Laravel where if statement

Question

I came to a problem with select from DB. Basically what I am trying to achieve is: I have a table with 3 columns

type | number | date

I need to do where based on column (type)

If(type = 1) then where number > 1 else where date today()

So if the type is equal to 1 then apply where to number else apply where to date. It is possible to do such a thing? And it is possible to do it with Laravel Eloquent? Thank you.

The original query and datatype are: type = string number = integer date = timestamp (Y-m-d H:i:s)

Query

SELECT * FROM table_name
WHERE CASE WHEN type = days THEN date > now() ELSE number > 0 END

Table schema

CREATE TABLE banners (
  id bigint(20) unsigned NOT NULL AUTO_INCREMENT,
  type varchar(255) COLLATE utf8_unicode_ci NOT NULL DEFAULT '0',
  number int(10) unsigned NOT NULL DEFAULT '0',
  finish_at timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=12 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci

Solution that worked:

select * from `banners` 
where `banners`.`block_id` = 1 and `banners`.`block_id` is not null 
and IF (`type` = 'days', `finish_at` > now(), `number` > 0)

Laravel Way:

ParentModel::banners()->whereRaw("IF (`type` = 'days', `finish_at` > now(), `number` > 0)")->get();

Answer 1

You can use it like as

SELECT * FROM tablename WHERE IF (`type` = 1, `number` > 1,`date` = today())

Using Laravel Eloquent you can use it like as

ModelName::whereRaw('IF (`type` = 1, `number` > 1,`date` = today())')->get();

Answer 2

You should use case for this,

SELECT * FROM tablename
WHERE  CASE WHEN type = 1 THEN number > 1 ELSE date = today() END