I have these following elements in a page and I need to select the element with the value 18 inside using Python and selenium script. That is the second link. Here is the HTML code of the page
<a class="ui-state-default ui-state-highlight ui-state-active" href="#">17</a>
<a class="ui-state-default ui-state-highlight ui-state-active" href="#">18</a>
<a class="ui-state-default ui-state-highlight ui-state-active" href="#">19</a>
I am trying to use the following Python and Selenium code to click
elem = driver.find_element_by_xpath('//a[#class="ui-state-default"]').click()
But that does not work. How do I fix it?
elem = driver.find_element_by_link_text("18")
elem.click()
That will find the first element matching that XPath, which is the one with the value 17. To select the element whose value is 18, try this:
driver.find_element_by_link_text('18').click()
PS: You don't need to set the click event to a variable. Only do this if you need to multiple actions on the element (send_keys(), click(), text, etc.)
Related
Trying to scrape a website, I created a loop and was able to locate all the elements. My problem is, that the next button id changes on every page. So I can not use the id as a locator.
This is the next button on page 1:
<a rel="nofollow" id="f_c7" href="#" class="nextLink jasty-link"></a>
And this is the next button on page 2:
<a rel="nofollow" id="f_c9" href="#" class="nextLink jasty-link"></a>
Idea:
next_button = browser.find_elements_by_class_name("nextLink jasty-link")
next_button.click
I get this error message:
Message: no such element: Unable to locate element
The problem here might be that there are two next buttons on the page.
So I tried to create a list but the list is empty.
next_buttons = browser.find_elements_by_class_name("nextLink jasty-link")
print(next_buttons)
Any idea on how to solve my problem? Would really appreciate it.
This is the website:
https://fazarchiv.faz.net/faz-portal/faz-archiv?q=Kryptow%C3%A4hrungen&source=&max=10&sort=&offset=0&_ts=1657629187558#hitlist
There are two issues in my opinion:
Depending from where you try to access the site there is a cookie banner that will get the click, so you may have to accept it first:
browser.find_element_by_class_name('cb-enable').click()
To locate a single element, one of the both next buttons, it doeas not matter, use browser.find_element() instead of browser.find_elements().
Selecting your element by multiple class names use xpath:
next_button = browser.find_element(By.XPATH, '//a[contains(#class, "nextLink jasty-link")]')
or css selectors:
next_button = browser.find_element(By.CSS_SELECTOR, '.nextLink.jasty-link')
Note: To avoid DeprecationWarning: find_element_by_* commands are deprecated. Please use find_element() import in addition from selenium.webdriver.common.by import By
You can't get elements by multiple class names. So, you can use find_elements_by_css_selector instead.
next_buttons = browser.find_elements_by_css_selector(".nextLink.jasty-link")
print(next_buttons)
You can then loop through the list and click the buttons:
next_buttons = browser.find_elements_by_css_selector(".nextLink.jasty-link")
for button in next_buttons:
button.click()
Try below xPath
//a[contains(#class, 'step jasty-link')]/following-sibling::a
I'm a amateur at using python, and I'm trying to scrape the url from the html below using selenium.
<a class="" href="#" style="text-decoration: none; color: #1b1b1b;" onclick="toDetailOrUrl(event, '1641438','')">[안내] 빗썸 - 빗썸 글로벌 간 간편 가상자산 이동 서비스 종료 안내</a>
In ordinary case, the link url i want to get is in just beside 'href=', but there is just "#" in that html.
When i run the code below that is usual way to using selenium to scrape the given html, it returns a https://cafe.bithumb.com/view/boards/43. But is just what i entered in 'driver.get()', and i don't want.
url = "https://cafe.bithumb.com/view/boards/43"
driver=webdriver.Chrome('chromedriver.exe')
driver.get(url)
driver.implicitly_wait(30)
bo =driver.find_element_by_xpath("//tbody[1]/tr[#style='cursor:pointer;border-top:1px solid #dee2e6;background-color: white']/td[2]/a")
print(bo.get_attribute('href'))
What i want is https://cafe.bithumb.com/view/board-contents/1641438. You can get this url when you click a item corresponding with the xpath i wrote above.
I want this url using selenium or other programmatic ways, no need to open a chrome and enter the url in addressbar, and click using mouse... like that.
good
You can use,
bo.click()
in order to click the element you want (I assumed you want to click bo)
print(driver.execute_script('return arguments[0].getAttribute("href")',bo))
selenium , bo.get_attribute('href') is actually doing document.getElementById("somelocaator").href which returns full href , as '#' indicates current page you get current URL you provided in get()
If you just need # you can use the execute_script
How to identify the link, I have inspected the elements which are as below :
<div class="vmKOT" role="navigation">
<a class="Ml68il" href="https://www.google.com" aria-label="Search" data-track-as="Welcome Header Search"></a>
<a class="WaidDw" href="https://mail.google.com" aria-label="Mail" data-track-as="Welcome Header Mail"></a>
<a class="a4KP9d" href="https://maps.google.com" aria-label="Maps" data-track-as="Welcome Header Maps"></a>
<a class="QJOPee" href="https://www.youtube.com" aria-label="YouTube" data-track-as="Welcome Header YouTube"></a>
</div>
I want to identify the class WaidDw or href and click it using python.
You can try
driver.find_element_by_class_name('WaidDw').click()
or
driver.find_element_by_xpath('//a[#href="https://mail.google.com" and #aria-label="Mail"]').click()
In your provided HTML all attribute's values are unique, you can locate easily that element by using their attribute value.
As your question points to locate this <a class="WaidDw" href="https://mail.google.com" aria-label="Mail" data-track-as="Welcome Header Mail"></a> element. I'm providing you multiple cssSelectors which can work easily to identify the same element as below :-
a.WaidDw
a.WaidDw[href='https://mail.google.com']
a.WaidDw[aria-label='Mail']
a.WaidDw[data-track-as='Welcome Header Mail']
a.WaidDw[href='https://mail.google.com'][aria-label='Mail']
a.WaidDw[href='https://mail.google.com'][aria-label='Mail'][data-track-as='Welcome Header Mail']
Note :- Keep in practice (priority) to use cssSelector instead xpath if possible, because cssSelectors perform far better than xpath
Locating Element by CSS Selectors using python :-
element = driver.find_element_by_css_selector('use any one of the given above css selector')
Clicks the element :-
element.click()
Reference link :-
https://www.w3schools.com/cssref/css_selectors.asp
https://developer.mozilla.org/en-US/docs/Web/CSS/CSS_Selectors
I'm trying to get Selenium to click on View All Companies button, but i'm not sure what am I doing wrong. It returns no element found
html code
<div class="screener-toggles">
<div class="buttons">
<span class="button selected" data-name="advanced-screener">Search by Screener<span data-name="advanced-screener" class="arrow selected"></span></span>
<span class="button" data-name="alpha-factors">Search by Alpha Factors<span data-name="alpha-factors" class="arrow"></span></span>
<span class="button" data-name="all-companies">View All Companies<span data-name="all-companies" class="arrow"></span></span>
</div>
</div>
python code I wrote
element1 = driver.find_elements_by_class_name('View All Companies')
element1.click()
# I have tried all-companies instead of View All Companies as well. But still doesn't work
Should I not be using find_elements_by_class_name?
Any advice on what I am doing wrong is greatly appreciated!
try xpath: "//span[contains(text(),'View All Companies')]"
View All Companies is text, not the class. Try looking by text with css_selector or xpath
element1 = find_element_by_css_selector('span:contains("View All Companies")')
element1 = find_element_by_xpath('//span[contains(text(), "View All Companies")]')
Or by the data-name attribute which contains all-companies
element1 = find_element_by_css_selector('span[data-name*="all-companies"]')
Yes, you should not use the find_elements_by_class_name instead of use find_element_by_class_name.
find_elements_by_class_name is used when your expecting your locator to return more than 1 element. for a specific element use only find_element_by_class_name.
Another thing is I am not able to see any class name as View All Companies in your HTML code. Please look into your HTML and select classname or other locator carefully
Hope it will help you
I'm pretty new to using python in selenium.
I have been trying to select a button on my web page. Here is the piece of HTML that appears after inspecting the element of the button:
<a class="btn col-xs-3 nav-btns" ui-sref="salt.dashboard.reports.minions" href="/dashboard/reports/minions/">
<span class="ssIcons-icon_reports salt-icon-3x ng-scope active" bs-tooltip="" data-title="Reports" container="body" placement="bottom" animation="none" data-trigger="hover" ng-class="{'active': state.current.name =='salt.dashboard.reports' … || state.current.name =='salt.dashboard.reports.minions'}">
::before
</span>
</a>
I have tried everything I can think of. Here are some of the things that I have tried:
element = driver.find_element_by_class_name("btncol-xs-3")
element = driver.find_element_by_name("Reports")
element = driver.find_element_by_id("Reports")
the error that I keep getting is:
selenium.common.exceptions.NoSuchElementException: Message: Unable to
locate element: {"method":"class
name","selector":"salt.dashboard.reports"} Stacktrace:
at FirefoxDriver.prototype.findElementInternal_ (file:///tmp/tmpoRPJXA/extensions/fxdriver#googlecode.com/components/driver-component.js:10299)
at FirefoxDriver.prototype.findElement (file:///tmp/tmpoRPJXA/extensions/fxdriver#googlecode.com/components/driver-component.js:10308)
at DelayedCommand.prototype.executeInternal_/h (file:///tmp/tmpoRPJXA/extensions/fxdriver#googlecode.com/components/command-processor.js:12282)
at DelayedCommand.prototype.executeInternal_ (file:///tmp/tmpoRPJXA/extensions/fxdriver#googlecode.com/components/command-processor.js:12287)
at DelayedCommand.prototype.execute/< (file:///tmp/tmpoRPJXA/extensions/fxdriver#googlecode.com/components/command-processor.js:12229)
root#chris-salt:/home/chris/Documents/projects/python-selenium#
Find the element by data-title:
driver.find_element_by_css_selector("span[data-title=Reports]")
Or, if you need to get to the a tag:
driver.find_element_by_xpath("//a[span/#data-title = 'Reports']")
Chris,
The span that you pasted doesn't has an attribute named id.
Also, your class selector is too wide, i'd suggest using a more explicit path following the dom structure. Bare in mind that there may be multiple elements that have that class name.
Also, you are trying to find by the attribute name, which you don't have in that element.
Finally, it seems that you might be using angular. Does the input that you are looking for is created with javascript dinamically ?
And also, why are you using root to do this tests ?
Before doing the asserts, can you store the resulting html and manually checking that you indeed have that element?.