I'm writing out a code that opens up all necessary tabs using the webbrowser module so that I don't have to open them all manually when I open my pc, but one of the links I want it to open in incognito mode (all the links are google links) is this possible? I don't want all the code to open on incognito mode I just want the Microsoft Teams URL to open in incognito. Thanks in advance.
my code:
import webbrowser
import time
webbrowser.open('https://www.youtube.com/')
time.sleep(0.3)
webbrowser.open('https://mail.google.com/')
time.sleep(0.3)
webbrowser.open('https://teams.microsoft.com')
The solution is to specify the browser executable with the appropriate command line arguments to start it in incognito mode. The exact flag may vary by browser. Firefox, for example, takes the -private-window flag to start in incognito mode.
import webbrowser
import time
firefox_path = 'C:/Program Files/Mozilla Firefox/firefox.exe'
incognito_browser = webbrowser.get(firefox_path + ' -private-window %s')
webbrowser.open('https://www.youtube.com/')
time.sleep(0.3)
webbrowser.open('https://mail.google.com/')
time.sleep(0.3)
incognito_browser.open('https://teams.microsoft.com')
# ...could use now incognito_browser to open additional sites as desired.
The '%s' is substituted with the URL during the call to open and must be present for get to recognize that a command line invocation is being supplied rather than a named browser instance, e.g. 'opera'.
I tried to use webbrowser to open a chrome tab.I tried it's every settings but it always opened url in new tab.
Is there any solution to open URL in same Tab?
I want to load a webpage in python and do something
import webbrowser
url = 'http://google.com'
webbrowser.open(url)
//do something
But I don't want the browser has been showed on my screen, I want to hide it so that I can do some other stuff,
please give me some advices,
thanks
In some cases, autoraise is working.
Example:
import webbrowser
webbrowser.open('www.yoursite.com', autoraise=False)
import webbrowser
webbrowser.open(url)
I am using this to open url in browser. But it opens only in 'Mozilla' why?
Just look at the docs. It uses the default browser. Look at webbrowser.get() for instructions on using a different browser.
import urllib
fun open():
return urllib.urlopen('http://example.com')
But when example.com opens it does not render CSS or JavaScript. How can I open the webpage in a web browser?
#error(404)
def error404(error):
return webbrowser.open('http://example.com')
I am using bottle. Giving me the error:
TypeError("'bool' object is not iterable",)
with the webbrowser module
import webbrowser
webbrowser.open('http://example.com') # Go to example.com
import webbrowser
webbrowser.open(url, new=0, autoraise=True)
Display url using the default browser. If new is 0, the url is opened in the same browser window if possible. If new is 1, a new browser window is opened if possible. If new is 2, a new browser page (“tab”) is opened if possible. If autoraise is True, the window is raised
webbrowser.open_new(url)
Open url in a new window of the default browser
webbrowser.open_new_tab(url)
Open url in a new page (“tab”) of the default browser
On Windows
import os
os.system("start \"\" https://example.com")
On macOS
import os
os.system("open \"\" https://example.com")
On Linux
import os
os.system("xdg-open \"\" https://example.com")
Cross-Platform
import webbrowser
webbrowser.open('https://example.com')
You have to read the data too.
Check out : http://www.doughellmann.com/PyMOTW/urllib2/ to understand it.
response = urllib2.urlopen(..)
headers = response.info()
data = response.read()
Of course, what you want is to render it in browser and aaronasterling's answer is what you want.
You could also try:
import os
os.system("start \"\" http://example.com")
This, other than #aaronasterling ´s answer has the advantage that it opens the default web browser.
Be sure not to forget the "http://".
Here is another way to do it.
import webbrowser
webbrowser.open("foobar.com")
I think this is the easy way to open a URL using this function
webbrowser.open_new_tab(url)