输入页码实现翻页跳转

try:
    self.browser.get(request.url)
    if page > 1:
        print(page)
        # input = self.wait.until(EC.presence_of_element_located((By.CSS_SELECTOR, '# J_bottomPage > span.p-skip > input')))
        # input.clear()
        # input.send_keys(page)
        # time.sleep(5)
        '''
        模拟点击下一页
        '''
        # fp_next = self.browser.find_element_by_xpath('//*[@id = "J_bottomPage"]/span[1]/a[9]/em')
        # # fp_next = browser.find_element_by_css_selector('a.fp-next')
        # # 点击下一页
        # fp_next.click()

        # 将网页中输入跳转页的输入框赋值给input变量 EC.presence_of_element_located，判断输入框已经被加载出来
        #input = self.wait.until(EC.presence_of_element_located((By.CSS_SELECTOR, '# J_bottomPage > span.p-skip > input')))
        input = self.browser.find_element_by_css_selector('span.p-skip input')
        input.clear()
        input.send_keys(page)
        print('成功输入页码')
        print('************************')
        # 将网页中调准页面的确定按钮赋值给submit变量，EC.element_to_be_clickable 判断此按钮是可点击的
        #submit = self.wait.until(EC.element_to_be_clickable((By.CSS_SELECTOR, '# J_bottomPage > span.p-skip > a')))
        submit = self.browser.find_element_by_css_selector('span.p-skip a')
        submit.click()  # 点击按钮
        time.sleep(5)
        print('成功点击下一页')
        print('************************')
        time.sleep(15)

        # 判断当前页码出现在了输入的页面中，EC.text_to_be_present_in_element 判断元素在指定字符串中出现
        #self.wait.until(EC.text_to_be_present_in_element((By.CSS_SELECTOR, '#J_bottomPage > span.p-num > a.curr'),str(page)))
        # 等待 #J_goodsList 加载出来，为页面数据，加载出来之后，在返回网页源代码
        #self.wait.until(EC.text_to_be_present_in_element((By.CSS_SELECTOR, '#J_bottomPage > span.p-num > a.curr'),str(page)))
    print(request.url)
    return HtmlResponse(url=request.url, body=self.browser.page_source, request=request, encoding='utf-8',status=200)
except TimeoutException:
    return HtmlResponse(url=request.url, status=500, request=request)