正则表达式：以“％”开头的单词

Question

I've been trying to figure this out myself but I'm afraid Regular Expressions just aren't my thing. In this sample string: "My name is %name and I live in $address." I'm trying to get the words which begin with "%" then replace them with their values in $name or $address depending on what word was found in the regex. The new string is then returned complete with the replaced values.

Should not return words like: aaa%aaa (% isn't the first character) and \\%word (the % is escaped)

This is in PHP. I'm using this since I'm grabbing the data from a *.ini file where everything is a string. I remove the $_POST example so it's not misleading.

Answer 1

This one should do fine:

(?<!\\|\w)%\w+

However it's usually a bad thing to not sanitize $_POST/$_GET before display/whatever.

Answer 2

In javascript or similar REGEX engines you could use:

\B(%\w+)\b

Edit:
In this case, the final \\b is optional, since \\w will match all chars needed and stop in any non word char as @tchrist pointed out.

Edit 2: This will not match \\%name :

(?:^|\s)(%\w+)

Answer 3

This looks like it possibly works. This align's on even escapes. -

(?<!\\\\)(?:\\\\.)*(?<![%\\w])%(\\w+)

expanded -

(?<!\\)     # Not an escape behind us
(?:\\.)*    # 0 or many esc plus anything
(?<![%\w])  # not % nor \w behind us
%(\w+)      # % then capture grp1 bunch of word chars

Answer 4

Something like this would be helpful.

$string = "Hello %name from %city ";

$name = "Joe";
$city = "Boston";
$array = array();


preg_match("/%\w+/",$string,$array);

foreach($array as $string1)
               echo "match: ".$string1;