我想得到email，但他们会提供错误的输出，这些是页面链接https://zoekeenadvocaat.advocatenorde.nl/advocaten/soesterberg/mevrouw-mr-mm-strengers/11094237420

import scrapy
from scrapy.http import Request
from bs4 import BeautifulSoup
from selenium import webdriver
import time
from scrapy_selenium import SeleniumRequest
import re

class TestSpider(scrapy.Spider):
    name = 'test'
    page_number=1
    def start_requests(self):
        yield SeleniumRequest(
                    url = "https://zoekeenadvocaat.advocatenorde.nl/zoeken?q=&type=advocaten&limiet=10&sortering=afstand&filters%5Brechtsgebieden%5D=%5B%5D&filters%5Bspecialisatie%5D=0&filters%5Btoevoegingen%5D=0&locatie%5Badres%5D=Holland&locatie%5Bgeo%5D%5Blat%5D=52.132633&locatie%5Bgeo%5D%5Blng%5D=5.291266&locatie%5Bstraal%5D=56&locatie%5Bhash%5D=67eb2b8d0aab60ec69666532ff9527c9&weergave=lijst&pagina=1",
                    wait_time = 3,
                    screenshot = True,
                    callback = self.parse,
                    dont_filter = True
                )

    def parse(self, response):
        books = response.xpath("//span[@class='h4 no-margin-bottom']//a//@href").extract()
        for book in books:
            url = response.urljoin(book)
            yield Request(url, callback=self.parse_book)

    def parse_book(self, response):
        title=response.css(".title h3::text").get()
        advocaten=response.css(".secondary::text").get()
        detail=response.xpath("//section[@class='lawyer-info']")
        for i in range(len(detail)):
            if re.search("@",detail[i].get()):
                d1=detail[i].xpath("//div[@class='column small-9']//a//@href").get()
                print(d1)