Question

我有以下Python代码：

def getAddress(text):
    text = re.sub('\t', '', text)
    text = re.sub('\n', '', text)
    blocks = re.findall('<div class="result-box" itemscope itemtype="http://schema.org/LocalBusiness">([a-zA-Z0-9 ",;:\.#&_=()\'<>/\\\t\n\-]*)</span>Follow company</span>', text)
    name = ''
    strasse = ''
    locality = ''
    plz = ''
    region = ''
    i = 0

    for block in blocks:
        names = re.findall('class="url">(.*)</a>', block)
        strassen = re.findall('<span itemprop="streetAddress">([a-zA-Z0-9 ,;:\.&#]*)</span>', block)
        localities = re.findall('<span itemprop="addressLocality">([a-zA-Z0-9 ,;:&]*)</span>', block)
        plzs = re.findall('<span itemprop="postalCode">([0-9]*)</span>', block)
        regions = re.findall('<span itemprop="addressRegion">([a-zA-Z]*)</span>', block)

        try:
            for name in names:
                name = str(name)
                name = re.sub('<[^<]+?>', '', name)
                break

            for strasse in strassen:
                strasse = str(strasse)
                strasse = re.sub('<[^<]+?>', '', strasse)
                break

            for locality in localities:
                locality = str(locality)
                locality = re.sub('<[^<]+?>', '', locality)
                break

            for plz in plzs:
                plz = str(plz)
                plz = re.sub('<[^<]+?>', '', plz)
                break

            for region in regions:
                region = str(region)
                region = re.sub('<[^<]+?>', '', region)
                break
        except:
            continue
        print i
        i = i + 1

        if plz == '':
            plz = getZipCode(strasse, locality, region)
        address = '"' + name + '"' + ';' + '"' + strasse + '";' + locality + ';' + str(plz) + ';' + region + '\n'

        #saveToCSV(address)

我想过滤掉这个html代码段。这段剪辑重复几次。我希望函数为每个代码段返回一个条目。但相反，它返回一个包含两个片段的条目。我需要改变什么？

<div class="result-box" itemscope itemtype="http://schema.org/LocalBusiness">
        <div class="clear">
            <h2 itemprop="name"><a href="http://www.manta.com/c/mxlk5yt/belgium-jewelers-corp" class="url">Belgium Jewelers Corp</a></h2>           </div>
        <div itemprop="address" itemscope itemtype="http://schema.org/PostalAddress">               <span itemprop="addressLocality">Lawrenceville</span> <span itemprop="addressRegion">NJ</span>
        </div>          <a href="#" class="followCompany" data-emid="mxlk5yt" data-companyname="Belgium Jewelers Corp" data-location="ListingFollowButton" data-location-page="Megabrowse">
            <span class="followMsg"><span class="followIcon mrs"></span>Follow company</span>
            <span class="followingMsg"><span class="followIcon mrs"></span>Following</span>
            <span class="unfollowMsg"><span class="followIcon mrs"></span>Unfollow company</span>
        </a>            <p class="type">Jewelry Stores</p>      </div>
    </li>       <li>        <div class="icons">
        <ul>            </ul>
    </div>

Answer 1

请放下那把锤子; HTML 不是一个正则表达形状的指甲。用于解析HTML的正则表达式变得非常复杂，并且非常脆弱，当HTML发生微妙变化时很容易被破坏。

改为使用正确的HTML解析器。 BeautifulSoup会使您的任务变得微不足道：

from bs4 import BeautifulSoup

soup = BeautifulSoup(text)
for block in soup.find_all('div', class_="result-box", itemtype="http://schema.org/LocalBusiness"):
    print block.find('a', class_='url').string

    street = block.find('span', itemprop="streetAddress")
    if street:
        print street.string

    locality = block.find('span', itemprop="addressLocality")
    if locality:
        print locality.string

    # .. etc. ..

Answer 2

您应该查看HTMLParser（documentation）的Python。正则表达式解析HTML非常糟糕。

如何使用Python提取以下HTML片段

2 个答案: