Question

我有2个数组，两个数组都有共同点，就是Director_id要匹配

我正在尝试将一个数组的键与另一个数组的值映射。

$invest_details = DB::table('directors_invest')
    ->join('directors', 'directors.id', '=', 'directors_invest.director_id')
    ->select('directors_invest.*', 'directors.name as directors_name')
    ->get();
##

$expense_details = DB::table('receipts')
    ->join('directors', 'directors.id', '=', 'receipts.director_id')
    ->join('account_type', 'account_type.id', '=', 'receipts.type')
    ->join('expense_type', 'expense_type.id', '=', 'receipts.exp_earn_type')

    ->select('receipts.*', 'directors.name as directors_name', 'account_type.name as account_name', 'expense_type.name as exp_name')
    ->get();

//dd($expense_details);

$director  = DB::table('directors')->get();
$directors = [];
foreach ($director as $value) {
    $directors[$value->id] = $value->name;
}

// Here I will map Director_id and total invest
$investor_total_by_id = [];

foreach ($expense_details as $exp) {
    if (isset($investor_total_by_id[$exp->director_id])) {
        $investor_total_by_id[$exp->director_id] += $exp->amount;

    } else {
        $investor_total_by_id[$exp->director_id] = $exp->amount;

    }
}

dd($investor_total_by_id);
//this gives me out put
//
//      array:2 [▼
//  1 => 7100.0
//  2 => 444.0
//        ]

//** Match Convert Director_id to Director Name. so it will be
//     Director1 => 7100
//     Director2 => 440

$director_id_name_map = [];

foreach ($director as $value) {
    foreach ($investor_total_by_id as $key => $id_value) {

        if (key($investor_total_by_id) == $director[$value->id]) {

            $director_id_name_map[$directors->name] = $investor_total_by_id[$id_value->amount];

        }

    }
}

//dd($investor_total_by_id);
dd($$director_id_name_map);

return view('Accounts/Investment/add')
    ->with('invest_details', $invest_details)
    ->with('expense_details', $expense_details)
    ->with('directors', $directors)
    ->with('investor_total_by_id', $investor_total_by_id)

;
}

查看2个数组的输出

$ investor_total_by_id的输出

array:2 [▼
  1 => 7100.0
  2 => 444.0
]

输出$ director

Collection {#531 ▼
  #items: array:2 [▼
    0 => {#533 ▼
      +"id": 1
      +"name": "Directror1"
      +"phone": ""
      +"email": ""
      +"address": "hi"
      +"created_at": "2019-04-18 13:10:04"
      +"updated_at": "2019-04-18 07:10:04"
    }
    1 => {#534 ▼
      +"id": 2
      +"name": "Director2"
      +"phone": ""
      +"email": ""
      +"address": "hi"
      +"created_at": "2019-04-18 14:35:56"
      +"updated_at": "2019-04-18 07:10:04"
    }
  ]
}

我正在寻找将在Blade中创建表的地方

Director1 7100
Direcot2  440

表格将在

下方

<table border=1>
    <th> Director Name </th>
    <th> Amount</th>
    <td> Director1 <td>
    <td>
    <td> 7100 </td>
    <td> Director2 <td>
    <td>
    <td> 440 </td>
<table>

Answer 1

您需要遍历Directors数组，然后在循环的方式上，可以借助$investor_total_by_id进行握手并从中提取数据来满足director_id，如下所示。

<table border="1">
    <tr>
        <th>Director Name</th>
        <th>Amount</th>
    </tr>
    @foreach ($directors as $key => $value)
        <tr>
            <td>{{ $value->name }}</td>
            <td>{{ $investor_total_by_id[$value->id] }}</td>
        </tr>
    @endforeach
</table>

编辑

return view('foldername.viewname', compact('directors', 'investor_total_by_id'));

您可以像上面一样调用view并压缩两个数组。

编辑2

$director_id_name_map = [];
foreach ($directors as $key => $value) {
    $director_id_name_map[$value] = $investor_total_by_id[$key];
}

通过上述代码，您可以在控制器中映射您的期望。

compact-创建包含变量及其值的数组

Answer 2

好的，因此根据您的评论，我在下面考虑将其作为您的数据库表。

director_name（id，name）

收据（id，exp_earn_type，金额，director_id）

因此，首先，我们将两个表进行内部联接，并使用sum()来获得每位董事的总费用。
然后，我们内部将上一个子查询与director_name表联接在一起，以获取导演姓名及其其他详细信息。

代码：

<?php

    $director_expenses = DB::table('receipts')
                         ->join('director_name','receipts.director_id','=','director_name.id')
                         ->select('receipts.director_id', DB::raw('sum(receipts.amount) as total_expense'))
                         ->groupBy('receipts.director_id');

    $director_details = DB::table('director_name')
                        ->joinSub($director_expenses, 'director_expenses', function ($join) {
                            $join->on('director_name.id', '=', 'director_expenses.director_id');
                        })
                        ->select('director_name.id','director_name.name','director_expenses.total_expense')
                        ->get();

    dd($director_details);

如何将阵列键名称与另一个阵列键名称匹配

2 个答案: