spaniel | Time span handling for Go | Date Time Utils library

by senseyeio Go Version: v1.0.0 License: MIT

X-Ray Key Features Code Snippets Community Discussions(4)Vulnerabilities Install Support

kandi X-RAY | spaniel Summary

spaniel is a Go library typically used in Utilities, Date Time Utils applications. spaniel has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

Spaniel contains functionality for timespan handling, specifically for merging overlapping timespans and finding the intersections between multiple timespans. It lets you specify the type of interval you want to use (open, closed), and provide handlers for when you want to add more functionality when merging/intersecting.

Support

Quality

Security

License

Reuse

Support

spaniel has a low active ecosystem.

It has 170 star(s) with 6 fork(s). There are 15 watchers for this library.

It had no major release in the last 12 months.

There are 0 open issues and 3 have been closed. On average issues are closed in 57 days. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of spaniel is v1.0.0

Quality

spaniel has 0 bugs and 0 code smells.

Security

spaniel has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

spaniel code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

spaniel is licensed under the MIT License. This license is Permissive.

Permissive licenses have the least restrictions, and you can use them in most projects.

Reuse

spaniel releases are available to install and integrate.

Installation instructions, examples and code snippets are available.

It has 1387 lines of code, 66 functions and 9 files.

It has medium code complexity. Code complexity directly impacts maintainability of the code.

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spaniel Key Features

No Key Features are available at this moment for spaniel.

spaniel Examples and Code Snippets

No Code Snippets are available at this moment for spaniel.

Community Discussions

Trending Discussions on spaniel

How do i remove white spaces between borders in table

how to create url from an api in json

How to filter an array of javascript objects based on an array of strings

filter on a value in a column and remove rows that dont meet condition

QUESTION

How do i remove white spaces between borders in table

Asked 2022-Mar-12 at 03:35

I've seen this answered in other questions but it doesn't work for me. i'm also only 1 month in so im very new to this.

i want to remove the space between each cell so there is a collapsed border around the td and th.

...

ANSWER

Answered 2022-Mar-12 at 03:35

hello i think that will solve your problem try this

Source https://stackoverflow.com/questions/71446540

QUESTION

how to create url from an api in json

Asked 2021-Dec-22 at 23:33

i have this api, and i need to create url from what i get from json

https://dog.ceo/api/breeds/list

...

ANSWER

Answered 2021-Dec-22 at 23:19

You could use the base url you have there and use string replace function to fill in place holder with the bird breed. Something like this:

Source https://stackoverflow.com/questions/70456032

QUESTION

How to filter an array of javascript objects based on an array of strings

Asked 2021-Dec-08 at 21:56

Given the following object, I would like to search all of the keys for multiple strings. I've been trying and searching and have come up empty on this one. Can anyone provide me some help here?

Searching this:

...

ANSWER

Answered 2021-Dec-08 at 19:53

const array = [
    { name: "Blue Iron Chow Chow", status: "Complete", creator: "John" },
    { name: "Purple Steel Husky", status: "Error", creator: "Chris" },
    { name: "Purple Composite Husky", status: "Ready", creator: "Chris" },
    { name: "Aqua Zinc Spaniel", status: "Complete", creator: "Chris" },
    { name: "Fuschia Silver Corgi", status: "Complete", creator: "John" }, 
];


const query = ['chris', 'com'];

var result = array.filter(({creator, name, status}) => {
    return query.every(item => {
            item = item.toLowerCase();
            return creator.toLowerCase().includes(item) || name.toLowerCase().includes(item) || status.toLowerCase().includes(item)
        })
});

Source https://stackoverflow.com/questions/70280101

QUESTION

filter on a value in a column and remove rows that dont meet condition

Asked 2020-Nov-30 at 05:12

I'm relatively new in R. I have this problem. I have data of dogs example of useful part of data (columns age_month and rasnaam (breed) are used)

I have to look for all the breeds if they are small, medium, large etc. And if they are a small breed then all the rows where age_month is lower than 9 have to be removed, if they are a medium sized breed rows where age_month is lower than 13 have to be removed, (large, age_month < 24). I've tried some things but it won't work. I've added all dogs to a list (also tried it with vector) like this: (only for small dogs here)

...

ANSWER

Answered 2020-Nov-30 at 03:39

If you want to stick with using case_when, this is one way to achieve what you're looking for:

Source https://stackoverflow.com/questions/65065517

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install spaniel

This package is "go-gettable", just do:.

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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