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 $0 but I'm not sure what it means.The output I was looking for was
3 1 2 2 2 8 2 9
But it just gives me a blank line. Any suggestions?
LardPies
This probably isn't the right way to do it, but it works.
("31222829").GetEnumerator() -join " "
The .GetEnumerator method iterates over each character in the string
The -join operator will then join all of those characters with the " " space
tl;dr
PS> "31222829" -replace '[0-9]', '$& '
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.
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]
