Python: Selenium: find_elements_by_xpath max 10

Question

So, I right now have a working piece of code that looks like this;

if browser.find_elements_by_xpath("//div[contains(@class, 'dealspg_item_cell')]"):
    snipes = browser.find_elements_by_xpath("//div[contains(@class, 'dealspg_item_cell')]")
else:
    print "Cant find snipes.. Retrying..."
    browser.get("https://www.rolimons.com/deals")
    time.sleep(2)
    print "Reloaded browser... Retrying..."
    if browser.find_elements_by_xpath("//div[contains(@class, 'dealspg_item_cell')]"):
        snipes = browser.find_elements_by_xpath("//div[contains(@class, 'dealspg_item_cell')]")
    else:
        print "Shutting down engine..."
        browser.quit()
        checking = False
        print "Restarting script..."
        break

This all works fine. And it gets 60 elements. Now I only need the first 10. How can I make it that it will only find_elements_by_xpath with a limit of 10?

Is there any way to do this?

Thanks!

==== EDIT ====

This code takes at the moment 4 seconds. And I hope to reduce this time by putting the max on 10.

Answer 1

you can use:

//[yourXpath]/*[position()<=10]

in Your case

//div[contains(@class, 'dealspg_item_cell')]/*[position()<=10]

Answer 2

Although using xpath is good answer. You can also refer css selector as per performance respective. Personally i faced such issue where xpath is not working but CSS selector is

div.dealspg_item_cell:nth-child(-1n+10)

Explanation nth-child(-1n+10)

+10 indicates here the node position, from where we want to pick the elements -1n all nodes before the 10th node

Lets say there are total 11 element then selection would be something like :

Another example would be nth-child(1n+10) , It means all other nodes after 10th node

You can explore more from here