How to write sql 'like' statement in Laravel?

Question

How would you write the following query in Laravel style?

SELECT * FROM `job_details` WHERE `job_title` LIKE '%officer%' AND `category_id` = 1 AND `city_id` = 1

I tried something below but it does not work:

DB::(job_details)->where(job_title LIKE '%officer%')->and(category_id=1)->and(city_id=1)

Laerte · Accepted Answer · 2017-08-04 14:50:04Z

7

Try this, it should work:

DB::table('job_details')->where('job_title', 'like', '%officer%')
                      ->where('category_id', 1)
                      ->where('city_id', 1)
                      ->get();

answered Aug 4, 2017 at 14:50

Laerte

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6 Comments

Ajmal Razeel Over a year ago

I am trying to fetch the data but it is showing blank page..this is my code:

Ajmal Razeel Over a year ago

$check = DB::table('job_details')->where('job_title', 'like', '%$keyword%') ->where('category_id', $category_id) ->where('city_id', $city_id)->get(); foreach($check as $result) { echo $result->job_title."<br>"; }

Laerte Over a year ago

What's the result of dd($check)? Probably it is empty.

Ajmal Razeel Over a year ago

yes it shows emply..but i have jobs with title 'officer'..it is supposed to show me the result

Maraboc Over a year ago

Do it like this where('job_title', 'like', '%'.$title.'%') ;)

|

Arun pandian M · Accepted Answer · 2017-08-04 15:43:18Z

1

Like this :

$users = DB::table('users')
                ->where('name', 'like', 'T%')
                ->get();

in your case try like this :

  DB::table('job_details')
          ->where([
            ['job_title', 'like', '%officer%'],
            ['category_id', '=', 1], 
            ['city_id', '=', 1]
                 ])->get();

refer : laravel where clauses

edited Aug 4, 2017 at 15:43

answered Aug 4, 2017 at 14:44

Arun pandian M

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