string padded with optional blank with max length

I have a problem building a regex. this is a sample of the text:

text 123 12345 abc 12    def 67 i 89 o 0    t 2

The numbers are sometimes padded with blanks to the max length (3). e.g.:

• "1" can be "1" or "1 "
• "13" can be "13" or "13 "

My re开发者_高级运维gex is at the moment this:

\b([\d](\s*)){1,3}\b

The results of this regex are the following: (. = blank for better visibility)

123.
12....
67.
89.
0....
2

But I need this: (. = blank for better visibility)

123
12.
67.
89.
0..
2

How can I tell the regex engine to count the blanks into the {1,3} option?

---

Try this:

\b(?:\d[\d\s]{0,2})(?:(?<=\s)|\b)

This will also cover strings like text 123 1 23 12345 123abc 12 def 67 i 89 o 0 t 2 and results in:

123
1.
23.
12.
67.
89.
0..
2

---

Does this do what you want?

\b(\d){1,3}\s*\b

This will also include whitespace (if available) after the selection.

---

I think you want this

\b(?:\d[\d\s]{0,2})(?!\d)

See it here on Regexr

the word boundary will not work at the end, because if the end of the match is a whitespace, there is no word boundary. Therefor I use a negative lookahead (?!\d) to ensure that there is no digit following.

But if you have a string like this "1 23". It will match only the "2" and the "23", but not the whitespace after the first "2".

---

Assuming you want to use the padded numbers somewhere else, break the problem apart into two; (simple) parsing the numbers, and (simple) formatting the numbers (including padding).

while ( $text =~ /\b(\d{1,3})\b/g ) {
  printf( "%-3d\n", $1 );
}

Alternatively:

@padded_numbers = map { sprintf( "%-3d", $_ ) } ( $text =~ /\b(\d{1,3})\b/g )