seasalt | A simple Javascript wrapper for Libsodiumjs

by jakcodex JavaScript Version: Current License: BSD-3-Clause

X-Ray Key Features Code Snippets Community Discussions(1)Vulnerabilities Install Support

kandi X-RAY | seasalt Summary

seasalt is a JavaScript library. seasalt has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

A simple Javascript class making Libsodium easier to use. See it in action here: You can find a full reference guide available in the docs.

Support

Quality

Security

License

Reuse

Support

seasalt has a low active ecosystem.

It has 1 star(s) with 1 fork(s). There are no watchers for this library.

It had no major release in the last 6 months.

There are 0 open issues and 1 have been closed. On average issues are closed in 58 days. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of seasalt is current.

Quality

seasalt has 0 bugs and 0 code smells.

Security

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

seasalt code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

seasalt is licensed under the BSD-3-Clause License. This license is Permissive.

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

Reuse

seasalt releases are not available. You will need to build from source code and install.

Installation instructions are not available. Examples and code snippets are available.

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seasalt Key Features

No Key Features are available at this moment for seasalt.

seasalt Examples and Code Snippets

No Code Snippets are available at this moment for seasalt.

Community Discussions

Trending Discussions on seasalt

Finding the most common word in Javascript

QUESTION

Finding the most common word in Javascript

Asked 2021-Jul-19 at 15:38

It currently finds the first most repeated word, now I need it to grab multiple if more a word is tied for being used. The part where my code fails is when I split() to separate the words then Set() to delete duplicates. I think my problem comes from switching from string to array? I just need a simple way for this to work, any suggestions are appreciated. Thank You!

...

ANSWER

Answered 2021-Jul-19 at 15:16

The easiest way in my eyes would be just using an array to begin with:

Source https://stackoverflow.com/questions/68442800

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

Vulnerabilities

No vulnerabilities reported

Install seasalt

You can download it from GitHub.

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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