Question

我确实有这个laravel集合

    Collection {#239 ▼
      #items: array:2 [▼
        0 => array:21 [▼
          "dueDate" => "2017-09-29"
          "groupByCode" => "REMINDER"
          "date" => "2017-09-29"
          "number" => "9030014"
          "status" => "Reminder"
          "currencyCode" => "kr"
          "amount" => 1745.0
          "remainingAmount" => 1745.0

        ],
  2 => array:21 [▼
          "dueDate" => "2017-09-29"
          "groupByCode" => "REMINDER"
          "date" => "2017-09-29"
          "number" => "9030014"
          "status" => "Reminder"
          "currencyCode" => "kr"
          "amount" => 1345.0
          "remainingAmount" => 1745.0

        ],
        3 => array:21 [▼
          "dueDate" => "2017-09-29"
          "groupByCode" => "INVOICE"
          "date" => "2017-09-29"
          "number" => "9030026"
          "status" => "invoice"
          "currencyCode" => "kr"
          "amount" => 2389.0
          "remainingAmount" => 2389.0

        ]
      ]
    }

现在我的问题是如何根据状态获取唯一的项目，上面的示例我有两个状态为Reminder的项目，而我希望保持相同的集合格式，但作为回报，只有一个项目的状态为“Reminder”with所有现有的密钥和总和将是两个提醒的总和....

我曾尝试使用`where（'status'，''Reminder'） - ＆gt; first（）但不起作用，因为它只需要firts元素。

所以输出应该是这样的：

    Collection {#239 ▼
      #items: array:2 [▼
        0 => array:21 [▼
          "dueDate" => "2017-09-29"
          "groupByCode" => "REMINDER"
          "date" => "2017-09-29"
          "number" => "9030014"
          "status" => "Reminder"
          "currencyCode" => "kr"
          "amount" => TOTAL OF BOTH REMINDERS
          "remainingAmount" => 1745.0

        ],
  2 => array:21 [▼
          "dueDate" => "2017-09-29"
          "groupByCode" => "INVOICE"
          "date" => "2017-09-29"
          "number" => "9030014"
          "status" => "Invoice"
          "currencyCode" => "kr"
          "amount" => 2312 
          "remainingAmount" => 1745.0

            ],
}

谢谢！

以下是代码：

$invoices = $this->boatService->getInvoices(cleanSsn(session('ssn')));

$invoices = collect($invoices)->whereIn('status', ['Reminder', 'Invoice']);

Answer 1

您可以进行原始查询以对项目求和并按状态对其进行分组。试试这个：

$collection = \DB::table('invoices')
                 ->select(\DB::raw('sum(amount) as amount_total, status'))
                 ->groupBy('status')
                 ->get();

从集合中仅获取唯一项目，并将其他值汇总到一个项目

1 个答案: