php 如果记录存在则 Laravel 更新，如果不存在则创建
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/33139076/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Laravel Update if record exists or Create if not
提问by ssuhat
I want to update my record if it exists but create new one if not exists. here is i got so far:
我想更新我的记录（如果存在），但如果不存在则创建新记录。这是我到目前为止：
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.
它工作得很好。
But as i know Laravel is very powerful for its eloquent model. Can I make it shorter? or i already doing it correctly?.
但据我所知，Laravel 的雄辩模型非常强大。我可以缩短它吗？还是我已经做对了？
I've tried using "updateOrCreate" method but my foreign key "merchant_branch_id" need to be fillable.
我试过使用“updateOrCreate”方法，但我的外键“merchant_branch_id”需要可填充。
回答by tuananh
Laravel provide method updateOrCreatefor that purpose
Laravel提供方法updateOrCreate为目的
If there's a flight from Oakland to San Diego, set the price to $99.
If no matching model exists, create one.
如果有从奥克兰飞往圣地亚哥的航班，请将价格设置为 99 美元。
如果不存在匹配的模型，则创建一个。
$flight = App\Flight::updateOrCreate(
['departure' => 'Oakland', 'destination' => 'San Diego'],
['price' => 99]
);
回答by Safoor Safdar
Laravel already using this methodology by savefunction
Laravel 已经按save功能使用了这种方法
$user->save()
Laravel Code
Laravel 代码
// If the model already exists in the database we can just update our record
// that is already in this database using the current IDs in this "where"
// clause to only update this model. Otherwise, we'll just insert them.
if ($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
https://github.com/laravel/framework/blob/5.1/src/Illuminate/Database/Eloquent/Model.php#L1491
->exists
->exists
All laravel models have a ->existsproperty.
所有 Laravel 模型都有一个->exists属性。
More specifically if the model is either loaded from the database, or has been saved to the database since being created the existsproperty will be true; Otherwise it will be false.
更具体地说，如果模型是从数据库加载的，或者自创建以来已保存到数据库中，则该exists属性将为真；否则会是假的。
If you understand the ->existsyou can use it but here is the another way to deal such requirement.
如果您了解->exists您可以使用它，但这是处理此类要求的另一种方法。
another way.
其它的办法。
/**
* 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;
}
回答by kamran karimi
Add new function code :
添加新的功能代码：
vendor/laravel/framework/src/Illuminate/Database/Eloquent/Builder.php :
供应商/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;
}
