问题
How do I optionally match the start ^ or end $ of a line in a regular expression.
For example:
/(?<=[\s\^])/ does not match starts with space character or start of line.
As requested my problem was in PHP matching the following.
$str = '**bold** **bold** **bold**';
echo preg_replace('/(?<=\s|^)\*\*(.+?)\*\*(?=\s|$)/', '<strong>\\1</strong>', $str);
My edge cases were the bold at the start and end of the string were not being matched. Some edge cases I came across with other variants were matching within strings, matching chains of asterisks, and countless other problems.
echo preg_replace('/(?<=^|\s|\>)[\*]{2,}(?=[^\s\*])(.+?)(?<=[^\s\*])[\*]{2,}(?=\s|\<|$)/', '<strong>\\1</strong>', $str);
回答1:
If you insist on matching after a whitespace character or the start of the line use an alternation (not the character class). Assuming your regex flavour supports alternations in lookbehind assertions.
/(?<=\s|^)\w+/m
but probably you are looking for a word boundary \b? They match on the start or end of a word. (Exactly on the change from a word character to a non word character or the other way round)
You should give us more information: the language, the input and the expected result (including corner cases).
回答2:
Just remove the anchors. Without the anchors, the regex will be matched anywhere in the string you are searching.
来源：https://stackoverflow.com/questions/15734322/regular-expression-optional-match-start-end-of-line