Add character in string counting from end

Question

I have a text file, where I need to check if a special character is present at a special position.

First I find out, if the character is present or not. If it is not, I want to add it at the position where it is missing. The problem is, that the number of characters in front of the character I am checking, are changing. Which means I can not simply add the character at the same position all the time using insert.

The end of the string looks always the same though. Is there a way, where I can add the character directly at the position where it is missing or insert it at a position after counting a number of characters backwards from the end of the string?

At the moment my code looks like this:

$content = Get-Content Path\to\file.txt
foreach ($line in $content) {
    if ($line[-6] -ne '\') {
        #Add character here
    }
    echo $line
}

EDIT:

This does what I want, but I think it is far away from a good solution:

$content = Get-Content path\to\file.txt
foreach ($line in $content) {
    if ($line[-25] -ne '\') {
        $new_line  = $line.Split(';')[0]
        $new_line1 = $line.Split(';')[1] + '\' | % {$_ -replace '"', ''} 
        $new_line2 = $line.Split(';')[2]
        $test = Write-Host $new_line';"'$new_line1'";'$new_line2
    }
}
echo $test

Ansgar Wiechers · Accepted Answer · 2016-05-23 16:21:26Z

2

You could do a regular expression replacement with a negative lookbehind and a positive lookahead:

$line -replace '(?<!\\)(?=.{6}$)', '\'

(?<!...) is a negative lookbehind, basically meaning "this does not exist before the match", "this" being a literal backslash here.
(?=...) is a positive lookahead, meaning "this does exist after the match", "this" being a sequence of 6 characters (.{6}) at the end of the string ($) here.

Both lookahead and lookbehind expressions don't become part of the match, so the above operation simply replaces the empty string between the negative lookbehind and the positive lookahead with a backslash (thus effectively inserting that character into the string), but only if the 6 characters at the end of the string are not already preceded by a backslash.

answered May 23, 2016 at 16:21

Ansgar Wiechers

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2 Comments

BallerNacken Over a year ago

That is exactly what I was looking for! Thx Now I only need to find out how to replace that character directly into the file.

Ansgar Wiechers Over a year ago

(Get-Content filename) -replace ... | Set-Content filename

Martin Brandl · Accepted Answer · 2016-05-23 16:17:02Z

1

This can be done with regex replace too (one-liner):

Assume your special character is a % and you always want to put it at the third last position:

"12345689" -replace '^(.*?)%?(.{2})$', '$1%$2'

Result:

1234567%89

If you already have the % in place, it won't change:

"123456%89" -replace '^(.*?)%?(.{2})$', '$1%$2'

Result:

1234567%89

Regex Demo.

edited May 23, 2016 at 16:17

answered May 23, 2016 at 16:04

Martin Brandl

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