从 HTML 字符串中提取两个字符串之间的字符串

Question

我有以下 python 代码，并在尝试打印 user 和 its number 时尝试regex 我做了以下操作：

import re


txt = '''Element.update("to_users2", "\n\n\n<div class=\"label-field-pair\">\n  <div class=\"label-field-pair11\">\n    <label for=\"student_grade\">Select member</label>\n    <div class =\"scrolable\" >\n      <div class=\"scroll-inside\">\n        <div class=\"hover\"><a href=\"#\" class=\"all\" onClick=\"add_all_recipient('0000000,1111111,2222222,3333333,4444444,5555555,6666666,7777777,8888888,9999999')\">Select All  <span> Add </span></a>\n\n        </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(0000000)\" success=\"Element.hide('loader')\">user zero M ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(1111111)\" success=\"Element.hide('loader')\">user One S ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(2222222)\" success=\"Element.hide('loader')\">user Two A ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(3333333)\" success=\"Element.hide('loader')\">user three H ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(4444444)\" success=\"Element.hide('loader')\">user four M ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(5555555)\" success=\"Element.hide('loader')\">user Five O ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(6666666)\" success=\"Element.hide('loader')\">user six F ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(7777777)\" success=\"Element.hide('loader')\">user Seven Mo ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(8888888)\" success=\"Element.hide('loader')\">user eight ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(9999999)\" success=\"Element.hide('loader')\">\u0650user nine M ...<span> Add </span></a>\n\n          </div>\n        \n      </div>\n    </div>\n  </div>\n</div>\n\n\n");'''


regexp = re.findall(
            r"add_recipient\(([0-9]+)\)\" success=.+>([a-zA-Z0-9\w]+) ", txt)

for x in regexp:
    print(x[1],  x[0])

执行上面的python code会打印如下：

user 0000000
user 1111111
User 2222222
user 3333333
user 4444444
user 5555555
user 6666666
user 7777777
user 8888888

我需要得到如下输出：

user Zero 0000000
user One 1111111
...

我怎样才能得到这样的输出？在某些情况下，re.findall 只返回user 8888888，我不知道为什么。但是我怎样才能得到完整的匹配呢？

Answer 1

使用正则表达式解析 XML/HTML 是不好的做法，使用解析器（借助一些正则表达式帮助）：

from bs4 import BeautifulSoup
import re

soup = BeautifulSoup(txt)

out = []
for e in soup.find_all('a', onclick=True):
    m = re.search('(?<=add_recipient\().*(?=\))', e['onclick'])
    if m:
        a = m.group()
        out.append((e.contents[0], a))

输出：

[('user zero M ...', '0000000'),
 ('user One S ...', '1111111'),
 ('user Two A ...', '2222222'),
 ('user three H ...', '3333333'),
 ('user four M ...', '4444444'),
 ('user Five O ...', '5555555'),
 ('user six F ...', '6666666'),
 ('user Seven Mo ...', '7777777'),
 ('user eight ...', '8888888'),
 ('ِuser nine M ...', '9999999')]

替代输出（仅名称的前 2 个单词），将最后一行替换为：

out.append((' '.join(e.contents[0].split(maxsplit=2)[:2]), a))

输出：

[('user zero', '0000000'),
 ('user One', '1111111'),
 ('user Two', '2222222'),
 ('user three', '3333333'),
 ('user four', '4444444'),
 ('user Five', '5555555'),
 ('user six', '6666666'),
 ('user Seven', '7777777'),
 ('user eight', '8888888'),
 ('ِuser nine', '9999999')]

Answer 2

您可以添加额外的捕获组，并更改打印组值的顺序。

请注意，您可以将[a-zA-Z0-9\w]+ 写为\w+，因为它也匹配a-zA-Z0-9。

您可以使用[^<>]*> 代替.+> 来防止某些回溯，而不是使用否定字符类跨越尖括号。

import re

txt = '''Element.update("to_users2", "\n\n\n<div class=\"label-field-pair\">\n  <div class=\"label-field-pair11\">\n    <label for=\"student_grade\">Select member</label>\n    <div class =\"scrolable\" >\n      <div class=\"scroll-inside\">\n        <div class=\"hover\"><a href=\"#\" class=\"all\" onClick=\"add_all_recipient('0000000,1111111,2222222,3333333,4444444,5555555,6666666,7777777,8888888,9999999')\">Select All  <span> Add </span></a>\n\n        </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(0000000)\" success=\"Element.hide('loader')\">user zero M ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(1111111)\" success=\"Element.hide('loader')\">user One S ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(2222222)\" success=\"Element.hide('loader')\">user Two A ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(3333333)\" success=\"Element.hide('loader')\">user three H ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(4444444)\" success=\"Element.hide('loader')\">user four M ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(5555555)\" success=\"Element.hide('loader')\">user Five O ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(6666666)\" success=\"Element.hide('loader')\">user six F ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(7777777)\" success=\"Element.hide('loader')\">user Seven Mo ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(8888888)\" success=\"Element.hide('loader')\">user eight ...<span> Add </span></a>\n\n          </div>\n        \n          \n          <div class=\"hover\"><a href=\"#\" before=\"Element.show('loader')\" class=\"individual\" onClick=\"add_recipient(9999999)\" success=\"Element.hide('loader')\">\u0650user nine M ...<span> Add </span></a>\n\n          </div>\n        \n      </div>\n    </div>\n  </div>\n</div>\n\n\n");'''

for x in re.findall(r"add_recipient\(([0-9]+)\)\" success=[^<>]*>(\w+) (\w+)", txt):
    print(x[1], x[2], x[0])

输出

user zero 0000000
user One 1111111
user Two 2222222
user three 3333333
user four 4444444
user Five 5555555
user six 6666666
user Seven 7777777
user eight 8888888

Answer 3

我不是正则表达式专家

你可以试试：

out = re.findall(r"add_recipient\(([0-9]+)\)\" success=.+>(\w+\s+\w+)", txt)
print(*[' '.join(i[::-1]) for i in out], sep='\n')

# Output
user zero 0000000
user One 1111111
user Two 2222222
user three 3333333
user four 4444444
user Five 5555555
user six 6666666
user Seven 7777777
user eight 8888888