Use a function in Powershell replace

Question

I'm trying to replace part of a string in Powershell. However, the replacement string is not hardcoded, it's calculated from a function:

$text = "the image is -12345-"
$text = $text -replace "-(\d*)-", 'This is the image: $1'
Write-Host $text

This gives me the correct result: "This is the image: 12345"

Now, I want to include the base64 encoded image. I can read the image from the id. I was hoping the following would work, but it doesn't:

function Get-Base64($path)
{
    [convert]::ToBase64String((get-content $path -encoding byte))
}
$text -replace "-(\d*)-", "This is the image: $(Get-Base64 '$1')"

The reason that it doesn't work, is because it first passes $1 (the string, not the value of $1) to the function, executes it and only then does it do the replace. What I want to do is

Find the occurrence of the pattern
Replace each occurence with the pattern
For each replace:
Pass the capture group to the function
Use the value of the capture group to get the base64 image
inject the base64 image into the replacement

Ansgar Wiechers · Accepted Answer · 2015-06-05 14:32:39Z

28

You can use the static Replace method from the [regex] class:

[regex]::Replace($text,'-(\d*)-',{param($match) "This is the image: $(Get-Base64 $match.Groups[1].Value)"})

Alternatively you can define a regex object and use the Replace method of that object:

$re = [regex]'-(\d*)-'
$re.Replace($text, {param($match) "This is the image: $(Get-Base64 $match.Groups[1].Value)"})

For better readability you could define the callback function (the scriptblock) in a separate variable and use that in the replacement:

$callback = {
  param($match)
  'This is the image: ' + (Get-Base64 $match.Groups[1].Value)
}

$re = [regex]'-(\d*)-'
$re.Replace($text, $callback)

edited Jun 5, 2015 at 14:32

Ansgar Wiechers

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answered Jun 5, 2015 at 12:13

user4003407

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1 Comment

Michael Sorens Over a year ago

For those looking to understand the derivation of this, it is using a sigature of the Replace method ( Regex.Replace Method (String, MatchEvaluator) ) that adds even more power to regexs by allowing computation on a matched argument. The neat thing is--which I did not realize until I saw this answer--is that a PowerShell script block is apparently compatible with a MatchEvaluator argument!

mklement0 · Accepted Answer · 2025-09-23 21:31:45Z

16

Note: As for why your own, string-based attempt didn't work: see the explanation in the bottom section of this answer.

PetSerAl's helpful answer, which relies on a direct call to a .NET API, is your only option in Windows PowerShell, the legacy, ships-with-Windows, Windows-only edition of PowerShell whose latest and last version is 5.1.

PowerShell (Core) 7, its modern, cross-platform, install-on-demand successor, now offers a native PowerShell solution via an enhancement to the -replace operator, which obviates the need to call [regex]::Replace():

Just as with [regex]::Replace(), you can now:

pass a script block as the -replace replacement operand, which must return the replacement string,
however, unlike with [regex]::Replace(), the match at hand (an instance of type [System.Text.RegularExpressions.Match]) is reflected in the automatic $_ variable, as is customary in PowerShell.

Applied to your case:

# PowerShell 7 only
$text -replace '-(\d*)-', { "This is the image: $(Get-Base64 $_.Groups[1].Value)" }

A simpler, self-contained example:

# PowerShell 7 only
# Increment the number embedded in a string:
# -> '43 years old'
'42 years old' -replace '\d+', { [int] $_.Value + 1 }

edited Sep 23 at 21:31

answered Aug 24, 2018 at 22:46

mklement0

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3 Comments

felixfbecker Over a year ago

This should be the accepted answer

Ωmega Over a year ago

PS> '42 years old' -replace '\d+', { [int] $_.Value + 1 } returns [int] 42 years old.Value + 1 years old

mklement0 Over a year ago

@Ωmega, that's what happens in Windows PowerShell, where using script blocks as substitution operand isn't supported (the script block is simply stringified there, meaning that its verbatim contents (sans { and }) are used. As the answer notes, you need PowerShell (Core), version 6.1 or above (current is v7.3.8).

js2010 · Accepted Answer · 2020-01-30 14:58:30Z

3

Here's another way. Use the -match operator, and then reference $matches. Note that $matches doesn't get set with arrays on the left side of the -match operator. $matches.1 is the first grouping formed by the ( ).

$text = "the image is -12345-"
function Get-Base64($path) { 
  [convert]::ToBase64String( (get-content $path -asbytestream) ) }  # ps 6 ver
if (! (test-path 12345)) { echo hi > 12345 }
$text -match '-(\d*)-'
$text -replace '-(\d*)-', "$(Get-Base64 $matches.1)"

the image is aGkNCg==

Or to break it up even more:

$text -match '-(\d*)-'
$result = Get-Base64 $matches.1
$text -replace '-(\d*)-', $result

edited Jan 30, 2020 at 14:58

answered Jan 29, 2020 at 20:38

js2010

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