以模型为参数的资源控制器方法不起作用 [英] Resource Controller methods with model as parameter not working

问题描述

就像基本的Laracasts.com教程(从零开始的Laracast 5.7)一样，我试图使用以下方法 public function show(prototypes $ prototypes)参数来构建视图.但是我的视图创建正确，但是 $ prototypes 为空.

As in the basic Laracasts.com tutorial (Laracast 5.7 from scratch) I'm trying to use the following methods public function show(prototypes $prototypes) parameter to construct a view. However my view is created correctly but $prototypes is null.

路由运行良好(/prototypes/1/edit)，我确保存在一个 ID 为 1 的 prototype 对象.我找到了一些较旧的解决方案，该解决方案声明使用诸如(integer $ id)之类的参数，但这会导致更多代码.它应该像这样工作:

The route works well(/prototypes/1/edit) and I ensured that a prototype object with the id 1 exists. I found some older solution which stated to use something like (integer $id) as parameter but this leads to some more code. It should work like this:

控制器:

public function edit(prototypes $prototypes)
{
    //
    return view('prototypes.edit', compact('prototypes'));
}

根据从头开始播放，这应该可以.

您知道我该如何解决吗?

Do you know how I could fix this?

prototypes.edit 方法知道如何使用正确的参数的背后机制是什么?

What mechanism is behind this that the prototypes.edit method knows how to use the correct parameter?

推荐答案

对于隐含使注入的变量名称有效的模型绑定应该与 route参数名称相匹配，在您的情况下，我认为您的参数名称可以为 {prototype}，您可以通过在控制台中发出命令 php artisan route:list 来进行验证.

For the Implicit Model Binding to works the injected variable name should match the route parameter name, in your case I think that your parameter name could be {prototype}, you can verify it by issuing the command php artisan route:list in the console.

如果是这样，则必须在控制器函数中将变量名称更改为 $ prototype (请注意单数形式)以匹配参数名称 {prototype} ，像这样:

If that is true you have to change the variable name to $prototype (please note the singular) in your controller function to match the parameter name {prototype}, like this:

public function edit(prototypes $prototype)
{
   return view('prototypes.edit', compact('prototype'));
}

更新:顺便说一句，关于Model名称的laravel约定是单数驼峰式，在您的情况下，您的Model应该命名为 Prototype 而不是 Prototypes ，即:

Update: BTW the laravel convention on Model's name is singular camel case, in your case your Model should be named Prototype not prototypes, i.e.:

public function edit(Prototype $prototype)
{
   return view('prototypes.edit', compact('prototype'));
}

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