stonehenge | Multi-project local development environment & toolset | Continuous Deployment library
kandi X-RAY | stonehenge Summary
kandi X-RAY | stonehenge Summary
Local development environment toolset on Docker supporting multiple projects.
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Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of stonehenge
stonehenge Key Features
stonehenge Examples and Code Snippets
git clone -b 3.x https://github.com/druidfi/stonehenge.git ~/stonehenge
cd ~/stonehenge
make up
Community Discussions
Trending Discussions on stonehenge
QUESTION
I am coding an app for a friend, and SQL is not my strong suit. I thought that I had laid this matter to rest with my previous question, which received an excellent answer.
However, my friend has moved the goal posts yet again (and swears that it is final this time).
Given these tables
...ANSWER
Answered 2020-Sep-03 at 11:56Example of working query in DB fiddle: https://www.db-fiddle.com/f/rQKazPgbtGS766WEiuiXyR/0
QUESTION
I can find all the places where the value string searchtheme occurs in my trainings by doing this:
...ANSWER
Answered 2019-Dec-03 at 04:09Yes, Bixby allows you to search the metadata using their Aligned NL syntax
QUESTION
I am new to coding. I am trying to create code that will substitute random words into preset patterns. I am stuck trying to insert a variable to call a list of words from which to grab a random word. Instead, Python pulls a random letter from the variable. I would like Python to call the list of words and pull a random element from the list.
I've tried creating a dict and calling the variable as a value but that produced the same result. Again, this is all new to me, so sorry if the answer is simple. There is probably a method call I don't know about or I am just on the wrong track here. Any help is appreciated. Thanks!
...ANSWER
Answered 2019-Feb-16 at 18:27As Bazingaa told in is comment, you can simply create a list of objects, instead of a list of strings!
QUESTION
With an image loaded into Python as shown below, how do I know which order the channels are in? (e.g. BGR or RGB)
Code
...ANSWER
Answered 2017-Oct-24 at 05:52Since you use PIL
and you don't specify any other mode to load the Image with, you get R G B.
You could verify that by checking the "mode" attribute on the Image instance:
QUESTION
I am writing a portfolio and part of it is in modals on the page. However, I CANNOT possibly get it to work for whatever reason. Here is the codepen: https://codepen.io/dmeskin/pen/GmvqxV (the modals are the Pottery & Design Buttons) Or, here is the plain html if you want it <3
...ANSWER
Answered 2017-May-12 at 02:04Nerd , instead of calling this $("#MyModal").modal() can you give a try like this $("#MyModal").modal('show') hopefully it should work.
QUESTION
I'm trying to get the wa tablle to display in HTML using javascript whenever certain values are entered in a textbox and button is clicked heres my code
...ANSWER
Answered 2017-Mar-24 at 11:03You are changing the visibility
in JavaScript whereas the display
is still none
.
So, Change the display
to ''
.
QUESTION
i want to code an algorithm, that can do an edge detection for an image. I already have a part of the code, which detects all edges in horizontal way. Example picture:
But i need an edge detection in horizontal, vertical and diagonal way. If i try it the same way like the horizontal way, i get an ArrayOutOfBoundaryException (e.g. if i do pixRechts= color(meinBild1.pixels[index1(x+1,y)]); )
Do u have any ideas how to do that? Im using Processing.
Thanks.
My Code until now.
...ANSWER
Answered 2017-Jan-15 at 23:29(e.g. if i do pixRechts= color(meinBild1.pixels[index1(x+1,y)]); )
This line would just find the already detected edges.
I would recommend you to do it like you mentioned by your example, but you should just detect for:
- x -1, y -1 for vertical
- x -1, y +1 for vertical
- x, y -1 for horizontal
You should also look at your two loops. The above mentioned cases will also run in an ArrayOutOfBoundsException with your code.
- will stop at the first pixel
- will stop at the last y pixel
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