如何抓取网站中的所有页面

Question

https://www.bestbuy.com/site/promo/health-fitness-deals

I want to loop through these 10 pages and scrape their names and hrefs Below is my code which only scrapes the 1st page continuously 10 times:

def name():
    for i in range(1, 11):
        tag = driver.find_elements_by_xpath('/html/body/div[4]/main/div[9]/div/div/div/div/div/div/div[2]/div[2]/div[3]/div/div[5]/ol/li[3]/div/div/div/div/div/div[2]/div[1]/div[2]/div/h4')
        for a in tag:
            for name in a.find_elements_by_tag_name('a'):
                links = name.get_attribute("href")
                names = name.get_attribute('text')
                watches_name.append(names)
                watches_link.append(links)
                # print(watches_name)
                # print(watches_link)


name()

Answer 1

If you want to get elements from next pages then you have to click() on link >

driver.find_element_by_css_selector('.sku-list-page-next').click()

---

Minimal working code with other changes.

I reduced xpath to something much simpler. And I keep name, link as pair because it is simpler to write in file CSV or in database or to filter and sort.

I had to use longer sleep - sometimes my browser needs more time to update elements on page.

from selenium import webdriver
import time

url = 'https://www.bestbuy.com/site/promo/health-fitness-deals'

driver = webdriver.Firefox()
driver.get(url)

time.sleep(2)

# page "Hello! Choose a Country" - selecting Unitet State flag
driver.find_element_by_class_name('us-link').click()

items = []

for page in range(1, 11):

    print('\n[DEBUG] wait 15 seconds to update page\n')
    time.sleep(15)

    print('\n--- page', page, '---\n')

    all_links = driver.find_elements_by_css_selector('#main-results h4 a')
    for a in all_links:
        link = a.get_attribute("href")
        name = a.get_attribute('text')
        items.append( [name, link] )
        print(name)

    print('\n[DEBUG] click next\n')
    driver.find_element_by_css_selector('.sku-list-page-next').click()
    
#print(items)

BTW:

This method could be done with while True and some method to recognize if there is link > - and exit loop when there is no > . This way it could work with any number of pages.

---

Other method.

When you manually visit few pages then you should see that second page has url with ?cp=2 , third with ?cp=3 , etc. so you could use it to load pages

driver.get(url + '?cp=' + str(page+1) )

---

Minimal working code.

from selenium import webdriver
import time

url = 'https://www.bestbuy.com/site/promo/health-fitness-deals'

driver = webdriver.Firefox()
driver.get(url)

time.sleep(2)

# page "Hello! Choose a Country" - selecting Unitet State flag
driver.find_element_by_class_name('us-link').click()

items = []

for page in range(1, 11):

    print('\n[DEBUG] wait 15 seconds to update page\n')
    time.sleep(15)

    print('\n--- page', page, '---\n')

    all_links = driver.find_elements_by_css_selector('#main-results h4 a')
    for a in all_links:
        link = a.get_attribute("href")
        name = a.get_attribute('text')
        items.append( [name, link] )
        print(name)

    print('\n[DEBUG] load next url\n')
    driver.get(url + '?cp=' + str(page+1) )
    
#print(items)

This method could also use while True and variable page to get any number of pages.

---

EDIT:

Versions with while True

from selenium import webdriver
import time

url = 'https://www.bestbuy.com/site/promo/health-fitness-deals'

driver = webdriver.Firefox()
driver.get(url)

time.sleep(2)

# page "Hello! Choose a Country" - selecting Unitet State flag
driver.find_element_by_class_name('us-link').click()

items = []

page = 1

while True:

    print('\n[DEBUG] wait 15 seconds to update page\n')
    time.sleep(15)

    print('\n--- page', page, '---\n')

    all_links = driver.find_elements_by_css_selector('#main-results h4 a')
    for a in all_links:
        link = a.get_attribute("href")
        name = a.get_attribute('text')
        items.append( [name, link] )
        print(name)

    page += 1

    print('\n[DEBUG] load next url\n')
    driver.get(url + '?cp=' + str(page) )

    if driver.title == 'Best Buy: Page Not Found':
        print('\n[DEBUG] exit loop\n')
        break
    
#print(items)

and

from selenium import webdriver
import time

url = 'https://www.bestbuy.com/site/promo/health-fitness-deals'

driver = webdriver.Firefox()
driver.get(url)

time.sleep(2)

# page "Hello! Choose a Country" - selecting Unitet State flag
driver.find_element_by_class_name('us-link').click()

items = []

page = 1

while True:

    print('\n[DEBUG] wait 15 seconds to update page\n')
    time.sleep(15)

    print('\n--- page', page, '---\n')

    all_links = driver.find_elements_by_css_selector('#main-results h4 a')
    for a in all_links:
        link = a.get_attribute("href")
        name = a.get_attribute('text')
        items.append( [name, link] )
        print(name)

    page += 1
    
    print('\n[DEBUG] click next\n')
    item = driver.find_element_by_css_selector('.sku-list-page-next')
    if item.get_attribute("href"):
        item.click()
    else:
        print('\n[DEBUG] exit loop\n')
        break        
    
#print(items)

Answer 2

I guess if your code is working right, you will just need to click the pagination button. I found it can be located with the help of css selector ('#Caret_Right_Line_Sm') . Try adding this line to your function:

def name():
    for i in range(1, 11):
        tag = driver.find_elements_by_xpath('/html/body/div[4]/main/div[9]/div/div/div/div/div/div/div[2]/div[2]/div[3]/div/div[5]/ol/li[3]/div/div/div/div/div/div[2]/div[1]/div[2]/div/h4')
        for a in tag:
            for name in a.find_elements_by_tag_name('a'):
                links = name.get_attribute("href")
                names = name.get_attribute('text')
                watches_name.append(names)
                watches_link.append(links)
                # print(watches_name)
                # print(watches_link)
        driver.find_elements_by_css_selector('#Caret_Right_Line_Sm')[1].click()

name()