OnlineFriend | JSP入门实战项目（javabeanJSPservletmysql）

by zifangsky Java Version: Current License: Apache-2.0

X-Ray Key Features Code Snippets Community Discussions(3)Vulnerabilities Install Support

kandi X-RAY | OnlineFriend Summary

OnlineFriend is a Java library. OnlineFriend has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. However OnlineFriend build file is not available. You can download it from GitHub.

OnlineFriend

Support

Quality

Security

License

Reuse

Support

OnlineFriend has a low active ecosystem.

It has 49 star(s) with 24 fork(s). There are 2 watchers for this library.

It had no major release in the last 6 months.

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

It has a neutral sentiment in the developer community.

The latest version of OnlineFriend is current.

Quality

OnlineFriend has 0 bugs and 0 code smells.

Security

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

OnlineFriend code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

OnlineFriend is licensed under the Apache-2.0 License. This license is Permissive.

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

Reuse

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

OnlineFriend has no build file. You will be need to create the build yourself to build the component from source.

OnlineFriend saves you 1045 person hours of effort in developing the same functionality from scratch.

It has 2371 lines of code, 152 functions and 50 files.

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

Top functions reviewed by kandi - BETA

kandi has reviewed OnlineFriend and discovered the below as its top functions. This is intended to give you an instant insight into OnlineFriend implemented functionality, and help decide if they suit your requirements.

Get http get
Continue processing for an article
Post post
Continue post processing
Handle post
Continue processing of a post
Handle GET
Process continue get
Login
Convert string to string

Get all kandi verified functions for this library.

OnlineFriend Key Features

No Key Features are available at this moment for OnlineFriend.

OnlineFriend Examples and Code Snippets

No Code Snippets are available at this moment for OnlineFriend.

Community Discussions

Trending Discussions on OnlineFriend

Component not updating even after state is changed

How do I place just one of my Mysqli bind results into an array to use the in_array() function for a match?

Storing multiple Models result in JSON in SequelizeJS

QUESTION

Component not updating even after state is changed

Asked 2021-May-04 at 13:51

I have this friends component where I search the database for the user's friends (stored in a string separated by commas) and display them in a scrollable div. You can also add a new friend. Everything is working fine in the database and the backend. However, the component isn't being re-rendered when the state changes. It's dependent on 'allFriends' so it should re-render when that gets updated. Do you guys have any idea what's going on here? I'm guessing it has something to do with asynchronicity but I can't seem to pinpoint it. I added a comment for each code block to try to make it a little easier to read.

...

ANSWER

Answered 2021-May-04 at 08:49

you need wait fetch response data

Source https://stackoverflow.com/questions/67380363

QUESTION

How do I place just one of my Mysqli bind results into an array to use the in_array() function for a match?

Asked 2019-Jan-17 at 02:11

I have an array and I want to match an array from a prepared statement results on $Fid only. How do I go about doing this with while($fr->fetch())?

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ANSWER

Answered 2019-Jan-17 at 02:11

As I understand your question, something like this should work. Note that as Fid is an integer in your table, and you are generating your comparison values from $_POST data (hence it will be string values), you cannot use strict matching in in_array. So remove the third parameter to change that.

Source https://stackoverflow.com/questions/54226847

QUESTION

Storing multiple Models result in JSON in SequelizeJS

Asked 2017-Oct-04 at 09:22

I am new to SequelizeJS. My question is how store multiple model's result in json. I have called 2 models Users and Friends and I want to get both model's result in JSON. I am trying to return return res.json({"OnlineUsers": onlineUsers, "onlineFriends": onlineFriends}); both result in a single JSON, but not getting result.

Here is how my controller looks like

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ANSWER

Answered 2017-Oct-04 at 09:22

You have to nest the queries, remember that node.js is asynchronous so after doing the first query you do the second one, and after that you return the response with both objects. Something like this:

Source https://stackoverflow.com/questions/46546933

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

Vulnerabilities

No vulnerabilities reported

Install OnlineFriend

You can download it from GitHub.
You can use OnlineFriend like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the OnlineFriend component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .

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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