将数据存储在Laravel控制器的2个不同表格中
问题描述:
请问我可以帮助一下吗?我正在尝试与可以访问此业务的企业和用户建立一个系统。用户与DB中的业务相关,并具有名为“id_business”的列,该列是该用户可以管理的业务的ID。起初,我创建了用户,我默认设置'id_business'的值为0.然后,当我创建新业务时,我设置业务名称并从下拉列表中(用户的值为'id_business = 0 ')我选择这个业务的用户。现在我的问题是,如何以一种形式存储新业务,并在用户表中设置'id_business'的价值,就像我刚刚创建的新业务的ID一样。我需要的用户,这是列表:将数据存储在Laravel控制器的2个不同表格中
<select name="id_business" class="form-control input-xlg">
<option value="">Select User </option>
@foreach($users as $user)
<option value="{{$user->id}}">{{$user->name}}</option>
@endforeach
</select>
在我控制我已经远远走了这一点,但它不工作:
public function store(Request $request)
{
$businesses = new Businesses();
$businesses->name = $request->name;
$businesses->user_create_id = Auth::user()->id;
$user= Businesses_Users::first();
if($businesses->save())
{
$request->id_business;
$user->id = Businesses_Users::find($request->id_business);
$user->id_business = $businesses->id;
return redirect()->route('businesses',$businesses->id)->with('success', sprintf('Business successfully created.'));
}
else
{
return redirect()->back()->with('error', sprintf('An unexpected error occurred.Please fill a report form on this issue.'));
}
}
在此先感谢。
答
缺少save()
功能
$user = Businesses_Users::find($request->id_business);
$user->id_business = $businesses->id;
$user->save();
什么错误? – JYoThI
@JYoThI它没有错误,它不工作。它不会将“id_business”保存到用户表中。 – User154584
缺少保存功能$ user = Businesses_Users :: find($ request-> id_business); $ user-> id_business = $ business-> id; $用户)>保存(; – JYoThI