正则表达式 - 匹配 [[[ 和 ]]] 之间任何东西的倍数

Question

I need to match anything between [[[ and ]]] using regex. I then need to put all the values found between the brackets into an array.

Example text:

here is some 'test text [[[media-2 large right]]], [[[image-0 large left]]] the another token [[[image-1]]

From the above text I need to match the first two:

1, [[[media-2 large right]]]
2, [[[image-0 large left]]]

but not the last as it only has two [ at the end.

Answer 1

This checks for:

1. [[[
2. Followed by:
   1. Anything but ] -or-
   2. One to two ] that are not followed by ]
3. Followed by ]]]

preg_match_all('/\[\[\[(?:(?:[^\]]*|]{1,2}(?!]))*)]]]/', $string, $matches);
print_r($matches[0]);

This regex has the benefit of matching ] inside of the triple-bracket wrapper (eg [[[foo]bar]]] .

Note: ] does not need to be escaped, except inside character classes.

Answer 2

A generic solution is this one:

\[{3}(?=.*?\]{3}(?!\]))((?:(?!\]{3}(?!\])).)*)

It reads

\[{3}         # 3 opening square brackets
(?=           # begin positive look-ahead ("followed by..."
  .*?\]{3}    #   ...3 closing brackets, anywhere ahead (*see explanation below)
  (?!\])      #   negative look-ahead: no more ] after the 3rd one
)             # end positive look-ahead
(             # begin group 1
  (?:         #   begin non-matching group (for atomic grouping)
    (?!       #     begin negative look-ahead ("not followed by"):
      \]{3}   #       ...3 closing square brackets
      (?!\])  #       negative look-ahead: no more ] after the 3rd one
    )         #     end negative look-ahead
    .         #     the next character is valid, match it
  )           #   end non-matching group
)             # end group 1 (will contain the wanted substring)

The positive look-ahead is a safeguard clause that allows the expression to fail fast when there is no "]]]" in a long input string.

Once it's established that a "]]]" will follow at some point ahead in the string, the negative look-ahead makes sure that the expression correctly matches strings like this one:

[[[foo [some text] bar]]]
                 ^
                 +-------- most of the other solutions would stop at this point

This expression checks at every character whether three ] follow or not, so in this example it would include the " bar" .

The "no more ] after the 3rd one" part of the expression makes sure that the match does not end prematurely, so that in this case:

[[[foo [some text]]]]

the match would still be "foo [some text]" .
Without it the expression would stop too early ( "foo bar [some text" ).

Side effect is that we do not need to actually match the "]]]" , since the positive look-ahead made clear that they are there. We only need to match up to them, which the negative look-ahead does nicely.

Note that you need to run the expression in "dotall" mode if your input contains newline characters.

See also: http://rubular.com/r/QFo9jHEh9d

Answer 3

更安全的解决方案：

\[{3}[^\]]+?\]{3}

Answer 4

I think this works:

\[\[\[(.*)\]\]\]

But it's probably the newb way to do it :)

Answer 5

If your string will always follow that format, subject , size , position , you can use this:

$string = "here is some 'test text [[[media-2 right]]], [[[image-0]]] the another [[[image-1 left large]]] and token [[[image-1]]";

preg_match_all('/[\[]{3}(.*?)(.*?)?(.*?)?[\]]{3}/', $string, $matches);
print_r($matches);

Answer 6

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