這是我的代碼，我想獲取位置的名稱和鏈接，變數“lugares”找到多個專案容器，但我只想要第一個 [0]；然后進入 for 回圈，但我找不到 span 類。
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)
uj5u.com熱心網友回復：
首先，您需要spans在第一個 Lugares 中獲取所有內容lugares[0]。
然后您需要對每個跨度進行迭代以獲取每個位置的鏈接和文本。
作業代碼：
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])
結果樣本：
['Huixquilucan', 'www.vivanuncios.com.mx/s-renta-inmuebles/huixquilucan/v1c1098l10689p1']
uj5u.com熱心網友回復：
有多種選擇可以實作目標，最好的選擇取決于您的期望以及在后續程序中想要使用此資訊做什么。
第一個選項
如果您只是在尋找第一個位置的資訊，您可以執行以下操作：
lugar = soup.select_one('div.items-container a')
b = [lugar.text, f'{i}{lugar["href"]}']
或者
lugar = soup.select('div.items-container a')[0]
b = [lugar.text, f'{i}{lugar["href"]}']
無論選擇第一個<a>在<div>帶班items-container。
輸出
['Huixquilucan','https://www.vivanuncios.com.mx/s-renta-inmuebles/huixquilucan/v1c1098l10689p1']
另類
如果您有興趣一次獲取所有資訊，則應該使用 的串列dicts，因此稍后您只需對其進行迭代并獲取所有資訊：
[{'name':x.text, 'link':f'{i}{x["href"]}'} for x in soup.select('div.items-container a')]
輸出
[{'name': 'Huixquilucan',
'link': 'https://www.vivanuncios.com.mx/s-renta-inmuebles/huixquilucan/v1c1098l10689p1'},
{'name': 'Naucalpan',
'link': 'https://www.vivanuncios.com.mx/s-renta-inmuebles/naucalpan/v1c1098l10710p1'},
{'name': 'Atizapán',
'link': 'https://www.vivanuncios.com.mx/s-renta-inmuebles/atizapan/v1c1098l10662p1'},
{'name': 'Metepec',
'link': 'https://www.vivanuncios.com.mx/s-renta-inmuebles/metepec-edomex/v1c1098l10707p1'},...]
示例（顯示兩者的結果）
from bs4 import BeautifulSoup
import requests
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")
lugar = soup.select_one('div.items-container a')
b = [lugar.text, f'{i}{lugar["href"]}']
print(f'First lugar:\n {b} \n')
## or alternative option
allLugaros = [{'name':x.text, 'link':f'{i}{x["href"]}'} for x in soup.select('div.items-container a')]
print(f'First lugar from lugaros (list of dict):\n {allLugaros[0]} \n')
print(f'All lugaros as list of dict:\n {allLugaros} \n')
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