Question

我按以下格式构建了文档：

123456789|XXX|1234567|05/05/2012 00:00|81900153|Signed|LASTNAME,FIRSTNAME, M.S.|024813|XXX|3410080|DNR Order Verification:Scanned|

xyz pqs 123

[report_end]

123456789|XXX|1234567|05/05/2012 00:00|81900153|Signed|LASTNAME,FIRSTNAME, M.S.|024813|XXX|3410080|A Note|

xyz pqs 123

[report_end]

每条记录的位置：

以|
有一个介入的自由文本块
以标记＆＃34; [report_end]＆＃34;

如何使用正则表达式捕获这三个元素？

我的方法是

搜索每个有11 |的行字符;
搜索[report_end]的每一行;
搜索这两行之间的内容。

但我不知道如何用正则表达式来实现这一目标。

Answer 1

您可以使用以下内容：

r"((?:.*?\|){11}\s+(?:.*)\s+\[report_end\])"

<强>输出：

Match 1.    [0-157] `123456789|XXX|1234567|05/05/2012 00:00|81900153|Signed|LASTNAME,FIRSTNAME, M.S.|024813|XXX|3410080|DNR Order Verification:Scanned|

xyz pqs 123

[report_end]


Match 2.    [159-292]   `123456789|XXX|1234567|05/05/2012 00:00|81900153|Signed|LASTNAME,FIRSTNAME, M.S.|024813|XXX|3410080|A Note|

xyz pqs 123

[report_end]

DEMO

https://regex101.com/r/xY5nI9/1

正则表达式解释

((?:.*?\|){11}\s+(?:.*)\s+\[report_end\])

Options: Case sensitive; Exact spacing; Dot doesn’t match line breaks; ^$ don’t match at line breaks; Regex syntax only

Match the regex below and capture its match into backreference number 1 «((?:.*?\|){11}\s+(?:.*)\s+\[report_end\])»
   Match the regular expression below «(?:.*?\|){11}»
      Exactly 11 times «{11}»
      Match any single character that is NOT a line break character «.*?»
         Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
      Match the character “|” literally «\|»
   Match a single character that is a “whitespace character” «\s+»
      Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
   Match the regular expression below «(?:.*)»
      Match any single character that is NOT a line break character «.*»
         Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
   Match a single character that is a “whitespace character” «\s+»
      Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
   Match the character “[” literally «\[»
   Match the character string “report_end” literally «report_end»
   Match the character “]” literally «\]»

根据您的评论进行更新

要获得3组，您可以使用：

r"((?:.*?\|){11})\s+(.*)\s+(\[report_end\])

循环所有群组：

import re
pattern = re.compile(r"((?:.*?\|){11})\s+(.*)\s+(\[report_end\])")

for (match1, match2, match3) in re.findall(pattern, string):
    print match1 +"\n"+ match2 +"\n"+ match3 +"\n"

现场演示

http://ideone.com/k8sA3k

Answer 2

您也可以尝试：

^(?P<fields>(?:[^|]+\|){11})(?P<text>[\s\S]+?)(?P<end>\[report_end\])

DEMO

使用正则表达式

2 个答案:

https://regex101.com/r/xY5nI9/1