如何从BeautifulSoup中的span获取数据？

Question

这是我的代码，我想获取位置的名称和链接，变量“lugares”查找多个项目容器，但我只想要第一个[0]；然后进行for循环，但我找不到span类

from bs4 import BeautifulSoup
import requests

b=[]
i="https://www.vivanuncios.com.mx"
url = "https://www.vivanuncios.com.mx/s-renta-inmuebles/estado-de-mexico/v1c1098l1014p1"

encabezado = {"User-Agent":"Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/94.0.4606.71 Safari/537.36",'Accept-Language': 'en-US, en;q=0.5'}

page =requests.get(url,headers=encabezado)

soup = BeautifulSoup(page.content,"html.parser")

lugares = soup.find_all("div",{"class":"items-container"})

lugares=lugares[0]
print(len(lugares))

for lugar in lugares:
    
    locationlink = i + str(lugar.find("span",{"class":"item"}).find("a")["href"])

    location= lugar.find("span",{"class":"item"}).text
    a=[location,locationlink]
    
    b.append(a)

Answer 1

首先，您需要在第一个Lugares lugares[0]中获得所有spans

然后需要对每个跨度进行迭代，以获得每个位置的链接和文本

代码：

from bs4 import BeautifulSoup
import requests

b=[]
i="https://www.vivanuncios.com.mx"
url = "https://www.vivanuncios.com.mx/s-renta-inmuebles/estado-de-mexico/v1c1098l1014p1"

encabezado = {"User-Agent":"Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/94.0.4606.71 Safari/537.36",'Accept-Language': 'en-US, en;q=0.5'}

page =requests.get(url,headers=encabezado)

soup = BeautifulSoup(page.content,"html.parser")

lugares = soup.find_all("div",{"class":"items-container"})

#lugares=lugares[0]
print(len(lugares))

# get all spans
spans = lugares[0].find_all("span",{"class":"item"})

# itreate throw each span
for span in spans: 
    # get location text
    location = span.find("a").text

    # locationlink builder
    site = "www.vivanuncios.com.mx"
    link = span.find("a")["href"]
    locationlink = f"{site}{link}"   

    a = [location,locationlink]
    b.append(a)

print (b[0])

输出：

['Huixquilucan', 'www.vivanuncios.com.mx/s-renta-inmuebles/huixquilucan/v1c1098l10689p1']