Reputation: 1949
I'm running the following command on my terminal:
firefox https://www.youtube.com/watch?v=TbvhGcf6UJU&feature=youtu.be&t=06m56s
However, when I run that, the time frame part gets truncated and firefox
opens at this link: firefox https://www.youtube.com/watch?v=TbvhGcf6UJU
This seems to happen for all videos that contain a time frame. Also this is the message that my terminal gives me:
g_slice_set_config: assertion 'sys_page_size == 0' failed
1429336854750 addons.xpi WARN Exception running bootstrap method shutdown on [email protected]: ReferenceError: sss is not defined (resource://gre/modules/addons/XPIProvider.jsm -> file:///usr/share/mozilla/extensions/%7Bec8030f7-c20a-464f-9b0e-13a3a9e97384%7D/[email protected]/bootstrap.js:72:4) JS Stack trace: shutdown@resource://gre/modules/addons/XPIProvider.jsm -> file:///usr/share/mozilla/extensions/%7Bec8030f7-c20a-464f-9b0e-13a3a9e97384%7D/[email protected]:72:5 < [email protected]:4305:9 < [email protected]:1991:13
How do I make firefox
open in the timeframe that is specified in the URL? Any leads appreciated.
I will eventually have to run this using python3
, in a os.system
call.
EDIT: I just checked this with google-chrome
as well, and chrome seems to truncate the url as well.
Upvotes: 0
Views: 149
Reputation: 168716
You have special characters in your url. You have to quote or escape them.
Try this:
firefox 'https://www.youtube.com/watch?v=TbvhGcf6UJU&feature=youtu.be&t=06m56s'
Upvotes: 1