milkyway | Milkyway sketch project | Frontend Framework library
kandi X-RAY | milkyway Summary
kandi X-RAY | milkyway Summary
This repo is an ongoing Milkyway implementation using Semantic UI, to be used with this framework's official react integration.
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Trending Discussions on milkyway
QUESTION
I am trying to write a code that fetches lines from a txt file and adds parsed out variables from those lines to an array. The txt file reading and the parsing works perfectly.
My text file "inventory.txt" looks like this:
...ANSWER
Answered 2021-Nov-21 at 15:48The problem is that all of your elements are simple references to item
whose content will ultimately be the last candy bar name you read.
Here is one way to solve this problem, by copying each parsed string to a newly-allocated string on the heap using strdup()
:
QUESTION
I am making a tool which calculates your weight on different planets. I dont' know why but my code is not working. I have attached it. The problem is with the javascript. Someone please help me.
I have put my code here : https://www.w3schools.com/code/tryit.asp?filename=GQKHM7XCL3KM
...ANSWER
Answered 2021-May-16 at 11:32You just need to change from onclick="Calculate" ();
to onclick="Calculate();"
QUESTION
I'm trying to find all rows with a certain value in an array. Currently use
...ANSWER
Answered 2021-Apr-22 at 11:38You can do something like this :
- If the JSON contains another JSON Object
QUESTION
I am trying to create a form with nested components where the data that I pass to the child component can be a nested value of an object.
The input fields with strings are working as expected, I just have an issue with the object value. Thanks in advance for your advice!
Here's the link to my sandbox: https://codesandbox.io/s/vue3-form-54xi8
My parent component VueVModel.vue
:
ANSWER
Answered 2021-Mar-15 at 18:14You're passing address.street
which cannot be mapped to a address.street
object in the props. Instead try passing the street only.
QUESTION
I am trying to replace a template string with values that are available in another dict. I am aware of how to do this using .findall
and then .replace
but I was trying to convert it to one operation.
This is what I tried:
...ANSWER
Answered 2020-Dec-16 at 14:53Ah, thanks to this SO answer, looks like when r'\1'
is used this way, it is passed as a string.
Reworking it this way works as intended:
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