php - 多行负向超前

Question

我对regex不太满意（我花了几个小时），而且我很难替换2个标识符（“ {|”和“ |}”）之间的所有空行

我的正则表达式看起来像这样（对不起，您的眼睛）：(\{\|)((?:(?!\|\}).)+)(?:\n\n)((?:(?!\|\}).)+)(\|\})


(\{\|)：字符“ {|”
((?:(?!\|\}).)+)：如果不是在“ |}”之后的所有内容（负向超前）
(?:\n\n)：我要删除的空行
((?:(?!\|\}).)+)：如果不是在“ |}”之后的所有内容（负向超前）
(\|\})：字符“ |}”


Demo

它可以工作，但是只删除最后一个空行，您能帮助我使它与所有空行一起工作吗？

我尝试在\ n \ n上添加一个否定的前瞻性，并在所有内容上重复一个组，但没有成功。

Answer 1

几种方法：

基于\G的模式：（仅需要一个模式）

$txt = preg_replace('~ (?: \G (?!\A) | \Q{|\E ) [^|\n]*+ (?s: (?! \Q|}\E | \n\n) . [^|\n]*)*+ \n \K \n+ ~x', '', $txt);

\G匹配字符串的开头或最后一次成功匹配之后字符串中的位置。这样可以确保多个匹配是连续的。

我所说的基于 \G的模式可以像这样被模式化：

(?: \G position after a successful match | first match beginning ) reach the target \K target

“到达目标”部分设计为永远不会匹配结束序列 |}。因此，一旦找到最后一个目标， \G部分将失败，直到第一个匹配部分再次成功。

~ 
### The beginning
(?:
    \G (?!\A) # contigous to a successful match
  |
    \Q{|\E # opening sequence
           #; note that you can add `[^{]* (*SKIP)` before to quickly avoid 
           #; all failing positions

           #; note that if you want to check that the opening sequence is followed by 
           #; a closing sequence (without an other opening sequence), you can do it
           #; here using a lookahead
)

### lets reach the target
#; note that all this part can also be written like that `(?s:(?!\|}|\n\n).)*`
#; or `(?s:[^|\n]|(?!\|}|\n\n).)*`, but I choosed the unrolled pattern that is
#; more efficient.

[^|\n]*+ # all that isn't a pipe or a newline

# eventually a character that isn't the start of |} or \n\n
(?s:   
    (?! \Q|}\E | \n\n ) # negative lookahead
    . # the character
    [^|\n]*
)*+
#; adding a `(*SKIP)` here can also be usefull if there's no more empty lines
#; until the closing sequence

### The target

\n \K \n+ # the \K is a conveniant way to define the start of the returned match
          # result, this way, only \n+ is replaced (with nothing)
~x

或 preg_replace_callback ：（更简单）

$txt = preg_replace_callback('~\Q{|\E .*? \Q|}\E~sx', function ($m) {
    return preg_replace('~\n+~', "\n", $m[0]);
}, $txt);

demos