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Get actual path from path using template

When used Test-Path

in a statement if

, I am looking to get the path at which the if statement is executed.

For example, these files exist in C:

C:\Test6_1_15.txt
C:\Test6_2_15.txt
C:\Test6_3_15.txt
C:\Test6_4_15.txt

what should I do in the "then" branch?

$Path = "C:\Test6_*_15.txt"
if (Test-Path $Path)
{
   # if test passes because there are 4 files that fit the test, but I want to be
   # able to output the file that made the if statement succeed. 
}
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3 answers


It sounds like you want Resolve-Path

:



if(($Paths = @(Resolve-Path "C:\Test6_*_15.txt"))){
    foreach($file in $Paths){
        # do stuff
    }
} else {
    # Resolve-Path was unable to resolve "C:\Test6_*_15.txt" to anything
}
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You can do get-item $path

that which will return the actual name of the file (s) in its result.



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You won't get it with Test-Path

. Test-Path

returns boolean value (s) representing the presence of the traversed path (s). Looking at the description from TechNet

It returns TRUE ($ true) if all elements exist and FALSE ($ false) if they are absent

If you only want the filenames that match, then use Get-Item

as it supports standard wildcards. You can get information from objects System.IO.FileInfo

returned Get-Item

.

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