How do I find the newest file in a directory using a PowerShell script?

Question

If, for example, I have a directory which contains the following files:

Test-20120626-1023.txt
Test-20120626-0710.txt
Test-20120626-2202.txt
Test-20120626-19         


        

Answer 1

If the name is the equivalent creation time of the file property CreationTime, you can easily use:

$a = Dir | Sort CreationTime -Descending | Select Name -First 1

then

$a.name

contains the name file.

I think it also works like this if name always have the same format (date and time with padding 0, for example, 20120102-0001):

$a = Dir | Sort Name -Descending | Select Name -First 1

Answer 2

I use this to get last updated files. Easy to write.

ls | sort LastWriteTime

Answer 3

If you're working with a large number of files, I'd suggest using the -Filter parameter as the execution time will be greatly reduced.

Based on the example James provided, here is how it would be used;

$dir = "C:\test_code"
$filter="*.txt"
$latest = Get-ChildItem -Path $dir -Filter $filter | Sort-Object LastAccessTime -Descending | Select-Object -First 1
$latest.name

Answer 4

If, for some reason, the files creation date is different than the one stamped in the file name then you can parse the file name into a datetime object and sort by expression:

$file = Get-ChildItem | 
Sort-Object { [DateTime]::ParseExact($_.BaseName,'\Te\s\t\-yyyyMMdd\-HHmm',$null) } |
Select-Object -Last 1

Answer 5

You could try something like this:

$dir = "C:\test_code"
$latest = Get-ChildItem -Path $dir | Sort-Object LastAccessTime -Descending | Select-Object -First 1
$latest.name