Laravel 如何遍历具有多条记录的雄辩对象？

我正在使用以下代码从表中获取记录：

namespace App\Http\Controllers;

use Illuminate\Http\Request;
use Illuminate\Database\Eloquent\Collection;
use App\Http\Requests;
//use App\tempLogin;

class loginController extends Controller
{
    public function checkCredentials(Request $request){

        $mobile = $request->mobile;

        $users = App\tempLogin::where('mobile','=',$mobile)->get();


        return $user->name;

    }

---

但这是一个有说服力的对象，所以它给了我一个错误。还有什么办法呢？

您试图使用未定义的

$user

变量。您应该迭代集合以获得单个用户：

foreach ($users as $user) {
    echo $user->name;
}

$user = App\tempLogin::where('mobile','=',$mobile)->first();
$name = $user->name;

---

通常，您会像这样循环收集项目：

$users = App\tempLogin::where('mobile','=',$mobile)->get();

if($users->count() > 0)
{
   forach($users as $user)
   {
      // do something with the App\User model
      $name = $user->name;
   }
}

foreach($users as $user)
{
    // now the $user contains an array containing each row data.
    $name = $user["name"];
}

但是，如果要以数组形式访问用户集合，请按如下方式更改查询：

$users = App\tempLogin::where('mobile','=',$mobile)->toArray();

然后您可以像这样访问它：

$users = App\tempLogin::where('mobile','=',$mobile)->get();

if($users->count() > 0)
{
   forach($users as $user)
   {
      // do something with the App\User model
      $name = $user->name;
   }
}

foreach($users as $user)
{
    // now the $user contains an array containing each row data.
    $name = $user["name"];
}

如果要访问单个用户：

foreach ($users as $user) {
    echo $user->name;
}

$user = App\tempLogin::where('mobile','=',$mobile)->first();
$name = $user->name;

这将获取第一行（偏移量0，限制1）。您需要指定获得所需第一行的顺序。

您有一个输入错误

$users = \App\tempLogin::where('mobile','=',$mobile)->get();//you are getting $users

&返回

return $user->name;

return $user->name;

所以把它改成

return $users->name;

return $users->name;

但我仍然怀疑它是否能解决您的问题，因为您正在使用

get

返回

集合

，除非您在

模板中有关系

因此，我假设您在DB中有一条记录，它的mobile值等于
something&如果是这样的话，您希望返回它的名称
然后你可以使用这个解决方案

然后返回

return $users->name;

return $users->name;

如果数据库中有多条记录&您想执行一个
获取
，那么
你可以这样做

这将仅从表中选择名称

现在你可以回来了

return $users;

return $users;

这将是一个
tempLogin模型的集合&如果您在视图中使用它作为api，它将使用如下laravel请求自动进行JSON编码

[{"name":'YOUR_NAME_OF_THE_RECORD'},{"name":'YOUR_NAME_OF_THE_RECORD'}]

$users = \App\tempLogin::select(['name'])->where('mobile','=',$mobile)->pluck('name')->all();
return $users;

[
 "YOUR_NAME_OF_THE_RECORD",
 "YOUR_NAME_OF_THE_RECORD",
]You have a typo

$users = \App\tempLogin::where('mobile','=',$mobile)->get();//you are getting $users

[{"name":'YOUR_NAME_OF_THE_RECORD'},{"name":'YOUR_NAME_OF_THE_RECORD'}]

$users = \App\tempLogin::select(['name'])->where('mobile','=',$mobile)->pluck('name')->all();
return $users;

[
 "YOUR_NAME_OF_THE_RECORD",
 "YOUR_NAME_OF_THE_RECORD",
]

或者，如果要将记录作为数组返回，可以这样使用

[{"name":'YOUR_NAME_OF_THE_RECORD'},{"name":'YOUR_NAME_OF_THE_RECORD'}]

$users = \App\tempLogin::select(['name'])->where('mobile','=',$mobile)->pluck('name')->all();
return $users;

[
 "YOUR_NAME_OF_THE_RECORD",
 "YOUR_NAME_OF_THE_RECORD",
]You have a typo

$users = \App\tempLogin::where('mobile','=',$mobile)->get();//you are getting $users

[{"name":'YOUR_NAME_OF_THE_RECORD'},{"name":'YOUR_NAME_OF_THE_RECORD'}]

$users = \App\tempLogin::select(['name'])->where('mobile','=',$mobile)->pluck('name')->all();
return $users;

[
 "YOUR_NAME_OF_THE_RECORD",
 "YOUR_NAME_OF_THE_RECORD",
]

这将返回如下数组

[{"name":'YOUR_NAME_OF_THE_RECORD'},{"name":'YOUR_NAME_OF_THE_RECORD'}]

$users = \App\tempLogin::select(['name'])->where('mobile','=',$mobile)->pluck('name')->all();
return $users;

[
 "YOUR_NAME_OF_THE_RECORD",
 "YOUR_NAME_OF_THE_RECORD",
]You have a typo

$users = \App\tempLogin::where('mobile','=',$mobile)->get();//you are getting $users

[{"name":'YOUR_NAME_OF_THE_RECORD'},{"name":'YOUR_NAME_OF_THE_RECORD'}]

$users = \App\tempLogin::select(['name'])->where('mobile','=',$mobile)->pluck('name')->all();
return $users;

[
 "YOUR_NAME_OF_THE_RECORD",
 "YOUR_NAME_OF_THE_RECORD",
]

&返回

return $user->name;

return $user->name;

所以把它改成

return $users->name;

return $users->name;

但我仍然怀疑它是否能解决您的问题，因为您正在使用
get返回
集合，除非您在
模板中有关系

因此，我假设您在DB中有一条记录，它的mobile值等于
something&如果是这样的话，您希望返回它的名称
然后你可以使用这个解决方案

然后返回

return $users->name;

return $users->name;

如果数据库中有多条记录&您想执行一个
获取
，那么
你可以这样做

这将仅从表中选择名称

现在你可以回来了

return $users;

return $users;

这将是一个
tempLogin模型的集合&如果您在视图中使用它作为api，它将使用如下laravel请求自动进行JSON编码

[{"name":'YOUR_NAME_OF_THE_RECORD'},{"name":'YOUR_NAME_OF_THE_RECORD'}]

$users = \App\tempLogin::select(['name'])->where('mobile','=',$mobile)->pluck('name')->all();
return $users;

[
 "YOUR_NAME_OF_THE_RECORD",
 "YOUR_NAME_OF_THE_RECORD",
]You have a typo

$users = \App\tempLogin::where('mobile','=',$mobile)->get();//you are getting $users

[{"name":'YOUR_NAME_OF_THE_RECORD'},{"name":'YOUR_NAME_OF_THE_RECORD'}]

$users = \App\tempLogin::select(['name'])->where('mobile','=',$mobile)->pluck('name')->all();
return $users;

[
 "YOUR_NAME_OF_THE_RECORD",
 "YOUR_NAME_OF_THE_RECORD",
]

或者，如果要将记录作为数组返回，可以这样使用

[{"name":'YOUR_NAME_OF_THE_RECORD'},{"name":'YOUR_NAME_OF_THE_RECORD'}]

$users = \App\tempLogin::select(['name'])->where('mobile','=',$mobile)->pluck('name')->all();
return $users;

[
 "YOUR_NAME_OF_THE_RECORD",
 "YOUR_NAME_OF_THE_RECORD",
]You have a typo

$users = \App\tempLogin::where('mobile','=',$mobile)->get();//you are getting $users

[{"name":'YOUR_NAME_OF_THE_RECORD'},{"name":'YOUR_NAME_OF_THE_RECORD'}]

$users = \App\tempLogin::select(['name'])->where('mobile','=',$mobile)->pluck('name')->all();
return $users;

[
 "YOUR_NAME_OF_THE_RECORD",
 "YOUR_NAME_OF_THE_RECORD",
]

这将返回如下数组

[{"name":'YOUR_NAME_OF_THE_RECORD'},{"name":'YOUR_NAME_OF_THE_RECORD'}]

$users = \App\tempLogin::select(['name'])->where('mobile','=',$mobile)->pluck('name')->all();
return $users;

[
 "YOUR_NAME_OF_THE_RECORD",
 "YOUR_NAME_OF_THE_RECORD",
]You have a typo

$users = \App\tempLogin::where('mobile','=',$mobile)->get();//you are getting $users

[{"name":'YOUR_NAME_OF_THE_RECORD'},{"name":'YOUR_NAME_OF_THE_RECORD'}]

$users = \App\tempLogin::select(['name'])->where('mobile','=',$mobile)->pluck('name')->all();
return $users;

[
 "YOUR_NAME_OF_THE_RECORD",
 "YOUR_NAME_OF_THE_RECORD",
]

错误消息是什么？我只是返回Var_dump（$users），所以我得到了类似object的输出。你说“它给我带来了一个错误”？
$user
没有定义，但
$users
isI将其更改为：$mobile=$request->mobile$users=tempLogin:：where（'mobile'，'='，$mobile）->toArray（）；返回$users；但是我在db中有重复的手机号，所以它会获取多条记录，这就是为什么我想循环$users。这不起作用，因为->get（）将返回一个集合。不是模特儿。您需要迭代或使用索引访问来获取模型属性。但在文档中，我们可以循环访问集合。@ARIF…我更改了如下内容：$users=tempLogin:：where（'mobile'，'='，$mobile）->get（）；我现在可以访问名称吗？不，您不能，因为它是一个
集合
&您想访问集合中的模型属性，只有当它是一个模型实例时，您才能访问名称