How to transform link extraction into valid element with Selenium Python?-CodePudding

I need to transform the return of the URL I'm extracting into a valid element.

the code captures the URLs and then enters each one of them to extract data from the page

terminal exit

terminal error

links = []
classe = driver.find_elements(By. XPATH, "//*[@class='LinksShowcase_UrlContainer__kMj_n']/p")
for i in classe:
    sleep(0.5)
    links.append(i)
    print(links)
    sleep(2)
for linkAtual in links:
    driver.get(linkAtual)

I cannot share the link, as it is a platform that needs to create an account and be accepted, but the link is as text within the TAG 'P', follow the image of the page

CodePudding user response：

find_elements method return a list of WebElement objects.
These are not links (strings).
WebElement is a reference, a pointer to physical web element on the web page.
WebElement may containg href attribute that normally contains some link.
As mentioned by KunduK anchor elements are normally containing links, not p tag elements.
So, in case elements you collecting are containing links you can extract these links from the WebElement objects and use them later.
I can't debug this code since you did not share a link to page you working on as well as you did not share all your Selenium code, but I guess something like following can work:

links = []
classe = driver.find_elements(By. XPATH, "//*[@class='LinksShowcase_UrlContainer__kMj_n']/p")
for i in classe:
    link = i.get_attribute("href")
    print(link)
    links.append(link)
for linkAtual in links:
    driver.get(linkAtual)

UPD
In your case it is not href attribute but a text content. So, you can simply extract the text as following:

links = []
classe = driver.find_elements(By. XPATH, "//*[@class='LinksShowcase_UrlContainer__kMj_n']/p")
for i in classe:
    link = i.text
    print(link)
    links.append(link)
for linkAtual in links:
    driver.get(linkAtual)