string replace add in powershell

Question

hello guys i'm a beginner in powershell , i have an outpout file from a script that i createdto check all volume data (datax) on a netapp SVM(na0maa-euc01) level3 folder who have everyone permission on it and without active directory security groups(must positionned on the level3 folders) , so the outpout txt file is like that

\\na0maa-euc01\data1\Nissan Production Engineering\TC Facility\X11M 
\\na0maa-euc01\data1\Powertrain_DIP-A entry\IQMS\Data
\\na0maa-euc01\data1\Powertrain_DIP-A entry\PROJECT INFORMATION\Data
\\na0maa-euc01\data2\powertrain_emb_ice_algorithm_OLD\Data\05_DCM

those level 3 folder where copied from a 7 mode netapp named f-typhon-old were the acl are ok

\\f-typhon-old\data1\Nissan Production Engineering.mig\TC Facility\X11M
\\f-typhon-old\data1\Powertrain_DIP-A entry.mig\IQMS\Data
\\f-typhon-old\data1\Powertrain_DIP-A entry.mig\PROJECT INFORMATION\Data
\\f-typhon-old\data2\powertrain_emb_ice_algorithm_OLD.mig\Data\05_DCM

so i created a new file with a baseline to correct the acl like that

data1\Nissan Production Engineering\TC Facility\X11M
data1\Powertrain_DIP-A entry\IQMS\Data
data1\Powertrain_DIP-A entry\PROJECT INFORMATION\Data
data2\powertrain_emb_ice_algorithm_OLD.mig\Data\05_DCM

made a little script with a loop on the file with get-acl and set-acl to correct but the issue is i need a text manipulation to create a variable $level3old to store in it the path with .mig in the level1 folders i don't now how to do it or if it was possible in powershell

$Pth = "C:\temp\mig\mig_CDOT\0INCHE\Log\Robocopy\Extract\acl_user"
$filerd = "na0maa-euc01"
$filers = "f-typhon-old"
$N3 = gc $pth\N3_witheveryoneacess_base.txt

foreach ($level3 in $N3) {
    $level3old = how too replace datax\level1\level2\level3 with datax\level1.mig\level2\level3
    get-acl "\\$filers\$level3old " | set-acl "\\$filerd\$level3"
}

$level3old = $level3 -replace '^\\\\.*?\\(.*?\\[^\\]+)','$1.mig' will replace the system name and add the .mig — AdminOfThings
– AdminOfThings, Commented Jan 25, 2020 at 13:02

AdminOfThings · Accepted Answer · 2020-01-26 04:44:00Z

I am not completely sure what your variables contain at all times. So I will provide some examples.

Assuming $level3 contains the server name and no .mig

$level3 = '\\na0maa-euc01\data1\Nissan Production Engineering\TC Facility\X11M'
$level3old = $level3 -replace '^\\\\.*?\\(.*?\\[^\\]+)','$1.mig'

# Output
$level3old
data1\Nissan Production Engineering.mig\TC Facility\X11M

Assuming $level3 starts with the shared folder and contains no .mig

$level3 = 'data1\Nissan Production Engineering\TC Facility\X11M'
$level3old = $level3 -replace '^.*?\\[^\\]+','$&.mig'

# Output
$level3old
data1\Nissan Production Engineering.mig\TC Facility\X11M

Explanation:

-replace uses regex matching followed by a string replace (with some match references). Since \ is a regex special character, it has to be backslash escaped for a literal match. So to match \, you need \\. To match \\, you need \\\\, i.e. one \ escape for each literal \.

The initial ^ matches the beginning of the string. It is the first possible place to start matching a string.

.*? lazily matches any characters. This usually continues matching until it hits another matching condition. So .*?\\ will match any character up until and including the next \.

[^] is the negation character class. Anything following the ^ within the [] means match anything except those characters. [^\\] matches anything except \.

+ matches one or more of the previous match. [^\\]+ matches anything except \ one or more times.

In the replacement string, $1 refers to the matched characters in capture group 1. In the regex portion, () denotes a capture group. The first open ( begins capture group, which is given the name 1 unless you explicitly name it.

In the replacement string, $& refers to every matched character. $0 could also be used because an entire match defaults to capture group 0.

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