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Home Backend Development PHP Tutorial 请教 laravel view 怎么显示 controller 中传递的对象值。

请教 laravel view 怎么显示 controller 中传递的对象值。

Jun 06, 2016 pm 08:16 PM
laravel php

Controller 代码:

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<code>public function show(Request $request,$id){

$person = DB::table('persons')->where('id',$id)->get();

return view('showuser',compact('person'));

}

</code>

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我在 view/showuser.blade.php 怎么写才能显示出 person 这个对象的值?

必须这么写才行 {{ $person[0]->id }} ???

感觉这样显示的值是个数组啊。 不是对象啊。

view()->withXXX
view('xxx,compact('xxx'))
这些传递到 view 的数据 都应该怎么显示???
这些数据都是什么形式?
希望熟悉 laravel 的 指点下~

回复内容:

Controller 代码:

1

2

3

4

5

<code>public function show(Request $request,$id){

$person = DB::table('persons')->where('id',$id)->get();

return view('showuser',compact('person'));

}

</code>

Copy after login
Copy after login

我在 view/showuser.blade.php 怎么写才能显示出 person 这个对象的值?

必须这么写才行 {{ $person[0]->id }} ???

感觉这样显示的值是个数组啊。 不是对象啊。

view()->withXXX
view('xxx,compact('xxx'))
这些传递到 view 的数据 都应该怎么显示???
这些数据都是什么形式?
希望熟悉 laravel 的 指点下~

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<code>$person = DB::table('persons')->where('id',$id)->get();

</code>

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这句其实是从 Builder 对象中取回一个 Collection 对象,而该 Collection 对象中包含了 Person 对象。Collection 实现了数组相关的接口,所以也可以使用访问数组的方式获取 Person 对象。

如果想通过 id 直接获取 Person 对象,可以使用 find 方法:

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<code>$person = DB::table('person')->find($id);

</code>

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Blade 中可以使用 @for、@foreach 或 @forelse 等遍历对象集合,再显示对象的属性值:

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<code>@forelse($persons as $person)

    <li> {{ $person->name }} </li>

@empty

    <span>Nothing.</span>

@endforelse

</code>

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对于单个对象自然直接访问即可:

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<code>Hello, {{ $person->name }} !</code>

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楼主既然写了获取某个ID的用户,那根本用不到 foreach 了吧。只获取一个用户的可以直接用 first() 方法:

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<code>$person = DB::table('persons')->where('id',$id)->first();

// Or

$person = DB::table('persons')->where('id',$id)->firstOrFail();</code>

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这样的话在 view 中可以直接 $person->name 来显示,而不用 $person[0]->name 了。

你需要@foreach

楼上正解

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<code>@foreache($person as $temp)

 {$temp->foo}

@endforeach

</code>

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你查询出来的数据形式可以试着

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<code>dd();

</code>

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laravel中传递的数据一般都为JSON数组形式

同在学习laravel,如有错误的地方欢迎指出

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