I am trying to click a link by:
driver.find_element_by_css_selector("a[href='javascript:openhistory('AXS0077')']").click()
This works nice if the link opens in a new window but in this case the link actually opens a pop up window. When I try clicking the link with this method, using selenium it gives me an error:
Message: u"The given selector
a[href='javascript:openhistory('AXS0077')'] is either invalid or does
not result in a WebElement. The following error
occurred:\nInvalidSelectorError: An invalid or illegal selector was
specified"
Is this not the right way ? because
I think there may be some different way to deal with pop windows.
Your css selector could be more generic, perhaps:
driver.find_element_by_css_selector("a[href^='javascript']").click()
You've got all kinds of crazy overlapping quotation marks there. You're probably confusing it.
I have more success using find_by_xpath
Take this site as an example popups
I use firebug to inspect the element and get the xpath.
Then using the following works perfectly.
from selenium import webdriver
baseurl="http://www.globalrph.com/davescripts/popup.htm"
dr = webdriver.Firefox()
dr.get(baseurl)
dr.find_element_by_xpath("/html/body/div/center/table/tbody/tr[7]/td/div/table/tbody/tr/td[2]/div[1]/form/table/tbody/tr[4]/td[1]/a").click()
Related
python selenium driver:
No thanks
when i open a google page the, there is a small window asking if i want to sign in or not. i want to click on "No thanks" button, which is as shown above.
i have tried these methods so far, but i keep getting errors. None of the following is working.
#self.driver.find_element(By.CSS_SELECTOR, 'button.M6CB1c')
#button=self.driver.find_elements(By.XPATH, '//button')
#abc=self.driver.find_elements(By.NAME, 'ZUkOIc').click()
#self.driver.find_element(By.TAG_NAME, 'button').click()
error message for the 1st line of code:
selenium.common.exceptions.NoSuchElementException: Message: no such element: Unable to locate element: {"method":"css selector","selector":".M6CB1c.rr4y5c"}
selenium.common.exceptions.NoSuchElement is caused when the element is not in the page at the current time.
TLDR;
What you're looking for is Explicit Wait in selenium. You need to use WebDriverWait with expected condition element_to_be_clickable.
When we load a page, modern pages tend to load javascript that can often manipulate DOM (html page objects). The proper way to handle this is to wait for the page or the required element to load, and then try to locate it.
The selenium waits section explains this very well with an example.
You should try this :
driver.find_element(By.XPATH, '//button[#id="W0wltc"])
I'm trying to learn how to click around using selenium. I have tried some different websites like reddit, google etc without success.
driver.get('https://www.dropbox.com/login')
driver.find_element_by_xpath('//a[#href="' + 'https://www.dropbox.com/forgot?email_from_login=' + '"]').click()
and
continue_link = driver.find_element_by_partial_link_text('Sign in')
They both exist but neither works. What am I doing wrong?
You're being a little specific in your xpath, so there're many more ways you might bugger something up. Using the xpath you can simply do:
driver.find_element_by_xpath("//*[#class='forgot-password-link']").click()
I make no assumptions, but just in case you aren't already, there's a very handy tool in Chrome's inspect element which lets you click an element and jumps to its node in the inspector.
Can you try the following:
from selenium.webdriver.common.by import By
driver.findElement(By.cssSelector(".login-button.signin-button.button-primary")).click()
You can use a shorter single class selector to Sign In button
driver.find_element_by_css_selector(".login-button").click()
Sign in with Google
driver.find_element_by_css_selector(".sign-in-text").click()
For forgot password
driver.find_element_by_css_selector(".forgot-password-link").click()
Single class css selectors will be the fastest method (faster than xpath and compound class)
I'm trying to click on the webpage "https://2018.navalny.com/hq/arkhangelsk/" from the website's main page. However, I get this error
selenium.common.exceptions.ElementNotInteractableException: Message:
There's nothing after "Message:"
My code
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
import time
browser = webdriver.Firefox()
browser.get('https://2018.navalny.com/')
time.sleep(5)
linkElem = browser.find_element_by_xpath("//a[contains(#href,'arkhangelsk')]")
type(linkElem)
linkElem.click()
I think xpath is necessary for me because, ultimately, my goal is to click not on a single link but on 80 links on this webpage. I've already managed to print all the relevant links using this :
driver.find_elements_by_xpath("//a[contains(#href,'hq')]")
However, for starters, I'm trying to make it click at least a single link.
Thanks for your help,
The best way to figure out issues like this, is to look at the page source using developer tools of your preferred browser. For instance, when I go to this page and look at HTML tab of the Firebug, and look for //a[contains(#href,'arkhangelsk')] I see this:
So the link is located within div, which is currently not visible (in fact entire sub-section starting from div with id="hqList" is hidden). Selenium will not allow you to click on invisible elements, although it will allow you to inspect them. Hence getting element works, clicking on it - does not.
What you do with it depends on what your expectations are. In this particular case it looks like you need to click on <label class="branches-map__toggle-label" for="branchesToggle">Список</label> to get that link visible. So add this:
browser.find_element_by_link_text("Список").click();
after that you can click on any links in the list.
HTML of the webpage I am trying to automate
I am trying to click the Link called 'Network'. I used the following statements but I am not able to get to work as it throws me the following error:
Error I get when I run the py script
Here is the command I used:
eleme=driver.find_element_by_xpath("//div[#id='maincontainer']//div[#id='tabmenu']//div[#class='tabmenu1']//ul[#class='tabmenu l1']//li[#class='tabmenu-item-network']//a[#href='Network']")
eleme.click()
Your xpath query is not good (href attribute is incorrect) so element can't be found, try with driver.find_element_by_xpath("//div[#id='maincontainer']//div[#id='tabmenu']//div[#class='tabmenu1']//ul[#class='tabmenu l1']//li[#class='tabmenu-item-network']//a[text()='Network']")
Or even better, driver.find_element_by_xpath("//div[#id='maincontainer']//a[text()='Network']")
EDIT:
You can click on link with driver.find_element_by_link_text('Network')
You should really check WebDriver API http://selenium-python.readthedocs.io/api.html#module-selenium.webdriver.remote.webelement
I am very much new to selenium WebDriver and I am trying to automate a page which has a button named "Delete Log File". Using FireBug I got to know that, the HTML is described as
and also the css selector is defined as "#DeleteLogButton" using firepath
hence I used
browser.find_element_by_css_selector("#DeleteLogButton").click() in webdriver to click on that button but its now working and also, I tried,
browser.find_element_by_id("DeleteLogButton").click() to click on that button. Even this did not find the solution for my problem...
Please help me out in resolving the issue.
Most of the times im using By.xpath and it works specially if you use contains in your xpath. For example : //*[contains(text(),'ABC')]
This will look for all the elements that contains string 'ABC'
In your case you can replace ABC with Delete Log File
try to find it by name like :
browser.find_element_by_name("Delete Log File").click();