Leftpad doesn't work when using with regex in Powershell - powershell

Question

Here are two code blocks which show the strange behavior that Leftpad does.
$array = "z", "fg"
$array -replace "(\w{1,2})", '$1'.PadLeft(2, "0")
# output: z, fg
$array = "z", "fg"
$array -replace "(\w{1,2})", '$1'.PadLeft(3, "0")
# output: 0z, 0fg
How comes the pad length not fixed?
EDIT: #LotPings
I ask this another question is because of your way of doing it when applied to rename-item statement will not affect files with brackets in its names.
$file_names = ls "G:\Donwloads\Srt Downloads\15" -Filter *.txt
# the files are: "Data Types _ C# _ Tutorial 5.txt", "If Statements (con't) _ C# _ Tutorial 15.txt"
$file_names |
ForEach{
if ($_.basename -match '^([^_]+)_[^\d]+(\d{1,2})$')
{
$file_names |
Rename-Item -NewName `
{ ($_.BaseName -replace $matches[0], ("{0:D2}. {1}" -f [int]$matches[2],$matches[1])) + $_.Extension }
}
}
# output: 05. Data Types.txt
# If Statements (con't) _ C# _ Tutorial 15.txt
As for the .PadLeft, I thought that the regex replacement group is of string type, it should work with .PadLeft but it doesn't.

Answer 1

Ansgars comment to your last question should have shown that your assumption on the order of actions was wrong.
And Lee_Dailey proved you wrong another time.
My answer to your previous question presented an alternative which also works here:
("Data Types _ C# _ Tutorial 5", "If Statements (con't) _ C# _ Tutorial 15") |
ForEach{
if ($_ -match '^([^_]+)_[^\d]+(\d{1,2})$'){
"{0:D2}. {1}" -f [int]$matches[2],$matches[1]
}
}
Sample output:
05. Data Types
15. If Statements (con't)
The last edit to your question is in fact a new question...
Rename-Item accepts piped input, so no ForEach-Object neccessary when also using
Where-Object with the -match operator to replace the if.
the $Matches collection is supplied the same way.
I really don't know why you insist on using the -replace operator when building the NewName from scratch with the -format operator.
$file_names = Get-ChildItem "G:\Donwloads\Srt Downloads\15" -Filter *.txt
$file_names | Where-Object BaseName -match '^([^_]+)_[^\d]+(\d{1,2})$' |
Rename-Item -NewName {"{0:D2}. {1}{2}" -f [int]$matches[2],$matches[1].Trim(),$_.Extension} -WhatIf

Answer 2

Several days after asking this question, I happen to figure out the problem.
The $number capture group references in -replace syntax are merely literal strings!
Powershell never treats them as anything special, but the Regex engine does. Look at the example below:
$array = "z", "fg"
$array -replace "(\w{1,2})", '$1'.Length
#output: 2
# 2
Looks strange? How comes the $1 capture group has both Lengths of 2 with "z" and "fg"? The answer is that the length being calculated is the string $1 rather than "z","fg"!
Let's look at another example, this time lets replace a letter within the capture group, see what happens:
$array -replace "(\w{1,2})", '$1'.Replace("z", "6")
#output: z
# fg
The output shows the .replace didn't apply to the capture group 1.
$array -replace "(\w{1,2})", '$1'.Replace("1", "6")
#output: $6
# $6
See? The string being replaced is $1 itself.
Now the cause of the .padleft problem should be understood. PS pad the literal string $1 and show the result with the content of the group.
When I pad it with .Padleft(2, "0"), nothing happens because the "$1" itself has the length of 2.
$array -replace "(\w{1,2})", '$1'.PadLeft(2, "0")
# output: z
# fg
If instead, I pad it with .Padleft(3, "0"), this time the pad method does take effect, it applies the extra "0" to $1 but shows the result with the "0" preceding the content of $1.
$array -replace "(\w{1,2})", '$1'.PadLeft(3, "0")
#output: 0z
# 0fg