Laravel 更新不起作用
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Laravel Update not working
提问by Saurabh Parekh
I have table called admins and model called Admin.
我有名为 admins 的表和名为 Admin 的模型。
When i try to update the records it fails every time with one field.
当我尝试更新记录时，它每次都用一个字段失败。
$admindetail = Admin::find($id);
$admindetail->fill(['is_delete'=>1])->save();
Above code is not updating record in DB. datatype for is_deleteis tinyint.
上面的代码没有更新数据库中的记录。数据类型为is_deletetinyint。
If i update name field it works with the same code like
如果我更新名称字段，它可以使用相同的代码
$admindetail = Admin::find($id);
$admindetail->fill(['name'=>'abc'])->save();
Can you help me why first code is not working.
你能帮我为什么第一个代码不起作用。
回答by Ramzan Mahmood
Modal use to interact with table so all attributes which you need to store in DB you should need to define in your model as your Admin modal should look like
模态用于与表交互，因此您需要存储在数据库中的所有属性都需要在模型中定义，因为您的管理模态应该看起来像
<?php
namespace App;
use Illuminate\Database\Eloquent\Model;
class Admin extends Model
{
protected $fillable = ['name', 'is_delete'];
}
You missing is_deletefrom your Modal code.
你错过is_delete了你的模态代码。
回答by sathish R
Instead of passing value as array, you could try the code below:
您可以尝试以下代码，而不是将值作为数组传递：
$admindetail = Admin::find($id);
if(!$admindetail)
{
$admindetail->is_delete=1;
$admindetail->->save();
}
回答by Andrey Lutskevich
Use boolean type, no integer 1: $adminDetail->fill(['is_delete' => true])->save();
使用布尔类型，无整数 1： $adminDetail->fill(['is_delete' => true])->save();
And you can use as property without fillmethod and you do not need accept is_deletein the $fillablearray:
您可以在没有fill方法的情况下用作属性，并且不需要is_delete在$fillable数组中接受：
$adminDetail->is_delete = true;
$adminDetail->save();
回答by Gilbert Tuazon
You should check your fillable in your model like this:
您应该像这样检查模型中的可填充物：
protected $fillable = [
'name',
'is_delete'
];
