How to get href link from list using selenium python

Question

Here is the code of the web

The xpath of search-results-list container grid is

 //[@id="product_type_products_list"]/div/div[2]/div

and the xpath of result is

 //*[@id="product_type_products_list"]/div/div[2]/div/div[1]

I have try using :

elems = driver.find_elements_by_xpath('//*[@id="product_type_products_list"]/div/div[2]/div')
url = driver.find_element_by_link_text(elems[0].text).get_attribute("href")
print(url)

this give the link to the beginning of the web.

Thank you for your consideration.

Dmitri T · Accepted Answer · 2019-07-01 12:43:37Z

1

The code you've provided doesn't look like a valid HTML to me, however you can try the following XPath expression:

//div[@class='result']/descendant::a

More information:

answered Jul 1, 2019 at 12:43

Dmitri T

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1 Comment

Somphon Rueangari Over a year ago

I don't know whether i can post like this,i edited the question anywhere thank for your help.

Arjun Tandon · Accepted Answer · 2019-07-03 01:09:46Z

1

Try Narrowing it down to the <'A> Tag by appending the xpath like so:

elems = driver.find_elements_by_xpath('.//*[@id="product_type_products_list"]/div/div[2]/div/div[1]/a')

Then just retrieve the href attribute like you did earlier but using the same element:

url = elems[0].get_attribute("href")

edited Jul 3, 2019 at 1:09

answered Jun 30, 2019 at 13:57

Arjun Tandon

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2 Comments

Corey Goldberg Over a year ago

in your answer, elems is a list of elements, which does not have a get_attribute method.

Arjun Tandon Over a year ago

Thanks for pointing it out Corey. I added the index in there.

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