Laravel 4.2

Question

我有150个对象的非常大的集合

 ...
 App\almacen {#1679
     id: 124,
     emp_id: 1,
     dst_id: 13,
     hora: 0,
     numMesa: 0,
     event_id: 1,
     created_at: "2018-01-22 11:41:03",
     updated_at: "2018-01-22 11:41:03",
   },
   App\almacen {#1680
     id: 125,
     emp_id: 1,
     dst_id: 11,
     hora: 0,
     numMesa: 0,
     event_id: 1,
     created_at: "2018-01-22 11:41:03",
     updated_at: "2018-01-22 11:41:03",
   },
   App\almacen {#1681
     id: 126,
     emp_id: 1,
     dst_id: 12,
     hora: 0,
     numMesa: 0,
     event_id: 1,
     created_at: "2018-01-22 11:41:03",
     updated_at: "2018-01-22 11:41:03",
   },
   App\almacen {#1682
     id: 127,
     emp_id: 1,
     dst_id: 20,
     hora: 0,
     numMesa: 0,
     event_id: 1,
     created_at: "2018-01-22 11:41:03",
     updated_at: "2018-01-22 11:41:03",
   },
   App\almacen {#1683
     id: 128,
     emp_id: 1,
     dst_id: 7,
     hora: 0,
     numMesa: 0,
     event_id: 1,
     created_at: "2018-01-22 11:41:03",
     updated_at: "2018-01-22 11:41:03",
   },
  ...

我必须比较密钥emp_id和dst_id是否等于例如8的数字并返回＆＃39; true＆＃39;如果它存在有没有办法做到这一点，而不使用'foreach＆＃39;直接看对象？已经使用contains（），search（）但总是返回＆＃39; false＆＃39;即使价值确实存在

Answer 1

Firstwhere：

$exists = $items->firstWhere(['emp_id' => 8, 'dst_id' => 8]) !== null;

首先：

$needle = 8;
$exists = $items->first(function($item) use($needle) { 
      return $item->emp_id === $needle && $item->dst_id === $needle ;
 }) !== null;

过滤器：

$needle = 8;
$exists = $items->filter(function($item) use($needle) { 
      return $item->emp_id === $needle && $item->dst_id === $needle ;
 })->count() > 0;

Answer 2

您可以像这样在where method上使用collections：

$collection = collect([
    ['product' => 'Desk', 'price' => 200],
    ['product' => 'Chair', 'price' => 100],
    ['product' => 'Bookcase', 'price' => 150],
    ['product' => 'Door', 'price' => 100],
]);

$filtered = $collection->where('price', 100);

$filtered->all();

Answer 3

Laravel 4.2

firstWhere （不起作用）

first:

$needle = 8;
$exists = $items->first(function($key, $item) use ($needle) { 
      return $item->emp_id === $needle && $item->dst_id === $needle ;
 }) !== null;

filter ：

$needle = 8;
$exists = $items->filter(function($key, $item) use ($needle) { 
      return $item->emp_id === $needle && $item->dst_id === $needle ;
 })->count() > 0;

Answer 4

就用

return $collection->where('price', '100')->toArray();

在集合中搜索laravel

4 个答案:

Laravel 4.2