I have created a button in debugging mode of OpenERP and want to display a message on click of that button.
This is the button Ex. then How to write an action on click of this button mainly the popup and message is required
You can write a small wizard to do so.refer to openerp documentation for how to.
Related
I am working on a project and I've created a virtual keyboard using python library Tkinter in which if I click the button the key value gets entered onto the entry field in the form of password. I want that value to be shown on some other site like facebook or any other logging site in real-time.
I tried using pynput and tried getting value on notepad but it is not happening as whenever I click on the button, the notepad file goes out of focus and the cursor moves to my python file.
I can't share my code as I am prohibited to do that but in simple words I want the value getting printed on entry field also gets printed to the input field of some other site.
I am using selenium with python and i am clicking buttons with this command
loginbtn= driver.find_element_by_id('btnLogin').click()
My problem now is that i got 2 buttons with no name or id and with exactly same code as the photo below.
Is there any way to define that i want to click on Submit button or Cancel button?
Use the below xpath.
For submit:
driver.find_element_by_xpath('//button[./span[text()="Submit"]]').click()
For cancel.
driver.find_element_by_xpath('//button[./span[text()="Cancel"]]').click()
I am new to Odoo 8 and I work on a module, I create a view with a button to send an appeal and I want that when I click the button it prints me a PDF or use it opens a box sends message I want to know what is the easiest solution and if possible send me some examples that I can use the two solutions ..
Thank you again to you
I need to make such feature: when user clicks "edit" button, the Django should open admin page for editing this object. I have made something like this:
<a onclick="window.open('/admin/posts/post/{{post.pk}}/', 'newwindow', 'width=auto, height=auto'); return false;">
And this will open window for editing this post. But when I click on "Save" button - object saves and nothing happens else. The window stays white...
But I need to close automatically window after "Save" button and refresh page from which user clicks "edit" button.
We can also see such window, when clicking on the icon "Change selected objectname", near foreign key in Django admin page.
Do you understand what I need? What should I do?
Thanks a lot.
I am working on a project. Each day I need to login to a website. Navigate to a report page. After entering the require information, and clicking on the "gererate report" icon, there's a pop up windown asking me to save the file. I want to do it automatically by python.
I search the internet for a couple of days, and now I can do it by Autoit, but can't find a way in python.
By using urllib2, I can process up to clicking the "generate report" icon. But how can I get the name and location of the file and download it?
I want to make clear that the "generate report" button is actuary the submit button of a form. There is no Href associate with the generate button. After submitting the form, a file-save dialog popup asking me where to save the file.
thanks in advance!
If it's a save file dialog from Fx it probably means that it directly links to a file. If you manage to fill all the information and click the generate_report icon, can't you check which href has the generate_button assigned to it?
import urlib
urlfile = "Href of the generate_button"
urllib.urlretrieve(urlfile,"filename")