rich-notifications 相关问题


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如何设置 Visual Studio 2022 .Net 7.0 以创建通知侦听器

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如何批量删除已合并的拉取请求的 Github 通知?

我正在查看我的 Github 通知 (https://github.com/notifications),并且有大量与已合并的拉取请求相关的未读通知。 我希望能够


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应用程序关闭时处理电容器本地通知上的点击事件

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jQuery Ajax 在 php 同一页面上传递值 - 更新

如何找回: 如何找回: <div id="test"> <?php if (isset($_POST['sweets'])) { ob_clean(); echo $_POST['sweets']; exit; } ?> </div> <form id="a" action="" method="post"> <select name="sweets" onchange="change()" id="select1"> <option >Chocolate</option> <option selected="selected">Candy</option> <option >Taffy</option> <option >Caramel</option> <option >Fudge</option> <option >Cookie</option> </select> </form> <!-- Script --> <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.6.0/jquery.min.js"></script> <script> function change() { var sweets = $("#select1").val(); $.ajax({ type: "POST", data: { sweets: sweets }, success: function(data) { $("#test").html(data); } }); } </script> 将值传递给 php 字符串: $string = $_POST['sweets']; <!-- I'm looking for this: --> 我希望这是可能的。我在 stackoverflow 和 google 上寻找答案,但找不到适合我的目的的答案。 对于同一个页面的ajax/PHP脚本,可以将PHP放在脚本的最前面,当有POST提交数据时以exit结束 为了使其更有意义,您应该返回与您通过 POST 提交的内容相关的内容(这是甜食的类型),作为示例,我们展示其一般定义。我们可以使用 switch,这是用于此目的的常用结构: switch ($string) { case "Chocolate": echo "Chocolate is made from cocoa beans, the dried and fermented seeds of the cacao tree"; break; case "Candy": echo "Candy is a sweet food made from sugar or chocolate, or a piece of this"; break; case "Taffy": echo "Taffy is a type of candy invented in the United States, made by stretching and/or pulling a sticky mass of a soft candy base"; break; case "Caramel": echo "Caramel is made of sugar or syrup heated until it turns brown, used as a flavouring or colouring for food or drink"; break; case "Fudge": echo "Fudge is a dense, rich confection typically made with sugar, milk or cream, butter and chocolate or other flavorings"; break; case "Cookie": echo "A cookie (American English) or biscuit (British English) is a baked snack or dessert that is typically small, flat, and sweet"; break; } exit; } ?> 所以以下是示例代码: <?php if (isset($_POST['sweets'])) { // ob_clean(); $string = $_POST['sweets']; switch ($string) { case "Chocolate": echo "Chocolate is made from cocoa beans, the dried and fermented seeds of the cacao tree"; break; case "Candy": echo "Candy is a sweet food made from sugar or chocolate, or a piece of this"; break; case "Taffy": echo "Taffy is a type of candy invented in the United States, made by stretching and/or pulling a sticky mass of a soft candy base"; break; case "Caramel": echo "Caramel is made of sugar or syrup heated until it turns brown, used as a flavouring or colouring for food or drink"; break; case "Fudge": echo "Fudge is a dense, rich confection typically made with sugar, milk or cream, butter and chocolate or other flavorings"; break; case "Cookie": echo "A cookie (American English) or biscuit (British English) is a baked snack or dessert that is typically small, flat, and sweet"; break; } exit; } ?> <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.6.0/jquery.min.js"></script> <select name="sweets" onchange="change()" id="select1"> <option value="">Please select</option> <option >Chocolate</option> <option >Candy</option> <option >Taffy</option> <option >Caramel</option> <option >Fudge</option> <option >Cookie</option> </select> <br><br> <div id="test"></div> <script> function change() { var sweets = $("#select1").val(); $.ajax({ type: "POST", data: { sweets: sweets }, success: function(data) { $("#test").html(data); } }); } </script>


Laravel 中的策略对我不起作用,这是我的代码

我无法让策略在我的 Laravel 项目中工作,我安装了一个新项目来从头开始测试,我有这个控制器: 我无法让策略在我的 Laravel 项目中工作,我安装了一个新项目来从头开始测试,我有这个控制器: <?php namespace App\Http\Controllers; use App\Http\Controllers\Controller; use Illuminate\Http\Request; use App\Models\User; class UserController extends Controller { public function index() { $this->authorize('viewAny', auth()->user()); return response("Hello world"); } } 本政策: <?php namespace App\Policies; use Illuminate\Auth\Access\Response; use App\Models\User; class UserPolicy { public function viewAny(User $user): bool { return true; } } 这是我的模型 <?php namespace App\Models; // use Illuminate\Contracts\Auth\MustVerifyEmail; use Illuminate\Database\Eloquent\Factories\HasFactory; use Illuminate\Foundation\Auth\User as Authenticatable; use Illuminate\Notifications\Notifiable; use Laravel\Sanctum\HasApiTokens; class User extends Authenticatable { use HasApiTokens, HasFactory, Notifiable; /** * The attributes that are mass assignable. * * @var array<int, string> */ protected $fillable = [ 'name', 'email', 'password', ]; /** * The attributes that should be hidden for serialization. * * @var array<int, string> */ protected $hidden = [ 'password', 'remember_token', ]; /** * The attributes that should be cast. * * @var array<string, string> */ protected $casts = [ 'email_verified_at' => 'datetime', 'password' => 'hashed', ]; } 我收到错误 403:此操作未经授权。我希望有人能帮助我解决我的问题。谢谢你 我也尝试过修改AuthServiceProvider文件,但没有任何改变。 必须在 App\Providers\AuthServiceProvider 中添加您的策略吗? protected $policies = [ User::class => UserPolicy::class ]; 您需要指定您正在使用的模型。具体来说,就是User。因此,传递当前登录的用户: $this->authorize('viewAny', auth()->user()); 此外,您正在尝试验证用户是否有权访问该页面。确保尝试访问该页面的人是用户,以便策略可以授权或不授权。 要在没有入门套件的情况下进行测试,请创建一个用户并使用它登录。 <?php namespace App\Http\Controllers; use Illuminate\Support\Facades\Auth; class UserController extends Controller { public function index() { $user = \App\Models\User::factory()->create(); Auth::login($user); $this->authorize('viewAny', auth()->user()); return response("Hello world"); } } 但是,如果您希望授予访客用户访问权限,您可以使用 ? 符号将 User 模型设为可选: public function viewAny(?User $user) { return true; }


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