Laravel Update if record exists or Create if not

Question

I want to update my record if it exists but create new one if not exists. here is i got so far:

 MerchantBranch.php
public function token()
{
   return $this->hasOne('App\MerchantBranchToken');
}

MerchantBranchToken.php
public function merchant_branch()
{
   return $this->belongsTo('App\MerchantBranch');
}

$find = MerchantBranchToken::find($id);

    if (!$find) {
        $branch = new MerchantBranchToken(['token' => $token]);
        MerchantBranch::find($id)->token()->save($branch);
    } else {
        $find->token = $token;
        $find->save();
    }

It's working perfectly.

But as i know Laravel is very powerful for its eloquent model. Can I make it shorter? or i already doing it correctly?.

I've tried using "updateOrCreate" method but my foreign key "merchant_branch_id" need to be fillable.

The Alpha · Accepted Answer · 2017-08-21 11:00:07Z

33

Laravel provide method updateOrCreate for that purpose

If there's a flight from Oakland to San Diego, set the price to $99.
If no matching model exists, create one.

$flight = App\Flight::updateOrCreate(
    ['departure' => 'Oakland', 'destination' => 'San Diego'],
    ['price' => 99]
);

edited Aug 21, 2017 at 11:00

The Alpha

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answered Nov 13, 2016 at 10:43

tuananh

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4 Comments

tuananh Over a year ago

If you define primary key in database, you must define it in model protected $primaryKey = 'flight_id';

Dr. House Over a year ago

what happens if there is a flight from Oakland to San Diego with the price of $99 ?

PoorBob Over a year ago

Actually, I'm pretty sure it would not send an update at all. Normally eloquent will check the isDirty() property before saving. Am I wrong about this?

Dylan Glockler Over a year ago

In that example, if there was already a flight OAK to SAN at 99, it would update the price to 99, from 99. If you set 'price' => 199, it would update the price to 199.

Community · Accepted Answer · 2020-06-20 09:12:55Z

8

Laravel already using this methodology by save function

$user->save()

Laravel Code

// If the model already exists in the database we can just update our record
// that is already in this database using the current IDs in this "where"
// clause to only update this model. Otherwise, we'll just insert them.
if ($this->exists)
{
    $saved = $this->performUpdate($query);
}

// If the model is brand new, we'll insert it into our database and set the
// ID attribute on the model to the value of the newly inserted row's ID
// which is typically an auto-increment value managed by the database.
else
{
    $saved = $this->performInsert($query);
}

https://github.com/laravel/framework/blob/5.1/src/Illuminate/Database/Eloquent/Model.php#L1491

`->exists`

All laravel models have a ->exists property.

More specifically if the model is either loaded from the database, or has been saved to the database since being created the exists property will be true; Otherwise it will be false.

If you understand the ->exists you can use it but here is the another way to deal such requirement.

another way.

/**
     * Create or update a record matching the attributes, and fill it with values.
     *
     * @param  array  $attributes
     * @param  array  $values
     * @return static
     */
    public static function updateOrCreate(array $attributes, array $values = array())
    {
        $instance = static::firstOrNew($attributes);

        $instance->fill($values)->save();

        return $instance;
    }

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Oct 15, 2015 at 5:29

Safoor Safdar

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5 Comments

ssuhat Over a year ago

Hi. Sorry but i don't get the point. Would you mind explain it a little bit more?

Safoor Safdar Over a year ago

I think you have to read some code documentation related to $this->exists how its work...

Safoor Safdar Over a year ago

@sstarlight please have a look to my updated answer.

ssuhat Over a year ago

I've tried second way using updateOrCreate but i got foreign key constraint fails because my merchant_branch_id at MerchantToken is not fillable. Exists i think its almost same way like i do right now. using "if"

Rolf Over a year ago

The database could already contain a record with the same primary key value (usually id). save() would then happily try to insert (instead of update) and fail with an error (unique constraint).

Ali Yazdanifar · Accepted Answer · 2022-06-15 11:06:15Z

0

very simple.

first find record then remove if exist and insert again

$record = Model::where(['id'=>1]);
if ($record->exists()) {
        $record->delete();
}
Model::create($request->all());

answered Jun 15, 2022 at 11:06

Ali Yazdanifar

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2 Comments

pedram shabani Over a year ago

it doesn't work because after delete record. id will be 2. you should use it Model::findOrFail(1)->update($request->all());

Phil Tune Over a year ago

Model::updateOrCreate([...]) is the Laravel standard way to do this. No need to make three separate calls for this single action. Most databases support this as a single call.

DjKarimi · Accepted Answer · 2016-12-14 19:19:57Z

-1

Add new function code :

vendor/laravel/framework/src/Illuminate/Database/Eloquent/Builder.php :

  public function updateOrInsert(array $attributes, array $values = [])
    {
        $instance = $this->where($attributes);
        if ($instance->count() != 0) {
            $instance->update($values);
        } else {
            $instance = $this->updateOrCreate($attributes, $values);
        }
        return $instance;
    }

answered Dec 14, 2016 at 19:19

DjKarimi

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1 Comment

Ethem Over a year ago

thank you, this is the code I'm looking for, the others are meaningless, I need to check for a single value as in this code.

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