正则表达式在前％后和后％前匹配

Question

I need a regular expression that can match what is inside the "%" marks in the following string:

1) Bla bla bla %yada yada yada% bla bla.

Even if the form is like this:

2) Bla bla bla %yada yada yada% bla bla %yada #1 yada%

I thought that /%([^%]*)%/ would do but in the second case it doesn't match the second part (%yada #1 yada%), only the first.

The expression has to match even if the substring between the "%" marks contains something like "#1". Thank you in advance

Regards

function replace($search, $replace, $subject, $link)
    {
        if (preg_match('/#[1-9]/', $search) == 1 && $link == null)
        {
            echo preg_replace($search, $replace, $subject);
        }
        if (preg_match('/%[a-zA-Z0-9# ]+%/', $search) == 1 && $link != null && $replace == null)
        {
            echo preg_replace('/%[a-zA-Z0-9# ]+%/', '<a href=$link>$1</a>', $subject, 1) . '<br>';
        }
    }

$string = 'bla bla #1 bla bla %bla bla% bla #2 %bla bla #3 bla%';
$newString = replace('/%[a-zA-Z0-9# ]+%/', null, $string, 'www.google.com');
echo $newString;
preg_match('/%[a-zA-Z0-9# ]+%/', $newString)); // 0 = no match

So on this second call I don't get a match. Hope this helps.

Answer 1

See your calls to preg_match ... The second argument should be $subject not $search ...

Also, use return to actually return something, not echo .

Answer 2

%.*%

This will match

1) a %

2) any number of any characters

3) a %

And since it will match greedily rather than lazily, it will match the widest %...% it can, rather than stop at the first.

For example, the lazy version of this would be %.*?% (putting ? after a * makes it lazy) and it would close after the first, not last %.

Answer 3

/%.+?%/ should do the trick.
See greedyness of regex expressions.
[^%] stops at the first % . But you want to include all characters between the two % s.

Answer 4

Try this expression

%[a-zA-Z0-9# ]*%

Hope it helps