Question

我有这个例子：

This is a simple test text.
Yet another line.
START: This is the part that
 needs match.
This part does not need
 capture.
Wherever else text.

我想匹配这部分：

START: This is the part that
     needs capture.

关键是我知道START:在那里，它以一条新行结束，除了后面有空格。

我从：START: (.*?)

开始尝试了很多组合

我和\ r \ n以及任何我想到的只有在没有空格的情况下才能匹配。

我不是一个小伙子，因为我很懒。我问了几个小时。

Answer 1

这个怎么样：

preg_match(
    '/^         # Start of line
    START:\     # Match "START: "
    .*          # Match any characters except newline
    \r?\n       # Match newline
    (?:         # Try to match...
     ^          # from the start of the line:
     \ +        #  - one or more spaces
     .*         #  - any characters except newline
     \r?\n      #  - newline
    )*          # Repeat as needed/mx', 
    $subject)

这假设所有行都是换行符。

Answer 2

此代码可以与您的示例测试一起正常运行。

解决方法是在preg_match之前替换新行的标记（在！之后恢复！）和正则表达式结束时的Ungreedy修饰符（U）

<?php

$token = '#####';

$text = <<<TXT
This is a simple test text.
Yet another line.
START: This is the part that
 needs match.
This part does not need
 capture.
Wherever else text.
TXT;

$text = str_replace("\n", $token, $text);

if (preg_match('/(?P<match>START:(.)*)(' . $token . '){1}[^ ]+/Uu', $text, $matches))
{
    $match = str_replace($token, "\n", $matches['match']);
    var_dump($match);
}

$text = str_replace($token, "\n", $text);

输出将是：

string(42) "START: This is the part that
 needs match."

正则表达式：匹配所有内容，直到新行在其后没有空格

2 个答案: