python selenium how to get link from google

Question

How do I get a link from Google? I tried the following ways, but none of them worked:

find_elements_by_xpath("//*[@class='r']/@href")
driver.find_element_by_xpath("//a").get_attribute("href")
driver.find_element_by_xpath("//a").get_attribute(("href")[2])

I am receiving "none"

I need to get google link, not link to the site (e.g. not www.stackoverflow.com). It's highlighted on this image:

Try as `driver.get_element_by_css_selector("h3.r > a").get_attribute("href")` and let me know..:) — Saurabh Gaur, Aug 28 '16 at 18:14
@SaurabhGaur 'WebDriver' object has no attribute 'get_element_by_css_selector' i tried "driver.find_element_by_css_selector" but response http://stackoverflow.com . i have to get other link , as i said — ömer g., Aug 29 '16 at 18:01
What do you mean other link??.. and yes it's just typo mistake it should be `driver.find_element_by_css_selector` — Saurabh Gaur, Aug 29 '16 at 18:10
@SaurabhGaur please look at the picture. i need the right link. /url?sa=t&rct=j&... — ömer g., Aug 29 '16 at 18:16

score 0 · Answer 1 · edited May 23 '17 at 12:32

You maye have multiple issues here: First and third options are a not a valid xpath. Second options may find more than one matches, so it will return the first that fits, which is not necessarily the one you want. So I suggest:

Make find specific enough to locate a proper element. I'd suggest find_element_by_link_text if you know the name of the link you are going to choose:
```
link = driver.find_element_by_link_text('Stack Overflow')
```
Given you chose the right link, you should be able to get the attribute:
```
href = link.get_attribute('href')
```
If the first statement throws an exception (most likely element not found), you may need to wait for the element to appear on the page, as described here

python selenium how to get link from google

1 Answers1