我有2个Array，两个Array的共同点是Director_id来匹配

我正在尝试将一个数组的键映射到另一个数组的值。

$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"
    }
  ]
}

我希望在 Blade 中创建一个表

Director1 7100
Direcot2  440

表格会在下面

<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>

最佳答案

好的，根据您的意见，我正在考虑将以下内容作为您的数据库表。

director_name(id,name)

receipts(id,exp_earn_type,amount,director_id)

• 因此，首先我们内部连接两个表并使用 sum() 获得每位董事的总费用。

• 然后我们将前面的子查询与 director_name 表进行内部连接，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);

关于php - 如何将数组键名称与另一个数组键名称匹配，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/55757079/