Python Can'；t使用Beauty soup仅打印文本

我正在努力在python3上创建我的第一个项目之一。当我使用以下代码时：

def scrape_offers():
    r = requests.get("https://www.olx.bg/elektronika/kompyutrni-aksesoari-chasti/aksesoari-chasti/q-1070/?search%5Border%5D=filter_float_price%3Aasc", cookies=all_cookies)
    soup = BeautifulSoup(r.text,"html.parser")
    offers = soup.find_all("div",{'class':'offer-wrapper'})

    for offer in offers:
        offer_name = offer.findChildren("a", {'class':'marginright5 link linkWithHash detailsLink'})
        print(offer_name.text.strip())

我得到以下错误：

Traceback (most recent call last):
  File "scrape_products.py", line 45, in <module>
    scrape_offers()
  File "scrape_products.py", line 40, in scrape_offers
    print(offer_name.text.strip())
  File "/usr/local/lib/python3.7/site-packages/bs4/element.py", line 2128, in __getattr__
    "ResultSet object has no attribute '%s'. You're probably treating a list of elements like a single element. Did you call find_all() when you meant to call find()?" % key
AttributeError: ResultSet object has no attribute 'text'. You're probably treating a list of elements like a single element. Did you call find_all() when you meant to call find()?

回溯（最近一次呼叫最后一次）：
文件“scrape_products.py”，第45行，在
scrape_offers（）
文件“scrape_products.py”，第40行，在scrape_offers中
打印（offer_name.text.strip（））
文件“/usr/local/lib/python3.7/site packages/bs4/element.py”，第2128行，在__
“ResultSet对象没有属性“%s”。您可能将元素列表视为单个元素。当您打算调用find（）时是否调用find_all（）？%key？”
AttributeError:ResultSet对象没有属性“text”。您可能将元素列表视为单个元素。当您打算调用find（）时，是否调用了find_all（）？

我读过很多关于StackOverFlow的类似案例，但我还是情不自禁。如果有人有任何想法，请帮助：）

---

注意：如果我在没有

的情况下运行代码。text

将显示整个

findchildren返回一个列表。有时会得到一个空列表，有时会得到一个包含一个元素的列表

您应该添加if语句来检查返回列表的长度是否大于1，然后打印文本

import requests
from bs4 import BeautifulSoup
def scrape_offers():
    r = requests.get("https://www.olx.bg/elektronika/kompyutrni-aksesoari-chasti/aksesoari-chasti/q-1070/?search%5Border%5D=filter_float_price%3Aasc")
    soup = BeautifulSoup(r.text,"html.parser")
    offers = soup.find_all("div",{'class':'offer-wrapper'})

    for offer in offers:
        offer_name = offer.findChildren("a", {'class':'marginright5 link linkWithHash detailsLink'})
        if (len(offer_name) >= 1):
            print(offer_name[0].text.strip())

scrape_offers()