How to replace dates in text file with Powershell using regular expressions

Question

I've done much searching on this and haven't worked out the answer, but I feel like I am close!

I have dates in a text file in the following format: 18/06/2012 23:00:43 (dd/mm/yyyy HH:MM:SS) which I want to convert to: 2012-18-06 23:00:43 (yyyy-dd-mm HH:MM:SS) using Powershell.

To perform the conversion in a text editor using regular expressions I would do this:

Find: ([0-9]+)/+([0-9]+)/+([0-9]+)

Replace with: \3-\2-\1

So I have tried using this same logic in a the following Powershell script:

(Get-Content C:\script\test.txt) | 
Foreach-Object {$_ -replace "([0-9]+)/+([0-9]+)/+([0-9]+)", "(\3-\2-\1)"} | 
Set-Content C:\script\test.txt

but that results in the following undesired change:

\3-\2-\1 23:00:43

Can anybody help me nail this?

Many thanks in advance!

Answer 1

This is what you want:

(Get-Content C:\script\test.txt) | 
Foreach-Object {$_ -replace "([0-9]+)/+([0-9]+)/+([0-9]+)", '$3-$2-$1'} | 
Set-Content C:\script\test.txt

Capturing group references are done with the $ symbol, not a backslash. Also, to reference a captured group by number, you must use single quotes around the replacement string; otherwise, PowerShell will interpret any $ symbol as a reference to a previously defined variable, which in this case will result in the string "--" since no such variables exist.

Answer 2

The -replace operator supports the same replacement text placeholders as the Regex.Replace() function in .NET. E.g. $& is the overall regex match, $1 is the text matched by the first capturing group, and ${name} is the text matched by the named group "name".

Instead of "(\3-\2-\1)" use '($3-$2-$1)'

'06/18/2012 23:00:43' -replace "(\d+)/(\d+)/(\d+)", '($3-$2-$1)'