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我正在将一本书从 PDF 转换为 calibre 的 epub。但标题不在标题标签内，因此尝试使用正则表达式替换它的 python 函数。

示例文本：

<p class="calibre1"><a id="p1"></a>Chapter 370: Slamming straight on</p>
<p class="softbreak"> </p>
<p class="calibre1">Hearing Yan Zhaoge’s suggestion, the Jade Sea City martial practitioners here were all stunned.</p>
<p class="calibre1"><a id="p7"></a>Chapter 372: Yan Zhaoge’s plan</p>
<p class="softbreak"> </p>
<p class="calibre1">Yan Zhaoge and Ah Hu sat on Pan-Pan’s back, black water swirling about Pan-Pan’s entire body, keeping away the seawater as he shot forward at lightning speed.</p>

我尝试使用正则表达式

def replace(match, number, file_name, metadata, dictionaries, data, functions, *args, **kwargs):
    
    pattern = r"</a>(?i)chapter [0-9]+: [\w\s]+(.*)<br>"
    list = re.findall(pattern, match.group())
    
    for x in list:
        x = "</a>(?i)chapter [0-9]+: [\w\s]+(.?)<br>"
        x = s.split("</a>", 1)[0] + '</a><h2>' + s.split("a>",1)[1]
        x = s.split("<br>", 1)[0] + '</h2><br>' + s.split("<br>",1)[1]
    return match.group()

和


def replace(match, number, file_name, metadata, dictionaries, data, functions, *args, **kwargs):
    pattern = r"</a>(?i)chapter [0-9]+: [\w\s]+(.*)<br>"
    s.replace(re.match(pattern, s), r'<h2>$0')

但仍然没有得到预期的结果。我想要的是...

输入

</a>Chapter 370: Slamming straight on</p>

输出

</a><h2>Chapter 370: Slamming straight on</h2></p>

h2 标签将被添加到所有类似的实例中

4

2 回答 2

1

regex不应该用于解析 xml。请参阅： 为什么不能使用正则表达式来解析 HTML/XML：外行术语的正式解释 （Why shouldn't you..会是一个更好的标题）

但是，您可以改用 BeautifulSoup：

from bs4 import BeautifulSoup
data = """<p class="calibre1"><a id="p1"></a>Chapter 370: Slamming straight on</p>
<p class="softbreak"> </p>
<p class="calibre1">Hearing Yan Zhaoge’s suggestion, the Jade Sea City martial practitioners here were all stunned.</p>
<p class="calibre1"><a id="p7"></a>Chapter 372: Yan Zhaoge’s plan</p>
<p class="softbreak"> </p>
<p class="calibre1">Yan Zhaoge and Ah Hu sat on Pan-Pan’s back, black water swirling about Pan-Pan’s entire body, keeping away the seawater as he shot forward at lightning speed.</p>
i t"""

soup = BeautifulSoup(data, 'lxml')


for x in soup.find_all('p', {'class':'calibre1'}):

    link = x.find('a')
    title = x.text
    corrected_title = soup.new_tag('h2')
    corrected_title.append(title)

    if link:
        x.string=''
        corrected_title = soup.new_tag('h2')
        corrected_title.append(title)
        link.append(corrected_title)
        x.append(link)

print(soup.body)

输出

<body>
    <p class="calibre1">
        <a id="p1">
            <h2>Chapter 370: Slamming straight on</h2>
        </a>
    </p>
    <p class="softbreak"> </p>
    <p class="calibre1">Hearing Yan Zhaoge’s suggestion, the Jade Sea City martial practitioners here were all stunned.</p>
    <p class="calibre1">
        <a id="p7">
            <h2>Chapter 372: Yan Zhaoge’s plan</h2>
        </a>
    </p>
    <p class="softbreak"> </p>
    <p class="calibre1">Yan Zhaoge and Ah Hu sat on Pan-Pan’s back, black water swirling about Pan-Pan’s entire body, keeping away the seawater as he shot forward at lightning speed.</p>
    i t
</body>
于 2019-06-17T22:05:24.983 回答
0

Jean-François的评论会更好地遵循，但如果我们可能不得不这样做，我的猜测是我们将从这个表达式开始：

(<\/a>)([^<]+)?(<\/p>)
(<\/a>)(chapter\s+[0-9]+[^<]+)?(<\/p>)

被替换为：

\1<h2>\2</h2>\3

演示 1

演示 2

测试

# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility

import re

regex = r"(<\/a>)(chapter\s+[0-9]+[^<]+)?(<\/p>)"

test_str = ("<p class=\"calibre1\"><a id=\"p1\"></a>Chapter 370: Slamming straight on</p>\n"
    "<p class=\"softbreak\"> </p>\n"
    "<p class=\"calibre1\">Hearing Yan Zhaoge’s suggestion, the Jade Sea City martial practitioners here were all stunned.</p>\n"
    "<p class=\"calibre1\"><a id=\"p7\"></a>Chapter 372: Yan Zhaoge’s plan</p>\n"
    "<p class=\"softbreak\"> </p>\n"
    "<p class=\"calibre1\">Yan Zhaoge and Ah Hu sat on Pan-Pan’s back, black water swirling about Pan-Pan’s entire body, keeping away the seawater as he shot forward at lightning speed.</p>")

subst = "\\1<h2>\\2</h2>\\3"

# You can manually specify the number of replacements by changing the 4th argument
result = re.sub(regex, subst, test_str, 0, re.MULTILINE | re.IGNORECASE)

if result:
    print (result)

# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.
于 2019-06-17T20:33:23.380 回答