; ...)
$text = Get-Content -Path "E:\myfile.txt"; for ($i = 1; $i -lt @($text).Count; $i+=2) { $text[$i] }

或者使用Select-String

# return only the even lines (for odd lines, remove the ! exclamation mark
(Select-String -Path "E:\myfile.txt" -Pattern '.*' | Where-Object {!($_.LineNumber % 2)}).Line

获取内容路径“~\桌面\字符串. txt”|选择字符串模式“字符串2|串4”

可以使用Where-Object cmdlet筛选对象流（本例中为字符串）：

Get-Content -Path "E:\myfile.txt" | Where-Object {$_ -match '[24]$'}
# or
Get-Content -Path "E:\myfile.txt" | Where-Object {$_ -like '*[24]'}
# or
Get-Content -Path "E:\myfile.txt" | Where-Object {$_.EndsWith('2') -or $_.EndsWith('4')'}

如果只需要文件中的偶数行：

Get-Content -Path "E:\myfile.txt" | Where-Object {$_.ReadCount % 2 -eq 0}