PHP 正则表达式中的这种反向引用条件如何工作？

Question

My requirement is as follows:

1. If a string contains the word "cat", "cat" must be matched.
   
2. However, if the word "cat" is preceded by the word "dog", then "cat" must also be succeeded by "dog" and all 3 words must be matched.
3. This means that the string "dog cat" must not be matched, since the 2nd "dog" isn't present.

Accordingly, I have written the following regex in PHP. It contains a backreference condition:

      $ptn = '@' .                    // PHP delimiter
             '(dog\s*)?' .            // dog
             'cat\s*' .               // cat
             '(?(1)dog)' .            // backreference cond
             '@';                     // PHP delimiter

The regex meets requirement 1:

     $str1b = 'cat';
     preg_match($ptn, $str1b, $matches);
     print_r($matches);

The O/P is:

Array ( [0] => cat )

The regex also meets requirement 2:

     $str1a = 'dog cat dog';
     preg_match($ptn, $str1a, $matches);
     print_r($matches);

The O/P is:

Array ( [0] => dog cat dog [1] => dog )

However, I'd like to ask why the array contains 2 elements? Is it because the regex has 2 consuming sub-expressions?

Now about requirement 3. The following data tests it:

      $str1c = 'dog cat';
      preg_match($ptn, $str1c, $matches);
      print_r($matches);

The O/P here is:

Array ( [0] => cat )

Here, I'd like to ask:

1. Why is "cat" matched? Since it is preceded by "dog", it should have been succeeded by "dog" too, which would have resulted in a match; else no match should have occurred.

2. Is this how the regex is supposed to work?

3. How can I achieve my 3 requirements?

Here's the code .

I am considering a solution only in PHP.

Answer 1

However, if the word "cat" is preceded by the word "dog", then "cat" must also be succeeded by "dog" and all 3 words must be matched

TheFourthBird has a good answer using PCRE

Based on my interpretation this regex may also work for you:

\b(?:dog cat dog|(?<!\bdog )cat)\b

RegEx Details:

• \b : Word boundary
• (?: : Start non-capture group
  • dog cat dog : Match dog cat dog
  • | :
  • (?<!\bdog )cat : Match cat if not preceded by word dog and a space
• ) : End non-capture group
• \b : Word boundary

Answer 2

You get the 2 matches as you are using a capture group.

With the pattern that you tried (dog\s*)?cat\s*(?(1)dog) you get a match for cat in dog cat

This is because the pattern optionally matches dog. If there is dog, it is captured and then tries to match cat.

Then in the if clause is states: if we have group 1 present, match dog. What happens is that if there is no match in group 1 , it can still match cat as the capture group 1 is optional.

So in dog cat it eventually can not match dog, but the following cat it can match when the attempt starts at cat.

---

If you want to match all 3 words dog cat dog or only a single cat and you don't want to match dog cat you might use

\b(?:dog cat dog|dog cat\b(*SKIP)(*F)|cat)\b

• \b A word boundary to prevent a partial match
• (?: Non capture group
  • dog cat dog Match literally
  • | Or
  • dog cat\b(*SKIP)(*F) In case of dog cat skip the match
  • | Or
  • cat Math only cat
• ) Close non capture group
• \b A word boundary

Regex demo | Php demo

For example

$strings = [
    "cat",
    "dog cat dog",
    "dog cat",
    "cat dog",
    "this cat cat is a test dog cat dog cat"
];
$pattern = "/\b(?:dog cat dog|dog cat\b(*SKIP)(*F)|cat)\b/";
foreach ($strings as $str) {
    preg_match_all($pattern, $str, $matches);
    print_r($matches[0]);
}

Output

Array
(
    [0] => cat
)
Array
(
    [0] => dog cat dog
)
Array
(
)
Array
(
    [0] => cat
)
Array
(
    [0] => cat
    [1] => cat
    [2] => dog cat dog
    [3] => cat
)

---

An alternative approach using a capture group could be matching what you want to avoid, and capture what you want to keep. For matching spaces, you could use \s but note that it could also match a newline.

\bdog cat\b(?! dog\b)|\b(dog cat dog|cat)\b

Regex demo

If a quantifier is available in a lookbehind assertion, you might also use

\bdog cat dog\b|(?<!dog *)\bcat\b|cat(?= *dog\b)

Regex demo

Answer 3

Here's a solution that works with standard syntax .

It matches "dog cat dog" and "cat" but NOT "dog cat".

I am unable to post the regex here, since SO claims it isn't indented properly (though it is). Please check the link for the regex.