Array to string conversion in Laravel using 'LIKE' query

Question

sorry for a newbie question. I run into a problem. I want to search the database for values which an in array. But I have an error:

Array to string conversion

And kinda get why but, I don't know how to make it right. Can you help me please? This is my code:

public function chassis($chassis){
    return $this->builder->whereIn('model_type_en', 'LIKE', (array)"%$chassis%");
}

P.S please don't laugh at me :)

Like and where in are completely separate things. Builder will probably have method called like with two parameters fieldname and string. Use that without casting to array (array). — Maciej Paprocki
– Maciej Paprocki, Commented Aug 17, 2017 at 12:45
hard, I need to make a solution. People in application want to input many chassis numbers at one time and result should display all of them or at least similar to them — Dmitry Malys
– Dmitry Malys, Commented Aug 17, 2017 at 12:47
The problem is mysql doesn't support in with array, instead you need to create multiple like statements joined by OR. However, please be aware, big amount of likes like that will make really inefficient query — Maciej Paprocki
– Maciej Paprocki, Commented Aug 17, 2017 at 12:53
@MaciejPaprocki yaa I kind of thinking to give up on LIKE, just use whereIN array — Dmitry Malys
– Dmitry Malys, Commented Aug 17, 2017 at 12:58

ch271828n · Accepted Answer · 2017-08-17 12:35:32Z

3

$collection = DB::table('your_table')->select('*');
foreach($chassis as $key=>$val) {
    if($key == 0) {
        $collection->where('model_type_en', 'like', "%$val%"));
    }
    $collection->orWhere('model_type_en', 'like', "%$val%"));
}
$name = $collection->get();

This may work. You can also look at the ref: laravel querybuilder how to use like in wherein function

(original wrong ans:)

If chassis is a string, you can do this:

$this->builder->where('model_type_en', 'LIKE', "%$chassis%");

You can read the docs: https://laravel.com/docs/5.4/queries

edited Aug 17, 2017 at 12:35

answered Aug 17, 2017 at 12:27

ch271828n

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4 Comments

DarkMukke Over a year ago

I fear he might still get the array to string conversion

Dmitry Malys Over a year ago

chassis is an array :)

DarkMukke Over a year ago

I like the pure laravel solution, although this should probably be in a nested where because of the amount of OR statements, to avoid confusion when adding more conditions

Dmitry Malys Over a year ago

Thank you. Give me some time to try it out

D Coder · Accepted Answer · 2017-08-17 12:29:12Z

0

you can use as like be sure $chassis is an array

public function chassis($chassis)
{
    return $this->builder->whereIn('model_type_en', $chassis);
}

answered Aug 17, 2017 at 12:29

D Coder

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3 Comments

DarkMukke Over a year ago

Although this solves the array to string conversion problem, he cannot use a wildcard like in w LIKE condition

D Coder Over a year ago

if you want to use whereIn() then no need to use LIKE

DarkMukke Over a year ago

but his question indicates he needs LIKE and IN

DarkMukke · Accepted Answer · 2017-08-17 12:38:38Z

0

What you are tying to do is a combination of an IN and a LIKE.

~~I would suggest RLIKE which is a LIKE but with REGEX. Although this is mysql specific, so I doubt there is a Laravel build in way to do this.~~

Update

Sadly I was mistaken, and though RLIKE could do it. Instead find your answer in this post here : Is there a way to combine IN and LIKE in MySQL?

edited Aug 17, 2017 at 12:38

answered Aug 17, 2017 at 12:33

DarkMukke

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