在 grep 中，如何拉取不以空格字符开头的字符串？

Question

I am trying to pull out percentage completion in curl output. The string is given as:

96 2001G 0 0 96 1920G 0 82.2M 6:17:53 6:01:25 0:12:35 97M

My grep -oP (?<=\s)[1-9][0-9]{0,2}(?=\s) command pulls out the middle 96 instead of the 96 to the left. How can I target the first 96 instead? I tried (?<=\s*)[1-9][0-9]{0,2}(?=\s) and (?<=^\s)[1-9][0-9]{0,2}(?=\s) to no avail. Thanks.

Answer 1

The wording of your question ("pull a string that does not start with whitespace") and the regular expressions you have tried (lookbehind assertions that are looking for white space before the string) do not match up , so it is not entirely clear what you are after.

If you are looking for all percentages including those separated by white space and those that are at the very beginning of the line (no white space before but white space after) and those at the very end of the line (white space before but no white space after), then:

Try:

grep -oP '(?<!\S)[1-9][0-9]{0,2}(?!\S)'

You need to use negative lookbehind and lookahead assertions : (?<!\S) and (?!\S)

The negative lookbehind assertion states that the percentage must not be preceded by a non-white space character. This is a double-negative that either white space or the start of the line or string satisfies. Likewise the negative lookahead assertion states that the percentage must not be followed by a non-white space character, again a double negative that either white space or the end of the line or string satisfies.

See Regex Demo

But if you are only interested in the percentage at the beginning of the line then:

Try:

grep -oP '^[1-9][0-9]{0,2}(?!\S)'