如果常见则使用日期字段合并表，如果日期字段不常见则添加新行

Question

我正在开发房屋应用程序。我使用查询从不同的表名分别作为费用和收入来检索费用和收入的数据。 如果我只使用费用，那么它会打印出来，如果我也尝试混合收入值，它会给我带来问题。

代码：

 $expenses = $expense->dailyExpense();
 $income = $income->dailyIncome();
 return response()->json(['data' => ['expenses' => $expenses , 'income' => $income] , 'msg' => 'Daily Expense']);

收入的查询部分是：

    public function dailyIncome()
    {

        return $this->makeModel()
            ->select(DB::raw("sum(cost) as income"),"date")
            ->groupBy('date')
            ->get();
    }

收入的查询部分是：

 public function dailyExpense()
    {

        return $this->makeModel()
            ->select(DB::raw("sum(cost) as cost") , "date" , DB::raw("dayname(date) as calendar"))
            ->groupBy('date')
            ->get();
    }

客户部分：

 $scope.genereateReport = function () {

    $scope.choices = ['Daily', 'Monthly', 'Yearly'];
    $scope.$watch('selection', function (newVal, oldVal) {
        switch (newVal) {
            case 'Daily':

                $scope.url = $scope.base_path + 'dailyExpenses';
                $http.get($scope.url).success(function (response) {
                    $scope.expenses = response.data.expenses;
                    $scope.income = response.data.income;
                    $scope.totalExpenses = response.data.totalExpenses;


                });

                $scope.displayedCollection = [].concat($scope.expenses,$scope.income);
                console.log("collection:" + $scope.displayedCollection);
                $scope.totalExpenses = [].concat($scope.$totalExpenses);
                // $scope.income = [].concat($scope.income);
                //

                $scope.itemsByPage = 10;
                break;
 }
);
};

查看部分：

 <tr>
                                <th class="table-header-check">S.N</th>
                                <th st-sort="date" class="table-header-repeat line-left minwidth-1">Date</th>
                                <th st-sort="date" class="table-header-repeat line-left minwidth-1">Calandar</th>
                                <th st-sort="cost" class="table-header-repeat line-left minwidth-1">
                                    Expense
                                </th> <th st-sort="cost" class="table-header-repeat line-left minwidth-1">
                                    Income
                                </th>

                            </tr>


                            </thead>
                            <tbody>

                            <tr data-ng-repeat="row in displayedCollection track by $index" >
                                <td><% $index+1 %></td>
                                <td><%row.date%></td>
                                <td><%row.calendar%></td>
                                <td><%row.cost%></td>

                                <td><%income['row.date']%></td>


                            </tr>
                            <tr>
                                <td colspan="4">
                                   Total Expense: <%totalExpenses%>
                                </td>


                            </tr>

上述json形式的查询结果为：

不显示收入的视图是这样的：

想要的输出是这样的..

]3

Answer 1

您可以通过使用 sql union 和 join 来获得预期的结果。您可以创建一个函数并使用原始查询，而不是创建两个单独的函数，如下所示：

select e.date,e.cost,'0' as income
from test.expenses as e
where e.date not in (select i.date from test.incomes as i)
union all
select e.date, e.cost as cost, i.income as income
from test.expenses as e
inner join test.incomes as i on e.date = i.date 
union all
select i.date,'0' as cost,i.income as income
from test.incomes as i
where i.date not in (select e.date from test.expenses as e);

在这里，我创建了两个名为expenses 和incomes 的表，其中包含您上面指定的表中的给定字段。我使用上面提到的查询，结果如下。

我希望这是你所期望的。