questionary | Python library to build pretty command line user prompts | Command Line Interface library
kandi X-RAY | questionary Summary
kandi X-RAY | questionary Summary
Questionary is a Python library for effortlessly building pretty command line interfaces . Used and supported by.
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Top functions reviewed by kandi - BETA
- Display a checkbox
- Return selected values
- Prompt the user for selection
- Prompt the user for a path
- Build a validator
- Autocomplete dialog
- Prompt the user for a dictionary style
- Prompt the user for unsafe questions
- Prompt the given questions
- Print formatted text
- Get completions
- Returns the number of characters that match the given choice
- Return the choices
- Display the text for a choice
- Create a layout for inquirer
- Fix blank lines in main window
- Function to ask pystyle
- Ask the user
- Safely ask the user
- Asynchronously ask the user
- Asynchronously ask for unsafe keys
- Return a dict of kwargs used in kwargs
- Return the arguments of a function
- Creates a new password
- Activates async mode
- Check if a prompt toolkit is supported
questionary Key Features
questionary Examples and Code Snippets
Lines of Code : 21License : Strong Copyleft (CC BY-SA 4.0)
Copy(rasa) C:\Users\>pip install --upgrade pip==20.2
Collecting pip==20.2
Downloading pip-20.2-py2.py3-none-any.whl (1.5 MB)
|████████████████████████████████| 1.5 MB 6.8 MB/s
Installing collected packages: pip
Attempting uninstallclass ChoiceOption(click.Option):
def __init__(self, param_decls=None, **attrs):
click.Option.__init__(self, param_decls, **attrs)
if not isinstance(self.type, click.Choice):
raise Exception('ChoiceOption typipenv install google-auth==1.10.1 prompt-toolkit==2.0.10 questionary==1.4.0 SQLAlchemy==1.3.12 urllib3==1.25.7
pip install google-auth==1.10.1 prompt-toolkit==2.0.10 questionary==1.4.0 SQLAlchemy==1.3.12 urllib3==import pdfkit
from django.template.loader import get_template
def render_to_pdf():
# prepare your context for html template, like you do for django templates
template = get_template('your_template.html')
html = template.rendclass TryApp(App):
def build(self):
self.root = Builder.load_file('questionario.kv')
Principale()
return Principale()
PongApp().run()
TryApp().rdf.rstrip('ipz')
Community Discussions
Trending Discussions on questionary
QUESTION
I'm not sure how to solve this, so hopefully, some of you have a great solution.
I've set up two contact forms - one for the initial information and a second form with relevant follow-up questions. Please note that it is on purpose that it sends two different forms.
When receiving the information by email, I would like a random ID, e.g.: #42432, to indicate that both emails received come from the same session/person.
...ANSWER
Answered 2021-Jul-15 at 13:12You can generate GUID or random number like mentioned in this answer as Global Variable that will be available in both forms if forms are loaded in same page without reloading. For example :
QUESTION
My Firebase looks like this.
I have QuestionModel class:
ANSWER
Answered 2021-May-18 at 21:50UPDATE: use PageView
Find out more about PageView: https://api.flutter.dev/flutter/widgets/PageView-class.html
QUESTION
i have been using rasa for the past few weeks without problems. But recently i had issues with the installation of Spacy, leading me to uninstall an reinstall python. The issue may have occurred because of some dualities between python3.8 and 3.9 which i wasnt abled to pinpoint.
After deleting all python version from my computer, i just reinstalled python 3.9.2. and reinstall rasa with:
...ANSWER
Answered 2021-Mar-21 at 14:59rasa 2.4 declares compatibility with Python 3.6, 3.7 and 3.8 but not 3.9 so pip is trying to find one compatible with 3.9 or at least one that doesn't declare any restriction. It finds such release at version 0.0.5.
To use rasa 2.4 downgrade to Python 3.8.
PS. Don't hurry up to upgrade to the latest Python — 3rd-party packages are usually not so fast. Currently Python 3.7 and 3.8 are the best.
QUESTION
I have just joined a project with Java (Hibernate I guess) and MySql in the backend. And I have this
...ANSWER
Answered 2021-Mar-17 at 19:42I'm not sure of the Hibernate version that you are using, but if its Hibernate 3 then you'd have to use the setMaxResults option to limit the number of responses that you need.
QUESTION
I have exactly the same problem as mentioned in PIP install rasa-x takes forever. In the Rasa installation guide they say, you have to create an environment first. Everytime I do: conda create --name rasa python==3.7.6 it automatically downloads pip-20.3.3. If I now try the pip install --upgrade pip==20.2 command it shows the following error: Error. What did I do wrong? Thanks for the help!
**Update: python -m pip install --upgrade pip==20.2 worked, but now there is another problem when trying to install Rasa-X:Rasa-X installation error
here is the code
...ANSWER
Answered 2021-Jan-25 at 13:34I had this issue as well and for me installing pip packages with python -m pip install worked. So python -m pip install --upgrade pip==20.2 should work for you.
See here:
QUESTION
I try to make an app as a questionary, I dont know how to figure this error out.
AttributeError: 'CustomItem' object has no attribute 'delete_item'
I know what does it mean, that CustomItem is another class so it hasnt that def, but when i copied that def into CustomItem class, it hasnt object panel. So is it possible to connect another class ?
main.py
...ANSWER
Answered 2021-Jan-05 at 10:27If you want to access HistoryScreen then you have to do it like self.parent.get_screen('history') then you can access other widgets or anything else you want from HistoryScreen
QUESTION
I would like to share my problem with you.
So I have a 3 entities that needs to be accessed on my query:
Evaluation:
...ANSWER
Answered 2020-Apr-03 at 22:26You should be able to use a nested $lookup by using uncorrelated sub-queries as described at https://docs.mongodb.com/manual/reference/operator/aggregation/lookup/#join-conditions-and-uncorrelated-sub-queries
In your case that would be:
QUESTION
I've made a simple Flask project with one input and this input has to be returned on the main page with AJAX.
my Html code:
...ANSWER
Answered 2020-Mar-17 at 13:50Work if you change:
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