preg_replace与单词之间的两个空格不匹配

问题描述：

i need to format uppercase words to bold but it doesn't work if the word contains two spaces

is there any way to make regex match only with words which end with colon?

$str = "BAKA NO TEST: hey";    
$str = preg_replace('~[A-Z]{4,}\s[A-Z]\s{2,}(?:\s[A-Z]{4,})?:?~', '<b>$0</b>', $str);

output: <b>BAKA NO TEST:</b> hey

but it returns <b>BAKA</b> NO TEST: hey

the original $str is a multiline text so there are many lowercase and uppercase words but i need to change only some

答

You can do it like this:

$txt = preg_replace('~[A-Z]+(?:\s[A-Z]+)*:~', '<b>$0</b>', $txt);

Explanations:

[A-Z]+      # uppercase letter one or more times
(?:         # open a non capturing group
    \s      # a white character (space, tab, newline,...)
    [A-Z]+  #
)*          # close the group and repeat it zero or more times

If you want a more tolerant pattern you can replace \s by \s+ to allow more than one space between each words.

答

Unless you have some good reason to use that regexp, try something simpler, like:

/([A-Z\s]+):/

Also, just so you know, you can use asterisk to specify none or more space characters: \s*

preg_replace与单词之间的两个空格不匹配

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