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)
Related
I am trying to click the search button in this link here
I would like to click the search button and then download all URL's on the next page but currently it is finding monthly list
The link below takes you to a screenshot of my code where it outputs the monthly list button instead of searcj/
Python code selenium
Welcome to Stack Overflow!
You can use the .click() method to click an element object, and the .find_element_by_xpath() method to find the element. I've located that the element's full XPATH is /html/body/div/div/div[3]/div[3]/div/form/fieldset/div[5]/input[2].
You can implement all the pieces together like so:
driver.find_element_by_xpath("/html/body/div/div/div[3]/div[3]/div/form/fieldset/div[5]/input[2]").click()
if you are starting I recommend you using some extension to help you find the xPath of the elements.
I used Xpath Helper from Chrome, but you can use any other.
The Xpath of the Search button is:
/html[#class='js']/body/div[#id='idox']/div[#id='pa']/div[#class='container'][2]/div[#class='content']/div[#class='tabcontainer']/form[#id='weeklyListForm']/fieldset/div[#class='buttons']/input[#class='button primary']
After you have the XPath you can use it in the selenium driver to find the elements.
This can be enough, but I recommend you to learn in depth how this works to know exactly what it is doing.
I'm searching for a tag in class.I tried many methods but I couldn't get the value.
see source code
The data I need is inside the "data-description".
How can i get the "data-description" ?
I Tried some method but didn't work
driver.find_element_by_name("data-description")
driver.find_element_by_css_selector("data-description")
I Solved this method:
icerisi = browser.find_elements_by_class_name('integratedService ')
for mycode in icerisi:
hizmetler.append(mycode.get_attribute("data-description"))
Thanks for your help.
I think css selector would work best here. "data-description" isn't an element, it's an attribute of an element. The css selector for an element with a given attribute would be:
[attribute]
Or, to be more specific, you could use:
[attribute="attribute value"]
Here's a good tip:
Most web browsers have a way of copying an elements Selector or XPATH. For example, in Safari if you view the source code then right-click on an element it will give you the option to copy it. Then select XPATH or Selector and in your code use driver.find_element_by_xpath() or driver.find_element_by_css_selector(). I am certain Google Chrome and Firefox have similar options.
This method is not always failsafe, as the XPATH can be very specific, meaning that slight changes to the website will cause your script to crash, but it is a quick and easy solution, and is especially useful if you don't plan on reusing your code months or years later.
for chrome, I install ChroPath to find elements on the page.
I want to find XPath for like elements on Instagram Page, but seems that not work :
//span[contains(#class,'glyphsSpriteHeart__outline__24__grey_9 u-__7')]
also, I try it :
/html[1]/body[1]/div[3]/div[1]/div[2]/div[1]/article[1]/div[2]/section[1]/span[1]/button[1]/span[1]
when selenium click :
elenium.common.exceptions.NoSuchElementException: Message: no such element: Unable to locate element: {"method":"css selector","selector":"div._2dDPU.vCf6V div.zZYga div.PdwC2._6oveC article.M9sTE.L_LMM.JyscU div.eo2As section.ltpMr.Slqrh span.fr66n button.coreSpriteHeartOpen.oF4XW.dCJp8 > span.glyphsSpriteHeart__outline__24__grey_9.u-__7"}
how can I find XPath? any good extension or something?
how can I find XPath? any good extension or something?
You cannot "find" the Xpath of an element. There are many, many XPath's that will find any element. Some will be stable, others will be unstable. The decision on which Xpath to use is based upon your understanding and experience of Selenium, and you understanding of how the Application Under Test is written and behaves.
If you are looking for a tool to experiment with different XPaths, then Chrome's built-in Developer Tools Console allows you to test both Xpath & CSS Selectors;
In your specific scenario about finding elements by class name, then CSS Selector is a much better choice than XPath as CSS selectors will treat multiple classes as an array where as XPath sees "class" as a literal string, hence why you needed to use "contains".
This might help:
https://selectorgadget.com/
This as well, to understand what you are manipulating:
https://www.w3schools.com/xml/xpath_syntax.asp
As for your example where you go down the tree using index numbers (ie: /html[1]/body[1]), A slight change in the site will make your script to fail. Find a way to build something more robust! Also have a look at CSS selectors if you object's appearance is known in advance.
To get all like buttons on instagram use css selector below:
span[aria-label="Like"]
You can get some helpful details here: https://www.w3schools.com/cssref/css_selectors.asp
I just started using the selenium, but after reading the docs I wasn't able to understand how to properly perform a wait (using EC) until webdriver identifies presence of an element which is a child of another element.
Let me explain. I want a specific element which I can access in 2 steps:
#find major group
listings = driver.find_element_by_id("new-listings");
#find subelement
checkbox = listings.find_element_by_class_name("listing-category")
That's OK, but I want to use EC to ensure that the checkbox is present. I cannot use smth like:
checkbox = driver.wait.until(EC.presence_of_element_located(
(By.CLASS_NAME, "listing-category")))
Just because there is a bunch of other similar listing-category elements. And sadly the only way to locate checkbox is through the nested request like illustrated earlier (it doesn't have any id or whatever).
How to properly express it?
Small extra question: in general, is it very bad for performance to use EC instead of just hitting elements and hope that they are already available? 99% of the time I will not need to wait for element to appear, so I just try to handle rare situation when browser is less responsive than usual. But I am not sure if EC introduces significant overhead for just creating the event handler etc.
I would suggest that you use a CSS selector to accomplish this in one go. You can use the method you are describing but with a CSS selector you can find the child of an element with a single locator. In this case the CSS selector would be #new-listings .listing-category. In CSS selectors, # indicates an ID and . indicates a class name. The space between the two parts indicates a descendant. If you wanted a child (a direct descendant), you would use >, e.g. #new-listings > .listing-category.
To your question about EC and performance, no it doesn't slow things down. It always checks immediately to see if the condition is true, if it's not then it waits so it generally doesn't hurt to add a wait.
If you are new to CSS selectors, here are some links that you can read up on and learn about them. They are very powerful and quick locators.
CSS selector reference
CSS selector tips
Advanced CSS selectors
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();