TinyUrl | complete solution of TinyUrl including backend API

by OneSlashNinja JavaScript Version: Current License: No License

X-Ray Key Features Code Snippets Community Discussions(10)Vulnerabilities Install Support

kandi X-RAY | TinyUrl Summary

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

You may check the explaination on Wiki and try it at tinyurl.com.

Support

Quality

Security

License

Reuse

Support

TinyUrl has a low active ecosystem.

It has 4 star(s) with 3 fork(s). There are 1 watchers for this library.

It had no major release in the last 6 months.

TinyUrl has no issues reported. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of TinyUrl is current.

Quality

TinyUrl has 0 bugs and 0 code smells.

Security

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

TinyUrl code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

TinyUrl does not have a standard license declared.

Check the repository for any license declaration and review the terms closely.

Without a license, all rights are reserved, and you cannot use the library in your applications.

Reuse

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

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

TinyUrl saves you 17 person hours of effort in developing the same functionality from scratch.

It has 49 lines of code, 0 functions and 8 files.

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

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TinyUrl Key Features

No Key Features are available at this moment for TinyUrl.

TinyUrl Examples and Code Snippets

No Code Snippets are available at this moment for TinyUrl.

Community Discussions

Trending Discussions on TinyUrl

"blocks are indirectly lost in loss record.." valgrind error - cpp

After press back from splash screen app goes to previous open activity android

how i can create tinyURL with expiration date with node js

How to make a fetch request to TinyURL?

Give the ability to user to copy text in tkinter gui

Using Rails globbing for URL parameters: some URL types get cut off

How to read value from yaml file which are in dictionary and need to use in helm

RestSharp get url after redirect

Why does pycharm gives error for importing numpy

Python doesnt imports matplotlib, matplotlib doesnt work

QUESTION

"blocks are indirectly lost in loss record.." valgrind error - cpp

Asked 2021-Jun-08 at 19:57

I am trying to figure out what is wrong with my cpp code, and i need your help!

valgrind Output:

the error occurs here (in the test file): list = list.apply(getLen);

the function getLen:

...

ANSWER

Answered 2021-Jun-08 at 19:45

You allocated a SortedList object in main() which allocated memory and never freed it.

The other blocks belong to the std::string objects inside the list. They are only lost because the list holding them lost track of them.

When doing your analysis, don't worry about the 6 indirectly lost blocks, focus on the one directly lost one.

Almost certainly this is a violation of the "Rule of Five". Your SortedList does an allocation in its constructor and doesn't properly manage that allocation in its move-assignment operator.

Looking at your code in the copy assignment operator, this line is causing the leak

Source https://stackoverflow.com/questions/67893153

QUESTION

After press back from splash screen app goes to previous open activity android

Asked 2021-May-30 at 14:26

In my application when user open app by click "app icon" then screen open by this flow

SplashActivity -> WelcomeActivity -> SignInActivity

when user open app by click "some link" then screen flow is

Splashactivity -> VideoActivity

when I open app by click "some link" then screen flow is same as above

Splashactivity -> VideoActivity

but when i press back in splash activity in this condition app goes to "SignInActivity" instead of close the app

manifest for Splash, Welcome, video activity

...

ANSWER

Answered 2021-May-28 at 08:47

This is your activity stack:

Source https://stackoverflow.com/questions/67735348

QUESTION

how i can create tinyURL with expiration date with node js

Asked 2021-Apr-20 at 12:10

I'm generation tinyUrl and sending this via mail as well sms to customer ,i want to generate tinyURL from nodejs code so that if user access this URL after particular date then it should get expire . I can successfully generating tinyurl from bellow code but I want to generate link with expiry date so that Link will expire automatically once user click after particular date.

any way to solve this or is another alternative for this.

...

ANSWER

Answered 2021-Apr-20 at 12:10

From what I can gather, you're using the tinyurl npm package from https://www.npmjs.com/package/tinyurl which leverages the API from tinyurl.com to create shortened urls. This package doesn't support expiring links, which means that once you create a url using their service, it will be forever registered at tinyurl.com.

You can, however, use the tinyurl API directly, ignoring the tinyurl npm package. Tinyurl has great documentation at tinyurl .com /app /dev (added spaces because link shorteners are blacklisted on SO) which explains that you can send a PATCH request to the api at /update to change where the link is sent.

Since I can't see what is calling generateTinyUrl or how the response is used, I'm not sure what the best path forward is for your case. But, as I see it, you have two options.

Option 1: I would advise that you learn how to use an external API and remove the tinyurl npm package from your code. You could then schedule a cron job or similar to scan your urls, detect which ones have expired, and send a request to /update to change the url to a custom page that explains to the user that their link has expired.
Option 2: When you generate a tinyURL, have it point to a domain that you control. Generate that page so that when it is accessed, it gathers relevant data from your database, including its expiration date. When the expiration date is reached, that page could then display a custom expiration message which you write.

Source https://stackoverflow.com/questions/67177962

QUESTION

How to make a fetch request to TinyURL?

Asked 2021-Apr-07 at 20:55

I am trying to make a fetch request specifically a post request to tinyURL to shortern a url generated on my site. here is the tinyURL API

Currently, I am writing my code like this but it doesn't appear to be returning the short url.

the word tinyurl seems to be banned within links so all links containing the word tinyurl have been replaced with "SHORT"

here is the tinyURL API https://SHORT.com/app/dev

...

ANSWER

Answered 2021-Apr-07 at 19:01

The snippet below seems to work

Source https://stackoverflow.com/questions/66991259

QUESTION

Give the ability to user to copy text in tkinter gui

Asked 2021-Apr-01 at 01:50

In this code when the user enters the link, a short version of the link get displayed but it does not give the ability to the user to copy the link from the GUI. How do i fix this?

...

ANSWER

Answered 2021-Apr-01 at 01:50

You can't directly copy the text from a Tkinter Label widget with CTRL+C.

This is a simple Tkinter app to copy the text of a Label to the clipboard:

Source https://stackoverflow.com/questions/66897250

QUESTION

Using Rails globbing for URL parameters: some URL types get cut off

Asked 2021-Feb-14 at 15:59

I'm having an odd problem with Ruby on Rails globbing. My route looks like this:

...

ANSWER

Answered 2021-Feb-14 at 15:59

Routes have an implicit optional (.:format) segment at the end. You can use this option to prevent that:

Source https://stackoverflow.com/questions/66191332

QUESTION

How to read value from yaml file which are in dictionary and need to use in helm

Asked 2021-Jan-27 at 15:46

This is my string in values.yaml: selectorLabels: { app.kubernetes.io/name: tinyurl }

Yaml file looks like below:

...

ANSWER

Answered 2021-Jan-27 at 13:40

You can use the index function

Source https://stackoverflow.com/questions/65918338

QUESTION

RestSharp get url after redirect

Asked 2021-Jan-25 at 21:51

Is there a way to get the page url after redirect, with RestSharp?

For example, the code below will return page source of this thread, but how do I get the actual URL of the thread?

...

ANSWER

Answered 2021-Jan-25 at 21:51

I haven't use RestSharp in a while, bit if I remember correctly the response should have a property named ResponseUri or similar.

Source https://stackoverflow.com/questions/65892539

QUESTION

Why does pycharm gives error for importing numpy

Asked 2020-Dec-24 at 12:59

In pycharm, when i try to import numpy , i get a runtime error like this.

...

ANSWER

Answered 2020-Dec-24 at 12:59

try installing this version :

Source https://stackoverflow.com/questions/65438575

QUESTION

Python doesnt imports matplotlib, matplotlib doesnt work

Asked 2020-Dec-22 at 17:06

So i searched a lot, but i am not able to solve the problem. Everytime i try to use matplotlib or similar (matplotlib.something), python isn't able to import it. It says:

...

ANSWER

Answered 2020-Dec-22 at 16:53

I had the same problem, you can fix it with:

Source https://stackoverflow.com/questions/65412636

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

Vulnerabilities

No vulnerabilities reported

Install TinyUrl

First of all, please make sure you have installed the MySQL locally and Node locally and started MySQL service. Then please execute the db_init_script.sql to create the database and relative table for this project, and make sure the username and password is correctly configed in Server/mysqlUtil.js file. After that, run npm install to install all the dependencies needed for this project. And please install gulp globally by run command: npm install gulp -g. Now, you are all set. please direct your terminal to the Server folder and run gulp command. once you see the Runing on PORT: 8000 prompt, you are able to see the site at localhost:8000.

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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