##// END OF EJS Templates
Make running PYTHONSTARTUP optional...
Make running PYTHONSTARTUP optional It makes sense either to have PYTHONSTARTUP set up things for all interactive Python sessions, or to bring some convenience features to the default Python shell. We used to ignore it, but IPython 1.x executes the PYTHONSTARTUP file. In conversation with @minrk, we felt there should be a config option not to run it, but running it should remain the default.

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console.py
120 lines | 3.9 KiB | text/x-python | PythonLexer
"""Utility functions for interacting with the console"""
#-----------------------------------------------------------------------------
# Copyright (c) 2013, the IPython Development Team.
#
# Distributed under the terms of the Modified BSD License.
#
# The full license is in the file COPYING.txt, distributed with this software.
#-----------------------------------------------------------------------------
#-----------------------------------------------------------------------------
# Imports
#-----------------------------------------------------------------------------
# Used to determine python version
import sys
#-----------------------------------------------------------------------------
# Classes and functions
#-----------------------------------------------------------------------------
def input(prompt_text):
"""
Prompt the user for input.
The input command will change depending on the version of python
installed. To maintain support for 2 and earlier, we must use
raw_input in that case. Else use input.
Parameters
----------
prompt_text : str
Prompt to display to the user.
"""
# Try to get the python version. This command is only available in
# python 2 and later, so it's important that we catch the exception
# if the command isn't found.
try:
majorversion = sys.version_info[0]
except AttributeError:
majorversion = 1
# Use the correct function to prompt the user for input depending on
# what python version the code is running in.
if majorversion >= 3:
return input(prompt_text)
else:
return raw_input(prompt_text).decode(sys.stdin.encoding)
def prompt_boolean(prompt, default=False):
"""
Prompt the user for a boolean response.
Parameters
----------
prompt : str
prompt to display to the user
default : bool, optional
response to return if none is given by the user
"""
response = input(prompt)
response = response.strip().lower()
#Catch 1, true, yes as True
if len(response) > 0 and (response == "1" or response[0] == "t" or response[0] == "y"):
return True
#Catch 0, false, no as False
elif len(response) > 0 and (response == "0" or response[0] == "f" or response[0] == "n"):
return False
else:
return default
def prompt_dictionary(choices, default_style=1, menu_comments={}):
"""
Prompt the user to chose one of many selections from a menu.
Parameters
----------
choices : dictionary
Keys - choice numbers (int)
Values - choice value (str), this is what the function will return
default_style : int, optional
Choice to select if the user doesn't respond
menu_comments : dictionary, optional
Additional comments to append to the menu as it is displayed
in the console.
Keys - choice numbers (int)
Values - comment (str), what will be appended to the
corresponding choice
"""
# Build the menu that will be displayed to the user with
# all of the options available.
prompt = ""
for key, value in choices.items():
prompt += "%d %s " % (key, value)
if key in menu_comments:
prompt += menu_comments[key]
prompt += "\n"
# Continue to ask the user for a style until an appropriate
# one is specified.
response = -1
while (not response in choices):
try:
text_response = input(prompt)
# Use default option if no input.
if len(text_response.strip()) == 0:
response = default_style
else:
response = int(text_response)
except ValueError:
print("Error: Value is not an available option. 0 selects the default.\n")
return choices[response]