如何在laravel中插入多条记录而不循环

Posted 2023-02-16

【中文标题】如何在laravel中插入多条记录而不循环

【英文标题】：How to insert multiple record without loop in laravel

【发布时间】：2021-01-29 22:33:59

【问题描述】：

我需要在数据库中插入多条记录。目前我正在插入循环，当记录很大时会导致超时。有什么不使用循环的方法吗？

$consignments =   Consignment::select('id')->where('customer_id',$invoice->customer_id)->doesntHave('invoice_charges')->get();
       foreach($consignments as $consignment)
         InvoiceCharge::create(['invoice_id'=>$invoice->id,'object_id'=>$consignment->id,'model'=>'Consignment']);
       

寄售在模型中有hasOne 关系

public function invoice_charges()
    
        return $this->hasOne('App\Models\Admin\InvoiceCharge', 'object_id')->where('model', 'Consignment');
    

【参考方案1】：

这个怎么样：

$consignments = Consignment::select('id')->where('customer_id',$invoice->customer_id)->doesntHave('invoice_charges')->get();
       foreach($consignments as $consignment)
         $consignment_data[] = ['invoice_id'=>$invoice->id,'object_id'=>$consignment->id,'model'=>'Consignment'];
       
InvoiceCharge::insert($consignment_data);

通过这种方式，您可以使用一个查询而不是循环输入。只需检查 consignment_data 数组是否正常。

【讨论】：

给出这个错误Argument 1 passed to Illuminate\Database\Grammar::parameterize() must be of the type array, string given, called in /var/www/html/coldxlogistics/vendor/laravel/framework/src/Illuminate/Database/Query/Grammars/Grammar.php on line 869

你能在 InvoiceCharge::create($consignment_data) 之前 dd($consignment_data);

你的方法是正确的但是需要使用insert而不是create

【参考方案2】：

如果你想节省时间，但可以给更多内存，可以使用Cursor，

光标：您将使用 PHP 生成器逐一搜索您的查询项目。 1）花费更少的时间 2）使用更多的内存

$consignments =   Consignment::select('id')->where('customer_id',$invoice->customer_id)->doesntHave('invoice_charges')->cursor();
       foreach($consignments as $consignment)
         InvoiceCharge::create(['invoice_id'=>$invoice->id,'object_id'=>$consignment->id,'model'=>'Consignment']);
       

您可以参考here