Scraper 无法打印所有结果

Question

我用 python 编写了一个脚本来从 craigslist 中抓取五个项目的“姓名”和“电话”。我面临的问题是，当我运行我的脚本时，它只给出三个结果而不是五个结果。更具体地说，由于前两个链接在他们的页面中没有额外的链接（联系信息），所以他们不需要再次请求打开任何额外的页面。但是，没有（联系信息）链接的两个链接无法通过我的第二个函数中的“if ano_page_link：”语句渗透，并且永远不会被打印。如何修复这个漏洞，无论是否有电话号码，刮板都会打印所有五个结果。

我正在尝试的脚本：

import re ; import requests ; from lxml import html

base = "http://bangalore.craigslist.co.in"

url_list = [
'http://bangalore.craigslist.co.in/reb/d/flat-is-for-sale-at-cooke-town/6266183606.html',
'http://bangalore.craigslist.co.in/reb/d/prestige-sunnyside/6259128505.html',
'http://bangalore.craigslist.co.in/reb/d/jayanagar-2nd-block-4000-sft/6221720477.html',
'http://bangalore.craigslist.co.in/reb/d/prestige-ozone-type-3-r-villa/6259928614.html',
'http://bangalore.craigslist.co.in/reb/d/zed-homes-3-bedroom-flat-for/6257075793.html'
]

def get_link(medium_link):
    response = requests.get(medium_link).text
    tree = html.fromstring(response)
    try:
        name = tree.cssselect('span#titletextonly')[0].text
    except IndexError:
        name = ""
    try:
        link = base + tree.cssselect('a.showcontact')[0].attrib['href']
    except IndexError:
        link = ""
    parse_doc(name, link)

def parse_doc(title, ano_page_link):

    if ano_page_link:
        page = requests.get(ano_page_link).text            
        tel = re.findall(r'\d{10}', page)[0] if re.findall(r'\d{10}', page) else ""
        print(title, tel)

if __name__ == '__main__':
    for link in url_list:
        get_link(link)

我得到的结果：

Jayanagar 2nd Block, 4000 sft Plot for Sale 9845012673
PRESTIGE OZONE TYPE D 3 B/R VILLA FOR SALE 9611226364
T ZED HOMES 3 BEDROOM FLAT FOR SALE 9611226364

我期待的结果：

A Flat is for sale at  Cooke Town
Prestige Sunnyside
Jayanagar 2nd Block, 4000 sft Plot for Sale 9845012673
PRESTIGE OZONE TYPE D 3 B/R VILLA FOR SALE 9611226364
T ZED HOMES 3 BEDROOM FLAT FOR SALE 9611226364

Answer 1

请注意，例如，http://bangalore.craigslist.co.in/reb/d/flat-is-for-sale-at-cooke-town/6266183606.html 上没有与 'a.showcontact' 选择器匹配的链接，因此请遵循块

try:
    link = base + tree.cssselect('a.showcontact')[0].attrib['href']
except IndexError:
    link = ""

将返回link = ""

然后，当您调用 if ano_page_link: 时，if 块中的所有命令都将被忽略，因为条件 if "" 是 False 并且什么都不会打印出来

您可以尝试以下方法：

def parse_doc(title, ano_page_link):

    if ano_page_link:
        page = requests.get(ano_page_link).text            
        tel = re.findall(r'\d{10}', page)[0] if re.findall(r'\d{10}', page) else ""
        print(title, tel)
    else:
        print(title)

Answer 2

您可以通过将收集数据和打印数据这两个任务分开来获得更大的灵活性。以后想要扩展时添加更多信息会更容易。

def collect_info(medium_link):
    response = requests.get(medium_link).text
    tree = html.fromstring(response)

    title = get_title(tree)
    contact_link = get_contact_link(tree)
    tel = get_tel(contact_link) if contact_link else ''

    return title, tel


def get_title(tree):
    try:
        name = tree.cssselect('span#titletextonly')[0].text
    except IndexError:
        name = ""
    return name

def get_contact_link(tree):
    try:
        link = base + tree.cssselect('a.showcontact')[0].attrib['href']
    except IndexError:
        link = ""
    return link

def get_tel(ano_page_link):
    page = requests.get(ano_page_link).text
    tel = re.findall(r'\d{10}', page)[0] if re.findall(r'\d{10}', page) else ""
    return tel

def print_info(title, tel):
    if tel:
        fmt = 'Title: {title}, Phone: {tel}'
    else:
        fmt = 'Title: {title}'
    print(fmt.format(title=title, tel=tel))

if __name__ == '__main__':
    for link in url_list:
        title, tel = collect_info(link)
        print_info(title, tel)