正则表达式，用于匹配封装在％中的令牌

Question

I have user-entered text with potentially mistyped "tokens" I'm trying to find using PHP .

A valid "token" is any number of word characters wrapped in percent signs - so %blah% %blah_moreblah% . Basically I'm looking for tokens where the user may have forgotten to put a leading or trailing '%'. I'm also looking for tokens in the valid format - as at this point in my code, all replaceable tokens have already been replaced.

So, the 3 situations I'm looking for are (to borrow regex syntax): %\\w+ , %\\w+% , or \\w+% .

In English, what I'm looking for is, "a string that starts with a % and/or ends with a % and contains only word characters'

The regex I have this far is: (%*\\w+%*) , but you'll notice it matches every single word. I'm stuck on making a match require at least a leading or a trailing %.

Edit : Initially I tried to have all 3 situations found with their own regex. However, I was finding that the regex for finding tokens in the first situation would also find tokens in the second situation, just without the trailing %. For example, /(%\\w+)/ , when checked against %before %both% , would match %before and %both .

Answer 1

To match tokens enclosed with % , or having % on either side, use

(?=\w*%)%*\w+%*

See another regex demo .

This is your pattern that I added a positive lookahead to. The (?=\\w*%) restricts to only such matches where a % appears after a zero or more occurrences of word characters.

Note also that %* will match zero or more percent signs, it may match %%%word%% . If it is not what you need, and if you need to match 1 or 0 % s, just replace the * with ? quantifier.

Answer 2

Try this:

$input_lines = "Hello this is a %string% with %some_words in it just for demo% purposes.";

preg_match_all("/\s[\w_\-]+%\.?|%[\w_\-]+(%|\s|\.)/", $input_lines, $output_array);

That will output this:

array(
    0   =>  %string%
    1   =>  %some_words 
    2   =>   demo%
)

Note that this will catch the valid cases, as well as the typos you are looking for.