CODEWITHSUNDEEP
powershell
LardPies
LardPies

Reputation: 95

How to replace character in Powershell with initial character + a space

Hopefully this question isn't already answered on the site. I want to replace every number in a string with that number and a space. So here's what I have right now:

"31222829" -replace ("[0-9]","$0 ")

The [0-9] looks for any numbers, and replaces it with that character and the space. However, it doesn't work. I saw from another website to use $0 but I'm not sure what it means.The output I was looking for was

3 1 2 2 2 8 2 9

But it just gives me a blank line. Any suggestions?

LardPies

Upvotes: 1

Views: 743

Answers (3)

favo
favo

Reputation: 69

You can use a regex with positive lookahead to avoid the trailing space. Lookahead and lookbehind are zero-length assertions similar to ^ and $ that match the start/end of a line. The regex \d(?=.) will match a digit when followed by another character.

PS> '123' -replace '\d(?=.)', '$0 '
1 2 3

To verify there's no trailing space:

PS> "[$('123' -replace '\d(?=.)', '$0 ')]"
[1 2 3]

Upvotes: 0

mklement0
mklement0

Reputation: 437618

tl;dr

PS> "31222829" -replace '[0-9]', '$& '
3 1 2 2 2 8 2 9

Note that the output string has a trailing space, given that each digit in the input ([0-9]) is replaced by itself ($&) followed by a space.

As for what you tried:

"31222829" -replace ("[0-9]","$0 ")

  • While enclosing the two RHS operands in (...) doesn't impede functionality, it's not really helpful to conceive of them as an array - just enumerate them with ,, don't enclose them in (...).

  • Generally, use '...' rather than "..." for the RHS operands (the regex to match and the replacement operand), so as to prevent confusion between PowerShell's string expansion (interpolation) and what the -replace operator ultimately sees.

    • Case in point: Due to use of "..." in the replacement operand, PowerShell's string interpolation would actually expand $0 as a variable up front, which in the absence of a variable expands to the empty string - that is why you ultimately saw a blank string.

    • Even if you had used '...', however, $0 has no special meaning in the replacement operand; instead, you must use $& to represent the matched string, as explained in this answer.

To unconditionally separate ALL characters with spaces:

Drew's helpful answer definitely works.
Here's a more PowerShell-idiomatic alternative:

PS> [char[]] '31222829' -join ' '
3 1 2 2 2 8 2 9

Casting a string to [char[]] returns its characters as an array, which -join then joins with a space as the separator.

Note: Since -join only places the specified separator (' ') between elements, the resulting string does not have a trailing space.

Upvotes: 2

Drew
Drew

Reputation: 4020

This probably isn't the right way to do it, but it works.

("31222829").GetEnumerator() -join " "

The .GetEnumerator method iterates over each character in the string The -join operator will then join all of those characters with the " " space

Upvotes: 2

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