Question

我正在解析文本但是当空格丢失时我无法获得一块（这是正常的）
修改：我在自由文本中添加了冒号编辑好吧，这是一种可以写入键值对的任意文本格式。丢弃元素[0]，数组上的其余元素产生一系列键值。它接受多行值。

这是测试用例文本：

:part1  only one \s removed:OK
:part2 :text :with
new lines
on it
:noSpaceAfterThis
:thisShoudBeAStandAlongText but: here there are more text
:part4 :even more text

这就是我想要的：

Array
(
    [0] => 
    [1] => part1
    [2] =>  only one \s removed:OK
    [3] => part2
    [4] => :text :with
new lines
on it
    [5] => noSpaceAfterThis
    [6] => 
    [7] => thisShoudBeAStandAlongText
    [8] => but: here there are more text
    [9] => part4
    [10] => :even more text
)

这就是我得到的：

Array
(
    [0] => 
    [1] => part1
    [2] =>  only one \s removed:OK
    [3] => part2
    [4] => :text :with
new lines
on it
    [5] => noSpaceAfterThis
    [6] => :thisShoudBeAStandAlongText but: here there are more text
    [7] => part4
    [8] => :even more text
)

这是我的测试代码：

<?php
$text = '
:part1  only one \s removed:OK
:part2 :text :with
new lines
on it
:noSpaceAfterThis
:thisShoudBeAStandAlongText but: here there are more text
:part4 :even more text';

echo '<pre>';
// my effort so far:
$ret = preg_split('|\r?\n:([\w\d]+)(?:\r?\s)?|i', $text, -1, PREG_SPLIT_DELIM_CAPTURE);
print_r($ret);

// nor this one:
$ret = preg_split('|\r?\n:([\w\d]+)\r?\s?|i', $text, -1, PREG_SPLIT_DELIM_CAPTURE);
print_r($ret);

// for debuging, an extra capturing group
$ret = preg_split('|\r?\n:([\w\d]+)(\r?\s)?|i', $text, -1, PREG_SPLIT_DELIM_CAPTURE);
var_dump($ret);

Answer 1

使用preg_match_all的另一种方法：

$pattern = '~(?<=^:|\n:)\S++|(?<=\s)(?:[^:]+?|(?<!\n):)+?(?= *+(?>\n:|$))~';
preg_match_all($pattern, $text, $matches);
echo '<pre>' . print_r($matches[0], true);

模式细节：

# capture all the first word at line begining preceded by a colon #
(?<=^:|\n:)       # lookbehind, preceded by the begining of the string
                  # and a colon or a newline and a colon
\S++              # all that is not a space

# capture all the content until the next line with : at first position #
(?<=\s)           # lookbehind, preceded by a space
(?:               # open a non capturing group
   [^:]+?         # all character that is not a colon, one or more times (lazy)
  |               # OR
   (?<!^|\n):     # negative lookbehind, a colon not preceded by a newline
                  # or the begining of the string
)+?               # close the non capturing group, 
                  #repeat one or more times (lazy)
(?= *+(?>\n:|$))  # lookahead, followed by spaces (zero or more) and a newline 
                  # with colon at first position or the end of the string

这里的优点是避免空洞结果。

或使用preg_split：

$res = preg_split('~(?:\s*\n|^):(\S++)(?: )?~', $text, -1, PREG_SPLIT_DELIM_CAPTURE);

解释：

目标是在两种情况下分割文本：

:

在换行符上当行开始:
在该行的第一个空格处

因此，在一条线的开始处有两个分裂点:word。必须删除:和后面的空格，但必须保留该单词。这就是我使用PREG_SPLIT_DELIM_CAPTURE来保持这个词的原因。

模式细节：

(?:           # non capturing group (all inside will be removed)
   \s*\n      # trim the spaces of the precedent line and the newline
  |           # OR
   ^          # it is the begining of the string
)             # end of the non capturing group
:             # remove the first character when it is a :
(\S++)        # keep the first word with DELIM_CAPTURE
(?: )?        # remove the first space if present

使用正则表达式将字符串拆分为数组以获取键值对

1 个答案: