Strange behavior in a Regex

I was replacing blank lines for a <p> tag. The regex replace all blacklines allowing white-spaces (\s), for one <p> tag.

For example this string:

$string="with.\n\n\n\nTherefore";

But return 2 <p> tags.

So, i've done this test:(It's not for replace

, just for test)

$string="with.\n\n\n\nTherefore";
$string=preg_replace('/(^)(\s*)($)/m','[$1]|$2|($3)',$string);
echo $string;

And check what's return:

with.
[]|

|()[]||()
Therefore

Imagining:

with.\n
^\n
\n
$^\n$\n
Therefore

The regex add one \n, and the 4th one does not do what 'she' have to do.(jump to the another line).

Someone that can h开发者_C百科elp. basically explain rather than solve the problem. Thanks evryone.

Your regex should match at least one whitespace character. So replace \s* with \s+ or if it needs escaping \s\+

\s* will match every single character, that's because it matches any(*) whitespace(\s), and because of any it includes none. By none I mean that in the string 'abc' \s* will match the "empty" characters between '^' and 'a', 'a' and 'b', 'b' and 'c', 'c' and '$'.

It is really easy to test on a linux terminal like this:

$ echo "abc" | sed 's:\s*:\n:g'  # replace \s* with \n for the whole string 

a
b
c

$ # ^ the result

As you can see it matches every "empty" character and replaces it with \n

On the other hand \s+ will force the regex to match at least 1(+) whitespace(\s) character, so it works as excpected.