How to click a button from Python

Question

I am using Anaconda2, Jupiter, and Chrome browser.

I am writing below code which is running successfully but now I want to click on button.

<a href="#" action="exportSelected" class="btn btn-default">
    <i class="glyphicon glyphicon-download"></i> Export
</a>

What should I write in Python to access this?

find_element_by_partial_link_text('btn btn-default').click()

It's giving an error.

Possible duplicate of Selenium/python - fails to find partial link text — JeffC
– JeffC, Commented Oct 2, 2017 at 20:17
You should probably spend some time reading through the python/selenium docs to learn the different locators, how they work, and how to use them. selenium-python.readthedocs.io/locating-elements.html — JeffC
– JeffC, Commented Oct 2, 2017 at 20:17

Andersson · Accepted Answer · 2017-10-02 17:30:29Z

5

"btn btn-default" is not a link text, but class names.

You can use one of below solutions:

Locate by exact link text:

find_element_by_link_text('Export').click()

Locate by compound class name:

find_element_by_css_selector('a.btn.btn-default').click()

Locate by action attribute

find_element_by_css_selector('a[action="exportSelected"]').click()

answered Oct 2, 2017 at 17:30

Andersson

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Harish Kumar Garg · Accepted Answer · 2017-10-04 15:34:56Z

0

Thank you all replying me and suggesting pyhthon doc. appreciate it.

So i have used very simple lines to get my job done.

driver.find_element_by_xpath('copy your xpath'). which is very simple

driver.find_element_by_xpath('//[@id="app_content"]/div[2]/div/div/section/div/div[2]/div[1]/div[1]/div/div[2]/a[1]').click() driver.implicitly_wait(5) driver.find_element_by_xpath('//[@id="export_button"]').click()

answered Oct 4, 2017 at 15:34

Harish Kumar Garg

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