如何提取複雜的字符串字符串？

Question

我開始字符串是：

$grp = DL-Test1-Test2-RW" 

我的目標是有

$grp = "Test1\Test2" 

所以我需要保持第一和最後一個字符串之間 - 字符「」。和替換 - 通過\

修訂 我嘗試這樣做：

$grp = "DL-test1-test2-RW" 
$Descritpion = $grp.Split("-") #Split - to have an array 
$Descritpion = $Descritpion.Split($Descritpion[0]) #Cut first element 
$Descritpion = $Descritpion.Split($Descritpion[-1]) # Cut last element 
#Here replace ? 
Write-Host "Description:"$Descritpion 

Answer 1

假設串總是有這種形式，你有興趣在第二和第三部分：

# $grp.Split("x") - split string on character x, creating an array 
# $grp.Split("x")[n] - get the nth element of the array 
# x,y -join "\" join the array elements x and y into a string, with "\" inbetween 
($grp.Split("-")[1],$grp.Split("-")[2]) -join "\" 

編輯 - 對於通用元件數

$($grp.Split("-") | Select-Object -SkipLast 1 | Select-Object -Last ($grp.Split("-").count - 2)) -join "\" 

多行：

$Descritpion = $grp.Split("-") 
$Descritpion = $Descritpion | Select-Object -SkipLast 1 
$Descritpion = $Descritpion | Select-Object -Last ($grp.Split("-").count - 2) 
$Descritpion = $Descritpion-join "\" 

Answer 2

您還可以使用正則表達式做

$grp = "DL-Test1-Test2-RW" 
$regex = "-(.*)-(.*)-" 
if ($grp -match $regex){ 
    $Matches[1] + "\" + $Matches[2] 
} 

如果字符串可以有起點和終點之間的多個組，試試這個

$grp = "DL-Test1-Test2-Test3-Test4-RW" 
$regex = "-(.*-){1,999}(.*)-" 
if ($grp -match $regex){ 
    $Matches[1].replace('-','\') + $matches[2] 
} 

Answer 3

只是做：

$grp = "DL-test1-test2-RW" 
$arraygrp=$grp.Split("-") 
$arraygrp[1..($arraygrp.Count -2)] -join "\" 

or this

$grp.Substring($grp.IndexOf('-') +1, $grp.LastIndexOf('-')-$grp.IndexOf('-')-1).Replace('-', '\')