Yos*_*sef 4 php laravel laravel-5 laravel-query-builder
好的,所以Trying to get property of non-object当我尝试从数据库中获取数据时,我得到了$settings = AdminSettings::first();
这是控制器代码
<?php
namespace App\Http\Controllers\AdminSettings;
use App\AdminSettings\AdminSettings;
use Illuminate\Http\Request;
use App\Http\Controllers\Controller;
class AdminSettingsController extends Controller
{
public function index()
{
$settings = AdminSettings::first();
return view('admins.settings.settings', compact('settings'));
}
}
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这是模态代码
<?php
namespace App\AdminSettings;
use Illuminate\Database\Eloquent\Model;
class AdminSettings extends Model
{
protected $table = 'site_settings';
protected $fillable = [
'site_title', 'site_url', 'email_from', 'email_to', 'timezone', 'date_format', 'time_format',
];
}
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在这里,我试图把它 site_title into the input but I get Trying to get property of non-object
<div class="form-group">
<label for="site_title" class="form-label">Site Title</label>
<input type="text" class="form-control" name="site_title" id="site_title" value="{{ $settings->site_title }}"/>
</div>
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当我尝试dd($settings);,我得到null
你说表是空的,所以make setting对象optional:
{{ optional($settings)->site_title }}
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