How do I get the current IPython / Jupyter Notebook name

I am trying to obtain the current NoteBook name when running the IPython notebook. I know I can see it at the top of the notebook. What I am after something like

currentNotebook = IPython.foo.bar.notebookname()

I need to get the name in a variable.

Solution 1:

adding to previous answers,

to get the notebook name run the following in a cell:

%%javascript
IPython.notebook.kernel.execute('nb_name = "' + IPython.notebook.notebook_name + '"')

this gets you the file name in nb_name

then to get the full path you may use the following in a separate cell:

import os
nb_full_path = os.path.join(os.getcwd(), nb_name)

Solution 2:

I have the following which works with IPython 2.0. I observed that the name of the notebook is stored as the value of the attribute 'data-notebook-name' in the <body> tag of the page. Thus the idea is first to ask Javascript to retrieve the attribute --javascripts can be invoked from a codecell thanks to the %%javascript magic. Then it is possible to access to the Javascript variable through a call to the Python Kernel, with a command which sets a Python variable. Since this last variable is known from the kernel, it can be accessed in other cells as well.

%%javascript
var kernel = IPython.notebook.kernel;
var body = document.body,  
    attribs = body.attributes;
var command = "theNotebook = " + "'"+attribs['data-notebook-name'].value+"'";
kernel.execute(command);

From a Python code cell

print(theNotebook)

Out[ ]: HowToGetTheNameOfTheNoteBook.ipynb

A defect in this solution is that when one changes the title (name) of a notebook, then this name seems to not be updated immediately (there is probably some kind of cache) and it is necessary to reload the notebook to get access to the new name.

[Edit] On reflection, a more efficient solution is to look for the input field for notebook's name instead of the <body> tag. Looking into the source, it appears that this field has id "notebook_name". It is then possible to catch this value by a document.getElementById() and then follow the same approach as above. The code becomes, still using the javascript magic

%%javascript
var kernel = IPython.notebook.kernel;
var thename = window.document.getElementById("notebook_name").innerHTML;
var command = "theNotebook = " + "'"+thename+"'";
kernel.execute(command);

Then, from a ipython cell,

In [11]: print(theNotebook)
Out [11]: HowToGetTheNameOfTheNoteBookSolBis

Contrary to the first solution, modifications of notebook's name are updated immediately and there is no need to refresh the notebook.

Solution 3:

As already mentioned you probably aren't really supposed to be able to do this, but I did find a way. It's a flaming hack though so don't rely on this at all:

import json
import os
import urllib2
import IPython
from IPython.lib import kernel
connection_file_path = kernel.get_connection_file()
connection_file = os.path.basename(connection_file_path)
kernel_id = connection_file.split('-', 1)[1].split('.')[0]

# Updated answer with semi-solutions for both IPython 2.x and IPython < 2.x
if IPython.version_info[0] < 2:
    ## Not sure if it's even possible to get the port for the
    ## notebook app; so just using the default...
    notebooks = json.load(urllib2.urlopen('http://127.0.0.1:8888/notebooks'))
    for nb in notebooks:
        if nb['kernel_id'] == kernel_id:
            print nb['name']
            break
else:
    sessions = json.load(urllib2.urlopen('http://127.0.0.1:8888/api/sessions'))
    for sess in sessions:
        if sess['kernel']['id'] == kernel_id:
            print sess['notebook']['name']
            break

I updated my answer to include a solution that "works" in IPython 2.0 at least with a simple test. It probably isn't guaranteed to give the correct answer if there are multiple notebooks connected to the same kernel, etc.

Solution 4:

It seems I cannot comment, so I have to post this as an answer.

The accepted solution by @iguananaut and the update by @mbdevpl appear not to be working with recent versions of the Notebook. I fixed it as shown below. I checked it on Python v3.6.1 + Notebook v5.0.0 and on Python v3.6.5 and Notebook v5.5.0.

import jupyterlab
if jupyterlab.__version__.split(".")[0] == "3":
    from jupyter_server import serverapp as app
    key_srv_directory = 'root_dir'
else : 
    from notebook import notebookapp as app
    key_srv_directory = 'notebook_dir'
import urllib
import json
import os
import ipykernel

def notebook_path(key_srv_directory, ):
    """Returns the absolute path of the Notebook or None if it cannot be determined
    NOTE: works only when the security is token-based or there is also no password
    """
    connection_file = os.path.basename(ipykernel.get_connection_file())
    kernel_id = connection_file.split('-', 1)[1].split('.')[0]

    for srv in app.list_running_servers():
        try:
            if srv['token']=='' and not srv['password']:  # No token and no password, ahem...
                req = urllib.request.urlopen(srv['url']+'api/sessions')
            else:
                req = urllib.request.urlopen(srv['url']+'api/sessions?token='+srv['token'])
            sessions = json.load(req)
            for sess in sessions:
                if sess['kernel']['id'] == kernel_id:
                    return os.path.join(srv[key_srv_directory],sess['notebook']['path'])
        except:
            pass  # There may be stale entries in the runtime directory 
    return None

As stated in the docstring, this works only when either there is no authentication or the authentication is token-based.

Note that, as also reported by others, the Javascript-based method does not seem to work when executing a "Run all cells" (but works when executing cells "manually"), which was a deal-breaker for me.

How do I get the current IPython / Jupyter Notebook name

Solution 1:

Solution 2:

Solution 3:

Solution 4:

Related

Recent Posts