Scraper无法打印所有结果

Question

我用python编写了一个脚本，从craigslist中抓取五个条目的“name”和“phone”。我面临的问题是，当我运行脚本时，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其他链接（联系信息），因此他们不需要再次请求打开任何其他页面。然而，这两个链接没有（联系信息）链接不能渗透通过“如果没有网页链接：”在我的第二个功能声明，永远不会得到打印。如何修复此缺陷，以便无论是否有电话号码刮板将打印所有五个结果。你知道吗

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)

结果是：

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

我期待的结果是：

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

注意，例如，在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，因此下面的块

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

将返回link = ""

然后，当调用if ano_page_link:时，if块中的所有命令都会被忽略，因为条件if ""是False，不会输出任何内容

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

通过将收集数据和打印数据这两个任务分开，您可以获得更大的灵活性。以后当您想扩展时，添加更多信息会更容易。你知道吗

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)