How to refine my code for like button SQL syntax (mysql)

Question

1.00/5 (4 votes)

See more:

I want to create a like button just like that of social media, but I don't seem to find any answer that worked fine for me, I don't know if anyone has ever come across to help out. I tried the [ON DUPLICATE KEY] AND [STORE PROCEDURE]. All only update otherwise insert, what if the user like an article [STATUS = 1] then he unlike the article [now STATUS = 0], after some few minutes he then like the article again [now STATUS = 1].
I need any sample of MySQL statement too achieve that please.

What I have tried:

INSERT INTO if_exist('id','email','status') VALUES('','myname@yahoo.com','1') ON DUPLICATE KEY UPDATE status = '0'

Posted 16-Oct-21 3:42am

Member 13591430

Updated 18-Oct-21 23:01pm

Add a Solution

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Member 15329613 · Answer 1 · 2021-10-18T06:45:00

Solution 1

Check if exists first and if it does, do an update, otherwise do an insert.

Posted 18-Oct-21 6:45am

Member 15329613

Comments

CHill60 19-Oct-21 5:09am

That is essentially what the OP's code is already doing, or trying to do.

CHill60 · Answer 2 · 2021-10-18T23:01:00

A couple of things concern me - firstly you are using text for the status. Try changing the type of status to be an tinyint or a bit. You can then use the bitwise XOR operator ^ (caret) to "toggle" between the values on each update

SQL

INSERT INTO if_exist('id','email','status') VALUES('','myname@yahoo.com',1) ON DUPLICATE KEY UPDATE status = status ^ 1

I.e. on first insert the value of status = 1 (Liked)
On 2nd attempt status will be changed to 0 (Unliked)
On 3rd attempt status will be changed to 1 (Liked, again)