Question

I want to replace some text in every script file in folder, and I’m trying to use this PS code:

$pattern = '(FROM [a-zA-Z0-9_.]{1,100})(?<replacement_place>[a-zA-Z0-9_.]{1,7})'
Get-ChildItem -Path 'D:\Scripts' -Recurse -Include *.sql | ForEach-Object { (Get-Content $_.fullname) -replace $pattern, 'replace text' | Set-Content $_.fullname }

But I have no idea how to keep first part of expression, and just replace the second one. Any idea how can I do this? Thanks.

Answer 1

Not sure that provided regex for tables names is correct, but anyway you could replace with captures using variables $1, $2 and so on, and following syntax: 'Doe, John' -ireplace '(\w+), (\w+)', '$2 $1'

Note that the replacement pattern either needs to be in single quotes ('') or have the $ signs of the replacement group specifiers escaped ("`$2 `$1").

# may better replace with     $pattern = '(FROM) (?<replacement_place>[a-zA-Z0-9_.]{1,7})'
$pattern = '(FROM [a-zA-Z0-9_.]{1,100})(?<replacement_place>[a-zA-Z0-9_.]{1,7})'

Get-ChildItem -Path 'D:\Scripts' -Recurse -Include *.sql | % `
{
   (Get-Content $_.fullname) | % `
     { $_-replace $pattern, '$1 replace text' } | 
     Set-Content $_.fullname -Force
}

If you need to reference other variables in your replacement expression (as you may), you can use a double-quoted string and escape the capture dollars with a backtick

     { $_-replace $pattern, "`$1 replacement text with $somePoshVariable" } |