Scrapy/Selenium: How do I follow the links in 1 webpage? - selenium

Question

I am new to web-scraping.
I want to go to Webpage_A and follow all the links there.
Each of the link lead to a page where I can select some button and download the data in an Excel file.
I tried the below code. But I believe there is an error with
if link:
yield SeleniumRequest(
Instead of using "SeleniumRequest" to follow the links, what should I use?
If using pure Scrapy, I know I can use
yield response.follow(
Thank you
class testSpider(scrapy.Spider):
name = 'test_s'
def start_requests(self):
yield SeleniumRequest(
url='CONFIDENTIAL',
wait_time=15,
screenshot=True,
callback=self.parse
)
def parse(self, response):
tables_name = response.xpath("//div[#class='contain wrap:l']//li")
for t in tables_name:
name=t.xpath(".//a/span/text()").get()
link = t.xpath(".//a/#href").get()
if link:
yield SeleniumRequest(
meta={'table_name': name},
url= link,
wait_time=15,
screenshot=True,
callback=self.parse_table
)
def parse_table(self, response):
name = response.request.meta['table_name']
button_select=response.find_element_by_xpath("(//a[text()='Select All'])").click()
button_st_yr=response.find_element_by_xpath("//select[#name='ctl00$ContentPlaceHolder1$StartYearDropDownList'] /option[1]").click()
button_end_mth=response.find_element_by_xpath("//select[#name='ctl00$ContentPlaceHolder1$EndMonthDropDownList']/option[text()='Dec']").click()
button_download=response.find_element_by_xpath("//input[#id='ctl00_ContentPlaceHolder1_DownloadButton']").click()
yield{
'table_name': name
}