从Python中的字符串中删除HTML

Question

from mechanize import Browser
br = Browser()
br.open('http://somewebpage')
html = br.response().readlines()
for line in html:
  print line

在HTML文件中打印一行时，我试图找到一种方法来只显示每个HTML元素的内容而不是格式本身。如果它找到'<a href="whatever.com">some text</a>'，它只会打印'some text'，'<b>hello</b>'打印'hello'等等。怎么会这样做呢？

Answer 1

我总是使用此函数来去除HTML标记，因为它只需要Python stdlib：

在Python 2上

from HTMLParser import HTMLParser

class MLStripper(HTMLParser):
    def __init__(self):
        self.reset()
        self.fed = []
    def handle_data(self, d):
        self.fed.append(d)
    def get_data(self):
        return ''.join(self.fed)

def strip_tags(html):
    s = MLStripper()
    s.feed(html)
    return s.get_data()

对于Python 3

from html.parser import HTMLParser

class MLStripper(HTMLParser):
    def __init__(self):
        self.reset()
        self.strict = False
        self.convert_charrefs= True
        self.fed = []
    def handle_data(self, d):
        self.fed.append(d)
    def get_data(self):
        return ''.join(self.fed)

def strip_tags(html):
    s = MLStripper()
    s.feed(html)
    return s.get_data()

注意：这仅适用于3.1。对于3.2或更高版本，您需要调用父类的 init 函数。见Using HTMLParser in Python 3.2

Answer 2

我没有想过它会错过的案例，但你可以做一个简单的正则表达式：

re.sub('<[^<]+?>', '', text)

对于那些不了解正则表达式的人，会搜索字符串<...>，其内部内容由一个或多个（+）字符组成，而不是{{1} }。 <表示它将匹配它可以找到的最小字符串。例如，给定?，它将<p>Hello</p>和<'p>分别与</p>匹配。没有它，它将匹配整个字符串?。

如果非标签<..Hello..>出现在html中（例如<），那么它应该被写为转义序列2 < 3，因此&...可能是不必要的。

Answer 3

为什么你们所有人都这么做？ 您可以使用BeautifulSoup get_text()功能。

from bs4 import BeautifulSoup

html_str = '''
<td><a href="http://www.fakewebsite.com">Please can you strip me?</a>
<br/><a href="http://www.fakewebsite.com">I am waiting....</a>
</td>
'''
soup = BeautifulSoup(html_str)

print(soup.get_text()) 
#or via attribute of Soup Object: print(soup.text)

Answer 4

简短版本！

import re, cgi
tag_re = re.compile(r'(<!--.*?-->|<[^>]*>)')

# Remove well-formed tags, fixing mistakes by legitimate users
no_tags = tag_re.sub('', user_input)

# Clean up anything else by escaping
ready_for_web = cgi.escape(no_tags)

Regex source: MarkupSafe。他们的版本也处理HTML实体，而这个快速版本则没有。

为什么我不能剥离标签并留下它？

让人们远离<i>italicizing</i>事情是一回事，而不会让i漂浮在身边。但是接受任意输入并使其完全无害是另一回事。此页面上的大多数技术都会保留未关闭的注释（<!--）和不包含标记（blah <<<><blah）的角括号等内容。 HTMLParser版本甚至可以保留完整的标签，如果它们在未公开的评论中。

如果您的模板是{{ firstname }} {{ lastname }}怎么办？ firstname = '<a'和lastname = 'href="http://evil.com/">'将通过此页面上的每个代码剥离器（@Medeiros除外）通过，因为它们本身并不是完整的标记。剥离普通的HTML标签是不够的。

Django的strip_tags，这个问题的最佳答案的改进版（见下一个标题），给出了以下警告：

  

绝对不保证生成的字符串是HTML安全的。因此，永远不要在strip_tags调用的结果标记为安全，而不首先将其转义，例如使用escape()。

听从他们的意见！

要使用HTMLParser剥离标记，您必须多次运行它。

很容易绕过这个问题的最佳答案。

查看此字符串（source and discussion）：

<img<!-- --> src=x onerror=alert(1);//><!-- -->

HTMLParser第一次看到它时，它无法判断<img...>是一个标记。它看起来很破碎，所以HTMLParser并没有摆脱它。它只会取出<!-- comments -->，而只剩下

<img src=x onerror=alert(1);//>

这个问题已于2014年3月向Django项目披露。他们的旧版strip_tags基本上与此问题的最佳答案相同。 Their new version基本上在循环中运行它，直到再次运行它不会更改字符串：

# _strip_once runs HTMLParser once, pulling out just the text of all the nodes.

def strip_tags(value):
    """Returns the given HTML with all tags stripped."""
    # Note: in typical case this loop executes _strip_once once. Loop condition
    # is redundant, but helps to reduce number of executions of _strip_once.
    while '<' in value and '>' in value:
        new_value = _strip_once(value)
        if len(new_value) >= len(value):
            # _strip_once was not able to detect more tags
            break
        value = new_value
    return value

当然，如果你总是逃避strip_tags()的结果，这一切都不是问题。

更新2015年3月19日：在1.4.20,1.6.11,1.7.7和1.8c1之前的Django版本中存在一个错误。这些版本可以在strip_tags（）函数中进入无限循环。固定版本在上面复制。 More details here

复制或使用的好东西

我的示例代码不处理HTML实体 - Django和MarkupSafe打包版本。

我的示例代码是从优秀的MarkupSafe库中提取的，用于防止跨站点脚本编写。它方便快捷（C加速到其原生Python版本）。它包含在Google App Engine中，由Jinja2 (2.7 and up)，Mako，Pylons等使用。它可以轻松地使用Django 1.7中的Django模板。

Django的strip_tags和其他html实用程序来自最近的版本很好，但我发现它们不如MarkupSafe方便。它们非常独立，您可以从this file复制所需内容。

如果您需要删除几乎所有标记，Bleach库就可以了。你可以让它强制执行“我的用户可以用斜体表示事物，但他们不能制作iframe”这样的规则。

了解标签剥离器的属性！对它进行模糊测试！ Here is the code我曾经为这个答案做过研究。

羞怯的说明 - 问题本身是关于打印到控制台，但这是“python strip html from string”的Google最高结果，所以这就是为什么这个答案是关于网络的99％

Answer 5

我需要一种方法来剥离标签和将HTML实体解码为纯文本。以下解决方案基于Eloff的答案（我无法使用，因为它剥离了实体）。

from HTMLParser import HTMLParser
import htmlentitydefs

class HTMLTextExtractor(HTMLParser):
    def __init__(self):
        HTMLParser.__init__(self)
        self.result = [ ]

    def handle_data(self, d):
        self.result.append(d)

    def handle_charref(self, number):
        codepoint = int(number[1:], 16) if number[0] in (u'x', u'X') else int(number)
        self.result.append(unichr(codepoint))

    def handle_entityref(self, name):
        codepoint = htmlentitydefs.name2codepoint[name]
        self.result.append(unichr(codepoint))

    def get_text(self):
        return u''.join(self.result)

def html_to_text(html):
    s = HTMLTextExtractor()
    s.feed(html)
    return s.get_text()

快速测试：

html = u'<a href="#">Demo <em>(&not; \u0394&#x03b7;&#956;&#x03CE;)</em></a>'
print repr(html_to_text(html))

结果：

u'Demo (\xac \u0394\u03b7\u03bc\u03ce)'

错误处理：

• 无效的HTML结构可能会导致HTMLParseError。
• 无效的命名HTML实体（例如&#apos;，在XML和XHTML中有效，但不是纯HTML）将导致ValueError例外。
• 指定Python可接受的Unicode范围之外的代码点的数字HTML实体（例如，在某些系统上，Basic Multilingual Plane之外的字符）将导致ValueError异常。

安全说明：不要将HTML剥离（将HTML转换为纯文本）与HTML清理（将纯文本转换为HTML）混淆。这个答案将删除HTML并将实体解码为纯文本 - 这不会使结果在HTML上下文中安全使用。

示例：<script>alert("Hello");</script>将转换为<script>alert("Hello");</script>，这是100％正确的行为，但如果生成的纯文本按原样插入到HTML页面中，则显然不够。

规则并不难：任何时候您将纯文本字符串插入HTML输出中，您应该总是 HTML将其转义（使用cgi.escape(s, True)）即使你“知道”它不包含HTML（例如，因为你剥离了HTML内容）。

（但是，OP询问是否将结果打印到控制台，在这种情况下不需要HTML转义。）

Python 3.4+版本：（使用doctest！）

import html.parser

class HTMLTextExtractor(html.parser.HTMLParser):
    def __init__(self):
        super(HTMLTextExtractor, self).__init__()
        self.result = [ ]

    def handle_data(self, d):
        self.result.append(d)

    def get_text(self):
        return ''.join(self.result)

def html_to_text(html):
    """Converts HTML to plain text (stripping tags and converting entities).
    >>> html_to_text('<a href="#">Demo<!--...--> <em>(&not; \u0394&#x03b7;&#956;&#x03CE;)</em></a>')
    'Demo (\xac \u0394\u03b7\u03bc\u03ce)'

    "Plain text" doesn't mean result can safely be used as-is in HTML.
    >>> html_to_text('&lt;script&gt;alert("Hello");&lt;/script&gt;')
    '<script>alert("Hello");</script>'

    Always use html.escape to sanitize text before using in an HTML context!

    HTMLParser will do its best to make sense of invalid HTML.
    >>> html_to_text('x < y &lt z <!--b')
    'x < y < z '

    Unrecognized named entities are included as-is. '&apos;' is recognized,
    despite being XML only.
    >>> html_to_text('&nosuchentity; &apos; ')
    "&nosuchentity; ' "
    """
    s = HTMLTextExtractor()
    s.feed(html)
    return s.get_text()

请注意，HTMLParser在Python 3中得到了改进（意味着代码更少，错误处理更好）。

Answer 6

有一种简单的方法：

def remove_html_markup(s):
    tag = False
    quote = False
    out = ""

    for c in s:
            if c == '<' and not quote:
                tag = True
            elif c == '>' and not quote:
                tag = False
            elif (c == '"' or c == "'") and tag:
                quote = not quote
            elif not tag:
                out = out + c

    return out

这个想法在这里解释：http://youtu.be/2tu9LTDujbw

您可以在此处看到它：http://youtu.be/HPkNPcYed9M?t=35s

PS - 如果您对该课程感兴趣（关于使用python进行智能调试），我会给你一个链接：http://www.udacity.com/overview/Course/cs259/CourseRev/1。免费！

欢迎你！ ：）

Answer 7

如果您需要保留HTML实体（例如&），我将“handle_entityref”方法添加到Eloff's answer。

from HTMLParser import HTMLParser

class MLStripper(HTMLParser):
    def __init__(self):
        self.reset()
        self.fed = []
    def handle_data(self, d):
        self.fed.append(d)
    def handle_entityref(self, name):
        self.fed.append('&%s;' % name)
    def get_data(self):
        return ''.join(self.fed)

def html_to_text(html):
    s = MLStripper()
    s.feed(html)
    return s.get_data()

Answer 8

如果要删除所有HTML标记，我发现的最简单方法是使用BeautifulSoup：

from bs4 import BeautifulSoup  # Or from BeautifulSoup import BeautifulSoup

def stripHtmlTags(htmlTxt):
    if htmlTxt is None:
            return None
        else:
            return ''.join(BeautifulSoup(htmlTxt).findAll(text=True)) 

我尝试了接受答案的代码，但是我得到了“RuntimeError：超出了最大递归深度”，这在上面的代码块中没有发生。

Answer 9

基于lxml.html的解决方案（lxml是一个本机库，因此比任何纯python解决方案都快得多）。

from lxml import html
from lxml.html.clean import clean_html

tree = html.fromstring("""<span class="item-summary">
                            Detailed answers to any questions you might have
                        </span>""")

print(clean_html(tree).strip())

# >>> Detailed answers to any questions you might have

  

另请参阅http://lxml.de/lxmlhtml.html#cleaning-up-html了解lxml.cleaner的确切内容。

如果您需要在转换为文本之前更好地控制清理的内容，那么您可能希望通过在构造函数中传递lxml Cleaner来显式使用options you want，例如：

cleaner = Cleaner(page_structure=True,
                  meta=True,
                  embedded=True,
                  links=True,
                  style=True,
                  processing_instructions=True,
                  inline_style=True,
                  scripts=True,
                  javascript=True,
                  comments=True,
                  frames=True,
                  forms=True,
                  annoying_tags=True,
                  remove_unknown_tags=True,
                  safe_attrs_only=True,
                  safe_attrs=frozenset(['src','color', 'href', 'title', 'class', 'name', 'id']),
                  remove_tags=('span', 'font', 'div')
                  )
sanitized_html = cleaner.clean_html(unsafe_html)

Answer 10

美丽的汤包立即为您做到这一点。

from bs4 import BeautifulSoup

soup = BeautifulSoup(html)
text = soup.get_text()
print(text)

Answer 11

您可以使用不同的HTML解析器（like lxml或Beautiful Soup） - 提供仅提取文本的功能的解析器。或者，您可以在删除标记的行字符串上运行正则表达式。有关详情，请参阅http://www.amk.ca/python/howto/regex/。

Answer 12

HTML-Parser解决方案只有运行一次才能破解：

html_to_text('<<b>script>alert("hacked")<</b>/script>

结果：

<script>alert("hacked")</script>

你打算阻止什么。如果您使用HTML-Parser，请计算标签直到零被替换：

from HTMLParser import HTMLParser

class MLStripper(HTMLParser):
    def __init__(self):
        self.reset()
        self.fed = []
        self.containstags = False

    def handle_starttag(self, tag, attrs):
       self.containstags = True

    def handle_data(self, d):
        self.fed.append(d)

    def has_tags(self):
        return self.containstags

    def get_data(self):
        return ''.join(self.fed)

def strip_tags(html):
    must_filtered = True
    while ( must_filtered ):
        s = MLStripper()
        s.feed(html)
        html = s.get_data()
        must_filtered = s.has_tags()
    return html

Answer 13

这里的解决方案与当前接受的答案（https://stackoverflow.com/a/925630/95989）类似，不同的是它直接使用内部HTMLParser类（即没有子类），从而使其简洁得多：

def strip_html(text):
    parts = []                                                                      
    parser = HTMLParser()                                                           
    parser.handle_data = parts.append                                               
    parser.feed(text)                                                               
    return ''.join(parts)

Answer 14

对于一个项目，我需要这样剥离HTML，还需要css和js。因此，我改编了Eloffs的回答：

class MLStripper(HTMLParser):
    def __init__(self):
        self.reset()
        self.strict = False
        self.convert_charrefs= True
        self.fed = []
        self.css = False
    def handle_starttag(self, tag, attrs):
        if tag == "style" or tag=="script":
            self.css = True
    def handle_endtag(self, tag):
        if tag=="style" or tag=="script":
            self.css=False
    def handle_data(self, d):
        if not self.css:
            self.fed.append(d)
    def get_data(self):
        return ''.join(self.fed)

def strip_tags(html):
    s = MLStripper()
    s.feed(html)
    return s.get_data()

Answer 15

这是一个简单的解决方案，它基于惊人的快速lxml库来剥离HTML标签并解码HTML实体：

from lxml import html

def strip_html(s):
    return html.fromstring(s).text_content()

strip_html('Ein <a href="">sch&ouml;ner</a> Text.')  # Output: Ein schöner Text.

Answer 16

søren-løvborg答案的蟒蛇3改编

from html.parser import HTMLParser
from html.entities import html5

class HTMLTextExtractor(HTMLParser):
    """ Adaption of http://stackoverflow.com/a/7778368/196732 """
    def __init__(self):
        super().__init__()
        self.result = []

    def handle_data(self, d):
        self.result.append(d)

    def handle_charref(self, number):
        codepoint = int(number[1:], 16) if number[0] in (u'x', u'X') else int(number)
        self.result.append(unichr(codepoint))

    def handle_entityref(self, name):
        if name in html5:
            self.result.append(unichr(html5[name]))

    def get_text(self):
        return u''.join(self.result)

def html_to_text(html):
    s = HTMLTextExtractor()
    s.feed(html)
    return s.get_text()

Answer 17

这是一个快速修复，可以更加优化，但它可以正常工作。此代码将使用＆＃34;＆＃34;替换所有非空标记。并删除所有html标签形成给定的输入文本。您可以使用./file.py输入输出

运行它

    #!/usr/bin/python
import sys

def replace(strng,replaceText):
    rpl = 0
    while rpl > -1:
        rpl = strng.find(replaceText)
        if rpl != -1:
            strng = strng[0:rpl] + strng[rpl + len(replaceText):]
    return strng


lessThanPos = -1
count = 0
listOf = []

try:
    #write File
    writeto = open(sys.argv[2],'w')

    #read file and store it in list
    f = open(sys.argv[1],'r')
    for readLine in f.readlines():
        listOf.append(readLine)         
    f.close()

    #remove all tags  
    for line in listOf:
        count = 0;  
        lessThanPos = -1  
        lineTemp =  line

            for char in lineTemp:

            if char == "<":
                lessThanPos = count
            if char == ">":
                if lessThanPos > -1:
                    if line[lessThanPos:count + 1] != '<>':
                        lineTemp = replace(lineTemp,line[lessThanPos:count + 1])
                        lessThanPos = -1
            count = count + 1
        lineTemp = lineTemp.replace("&lt","<")
        lineTemp = lineTemp.replace("&gt",">")                  
        writeto.write(lineTemp)  
    writeto.close() 
    print "Write To --- >" , sys.argv[2]
except:
    print "Help: invalid arguments or exception"
    print "Usage : ",sys.argv[0]," inputfile outputfile"

Answer 18

我已成功使用Eloff的答案用于Python 3.1 [非常感谢！]。

我升级到Python 3.2.3，并遇到了错误。

感谢响应者Thomas K提供的here解决方案是将super().__init__()插入以下代码中：

def __init__(self):
    self.reset()
    self.fed = []

...为了使它看起来像这样：

def __init__(self):
    super().__init__()
    self.reset()
    self.fed = []

...它适用于Python 3.2.3。

再次感谢Thomas K的修复以及上面提供的Eloff的原始代码！

Answer 19

# This is a regex solution.
import re
def removeHtml(html):
  if not html: return html
  # Remove comments first
  innerText = re.compile('<!--[\s\S]*?-->').sub('',html)
  while innerText.find('>')>=0: # Loop through nested Tags
    text = re.compile('<[^<>]+?>').sub('',innerText)
    if text == innerText:
      break
    innerText = text

  return innerText.strip()

Answer 20

我就是这样做的，但我不知道我在做什么。我通过去除 HTML 标签从 HTML 表中获取数据。

这需要字符串“name”并返回不带 HTML 标签的字符串“name1”。

        x = 0
        anglebrackets = 0
        name1 = ""
        while x < len(name):
            
            if name[x] == "<":
                anglebrackets = anglebrackets + 1
            if name[x] == ">":
                anglebrackets = anglebrackets - 1
            if anglebrackets == 0:
                if name[x] != ">":
                    name1 = name1 + name[x]
            x = x + 1

Answer 21

2020更新

使用Mozilla Bleach library，它实际上使您可以自定义要保留的标签和要保留的属性，并根据值过滤出属性

这里有2种情况可以说明

1）不允许任何HTML标记或属性

获取示例原始文本

raw_text = """
<p><img width="696" height="392" src="https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-768x432.jpg" class="attachment-medium_large size-medium_large wp-post-image" alt="Ethereum Classic 51% Attack: Okex Crypto Exchange Suffers $5.6 Million Loss, Contemplates Delisting ETC" style="float:left; margin:0 15px 15px 0;" srcset="https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-768x432.jpg 768w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-300x169.jpg 300w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-1024x576.jpg 1024w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-696x392.jpg 696w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-1068x601.jpg 1068w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-747x420.jpg 747w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-190x107.jpg 190w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-380x214.jpg 380w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-760x428.jpg 760w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc.jpg 1280w" sizes="(max-width: 696px) 100vw, 696px" />Cryptocurrency exchange Okex reveals it suffered the $5.6 million loss as a result of the double-spend carried out by the attacker(s) in Ethereum Classic 51% attack. Okex says it fully absorbed the loss as per its user-protection policy while insisting that the attack did not cause any loss to the platform&#8217;s users. Also as part [&#8230;]</p>
<p>The post <a rel="nofollow" href="https://news.bitcoin.com/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc/">Ethereum Classic 51% Attack: Okex Crypto Exchange Suffers $5.6 Million Loss, Contemplates Delisting ETC</a> appeared first on <a rel="nofollow" href="https://news.bitcoin.com">Bitcoin News</a>.</p> 
"""

2）从原始文本中删除所有HTML标签和属性

# DO NOT ALLOW any tags or any attributes
from bleach.sanitizer import Cleaner
cleaner = Cleaner(tags=[], attributes={}, styles=[], protocols=[], strip=True, strip_comments=True, filters=None)
print(cleaner.clean(raw_text))

输出

Cryptocurrency exchange Okex reveals it suffered the $5.6 million loss as a result of the double-spend carried out by the attacker(s) in Ethereum Classic 51% attack. Okex says it fully absorbed the loss as per its user-protection policy while insisting that the attack did not cause any loss to the platform’s users. Also as part […]
The post Ethereum Classic 51% Attack: Okex Crypto Exchange Suffers $5.6 Million Loss, Contemplates Delisting ETC appeared first on Bitcoin News. 

3仅允许具有srcset属性的img标签

from bleach.sanitizer import Cleaner
# ALLOW ONLY img tags with src attribute
cleaner = Cleaner(tags=['img'], attributes={'img': ['srcset']}, styles=[], protocols=[], strip=True, strip_comments=True, filters=None)
print(cleaner.clean(raw_text))

输出

<img srcset="https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-768x432.jpg 768w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-300x169.jpg 300w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-1024x576.jpg 1024w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-696x392.jpg 696w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-1068x601.jpg 1068w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-747x420.jpg 747w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-190x107.jpg 190w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-380x214.jpg 380w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc-760x428.jpg 760w, https://news.bitcoin.com/wp-content/uploads/2020/08/ethereum-classic-51-attack-okex-crypto-exchange-suffers-5-6-million-loss-contemplates-delisting-etc.jpg 1280w">Cryptocurrency exchange Okex reveals it suffered the $5.6 million loss as a result of the double-spend carried out by the attacker(s) in Ethereum Classic 51% attack. Okex says it fully absorbed the loss as per its user-protection policy while insisting that the attack did not cause any loss to the platform&#8217;s users. Also as part [&#8230;]
The post Ethereum Classic 51% Attack: Okex Crypto Exchange Suffers $5.6 Million Loss, Contemplates Delisting ETC appeared first on Bitcoin News. 

Answer 22

hext是一个可以打包strip HTML的程序包。它是getLines的替代方法。以下内容已通过beautifulsoup进行了测试。

将其保存在实用程序模块中，例如hext==0.2.3：

util/hext.py

用法示例：

import hext

_HTML_TEXT_RULE = hext.Rule('<html @text:text />')


def html_to_text(text: str) -> str:
    # Ref: https://stackoverflow.com/a/56894409/
    return _HTML_TEXT_RULE.extract(hext.Html(f'<html>{text}</html>'))[0]['text']

HTML格式错误的使用示例：

>>> from .util.hext import html_to_text

>>> html_to_text('<b>Hello world!</b>')
'Hello world!'

>>> html_to_text('<a href="google.com">some text</a>')
'some text'

>>> html_to_text('<span class="small-caps">l</span>-arginine minimizes immunosuppression and prothrombin time and enhances the genotoxicity of 5-fluorouracil in rats')
'l-arginine minimizes immunosuppression and prothrombin time and enhances the genotoxicity of 5-fluorouracil in rats'

>>> html_to_text('Attenuation of diabetic nephropathy by dietary fenugreek (<em>Trigonella foenum-graecum</em>) seeds and onion (<em>Allium cepa</em>) <em>via</em> suppression of glucose transporters and renin-angiotensin system')
'Attenuation of diabetic nephropathy by dietary fenugreek (Trigonella foenum-graecum) seeds and onion (Allium cepa) via suppression of glucose transporters and renin-angiotensin system'

Answer 23

简单的代码！。这将删除其中的所有标记和内容。

def rm(s):
    start=False
    end=False
    s=' '+s
    for i in range(len(s)-1):
        if i<len(s):
            if start!=False:
                if s[i]=='>':
                    end=i
                    s=s[:start]+s[end+1:]
                    start=end=False
            else:
                if s[i]=='<':
                    start=i
    if s.count('<')>0:
        self.rm(s)
    else:
        s=s.replace('&nbsp;', ' ')
        return s

但是，如果文本中包含 <> 个符号，则不会给出完整的结果。

Answer 24

这是我对python 3的解决方案。

import html
import re

def html_to_txt(html_text):
    ## unescape html
    txt = html.unescape(html_text)
    tags = re.findall("<[^>]+>",txt)
    print("found tags: ")
    print(tags)
    for tag in tags:
        txt=txt.replace(tag,'')
    return txt

不确定它是否完美，但解决了我的用例，看起来很简单。

Answer 25

使用BeautifulSoup，html2text或来自@Eloff的代码，大多数情况下，它仍然是一些html元素，javascript代码......

因此，您可以使用这些库的组合并删除降价格式（Python 3）：

import re
import html2text
from bs4 import BeautifulSoup
def html2Text(html):
    def removeMarkdown(text):
        for current in ["^[ #*]{2,30}", "^[ ]{0,30}\d\\\.", "^[ ]{0,30}\d\."]:
            markdown = re.compile(current, flags=re.MULTILINE)
            text = markdown.sub(" ", text)
        return text
    def removeAngular(text):
        angular = re.compile("[{][|].{2,40}[|][}]|[{][*].{2,40}[*][}]|[{][{].{2,40}[}][}]|\[\[.{2,40}\]\]")
        text = angular.sub(" ", text)
        return text
    h = html2text.HTML2Text()
    h.images_to_alt = True
    h.ignore_links = True
    h.ignore_emphasis = False
    h.skip_internal_links = True
    text = h.handle(html)
    soup = BeautifulSoup(text, "html.parser")
    text = soup.text
    text = removeAngular(text)
    text = removeMarkdown(text)
    return text

它适合我，但它可以增强，当然......

Answer 26

我正在解析Github自述文件，我发现以下内容确实运行良好：

import re
import lxml.html

def strip_markdown(x):
    links_sub = re.sub(r'\[(.+)\]\([^\)]+\)', r'\1', x)
    bold_sub = re.sub(r'\*\*([^*]+)\*\*', r'\1', links_sub)
    emph_sub = re.sub(r'\*([^*]+)\*', r'\1', bold_sub)
    return emph_sub

def strip_html(x):
    return lxml.html.fromstring(x).text_content() if x else ''

然后

readme = """<img src="https://raw.githubusercontent.com/kootenpv/sky/master/resources/skylogo.png" />

            sky is a web scraping framework, implemented with the latest python versions in mind (3.4+). 
            It uses the asynchronous `asyncio` framework, as well as many popular modules 
            and extensions.

            Most importantly, it aims for **next generation** web crawling where machine intelligence 
            is used to speed up the development/maintainance/reliability of crawling.

            It mainly does this by considering the user to be interested in content 
            from *domains*, not just a collection of *single pages*
            ([templating approach](#templating-approach))."""

strip_markdown(strip_html(readme))

正确删除所有降价和HTML。

Answer 27

您可以编写自己的功能：

def StripTags(text):
     finished = 0
     while not finished:
         finished = 1
         start = text.find("<")
         if start >= 0:
             stop = text[start:].find(">")
             if stop >= 0:
                 text = text[:start] + text[start+stop+1:]
                 finished = 0
     return text

Answer 28

这种方法对我来说完美无缺，无需额外安装：

import re
import htmlentitydefs

def convertentity(m):
    if m.group(1)=='#':
        try:
            return unichr(int(m.group(2)))
        except ValueError:
            return '&#%s;' % m.group(2)
        try:
            return htmlentitydefs.entitydefs[m.group(2)]
        except KeyError:
            return '&%s;' % m.group(2)

def converthtml(s):
    return re.sub(r'&(#?)(.+?);',convertentity,s)

html =  converthtml(html)
html.replace(" ", " ") ## Get rid of the remnants of certain formatting(subscript,superscript,etc).