Use like operator and order by sub query

Question

Introduction

I have a project where we help the IT community to organize better IT meetups (just like meetup.com but only for IT community) called https://codotto.com/

We are developing a new feature where we would like to show the number of usages of certain tags. The algorithm should work just like Stackoverflow's one. You write javascript and you get a list with the tags that match your query sorted by the most used ones.

I'm currently using Laravel but I will post the raw query so that it's easier for mysql wizards to help me out if possible :)

Problem

I have the following tables

tags table

id  name
1   javascript
2   javascript-tools
3   javascript-security

group_has_tags table

group_id  tag_id
1         2
2         2
2         3

We have the tags javascript-tools being used two times, javascript-security is used one while javascript is not used at all.

Now, if a user search for javascript, he should get first javascript (because it is a direct match) followed by the rest of the tags sorted by their usage.

In Laravel this is the code that I have (simplified ofc)

$tags = Tag::withCount('groups')
  ->orderBy('groups_count', 'DESC')
  ->where('name', 'LIKE', 'javascript%')
  ->get(2);

The problem is that since I only return back 2 results, javascript is not being included in the results, even tho it's what the user literally wrote

For the mysql magicians, here is the raw query

select
  `tags`.*,
  (
    select
      count(*)
    from
      `meetups`
        inner join `meetup_has_tags` on `meetups`.`id` = `meetup_has_tags`.`meetup_id`
    where
        `tags`.`id` = `meetup_has_tags`.`tag_id`
  ) as `meetups_count`
from
  `tags`
where
    `title` LIKE 'javascript%'
order by
  `meetups_count` desc
limit
  2 offset 0

Question

The main objective here is to return the most relevant result to the user. He writes javascript and javascript shows up first followed by less "relevant" results. The way I found was to sort by the number of times a tag was used.

Is there a solution where I can do "please, fetch the results that match this query first, then return the most relevant results"?

By "most relevant" results, I simply mean "what the user is looking for". If he writes "javascript" it should return "javascript" followed by "javascript-tools" (because "javascript-tools" was used twice but the user is literally searching for "javascript")

Answer 1

So basically you want your program to end if the variable X is equal to the word "STOP" right? So in that case you need to change de comparison inside if. In java you can't compare strings using the operator ==. When you do that you're comparing if the objects reference are equal.

Solution: To compare Strings you can use YourVariable.equals(SomeString). In your case just change x=="STOP" to x.equals("STOP")

For reference:
Strings Method Equals: https://www.w3schools.com/java/ref_string_equals.asp