用python爬取小说时，拼接url类型错误

Question

问题遇到的现象和发生背景

用python爬取网页时，拼接url类型错误，运行后显示拼接的那段url需要转换为str类型，但是如果强制转换后输出的url就会带有[]，是无效的url，就不能继续进行对新的url的解析了

问题相关代码，请勿粘贴截图

import requests
from lxml import html
etree=html.etree
if __name__ == '__main__':
    url="https://www.shicimingju.com/book/sanguoyanyi.html"
    headers = {
    "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/96.0.4664.93 Safari/537.36"}
    page_text=requests.get(url,headers).content.decode('utf-8')
    tree=etree.HTML(page_text)
    li_link=tree.xpath('//div[@class="book-mulu"]/ul/li')
    fp = open('./三国.txt', 'w', encoding='utf-8')
    for link in li_link:
        #拼接错误！！！！！！
        #li_url="https://www.shicimingju.com"+str(link.xpath('./a/@href'))
        li_url = "https://www.shicimingju.com" + link.xpath('./a/@href')
        print(li_url)
        title=link.xpath('./a/text()')
        detail_text=requests.get(url=li_url,headers=headers).content.decode('utf-8')
        tree = etree.HTML(detail_text)

        content=tree.xpath('//div[@class="card bookmark-list"]/div/text()')
        fp.write(str(title+content))
        # print(content,"结束")
        print(title,'成功！！')

运行结果及报错内容

```

```li_url = "https://www.shicimingju.com" + link.xpath('./a/@href')
TypeError: can only concatenate str (not "list") to str

我的解答思路和尝试过的方法

加str强制转换

我想要达到的结果

拼接生成正确的url

Answer 1

li_url = "https://www.shicimingju.com" + link.xpath('./a/@href')[0]

这样子就行了