合并表使用日期字段，如果共同，并添加新行，如果日期字段不常见

我正在开发家app.I检索数据使用查询费用和来自不同表名称收入分别作为开支和收入。如果我只使用费用，那么它会打印，如果我尝试混合收入价值，它会给我问题。合并表使用日期字段，如果共同，并添加新行，如果日期字段不常见

代码：

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

收入查询部分是：

public function dailyIncome() 
    { 

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

收入查询部分是：

public function dailyExpense() 
    { 

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

客户端部分：

$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; 
} 
); 
};

视图部分：在JSON形式上述查询的

<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>

结果是：

View是这样，不显示收入：

希望的输出是这样的..

enter image description here ] 3

来源

2016-07-27 user2851129

您可以使用sql union和join来获得预期结果。而不是创建两个单独的功能，你可以创建一个功能，并使用原始查询象下面这样：

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，我创建了一个名为expenses和incomes与在表中给出的字段两个表指定above.And我用的是上面提到的查询和结果如下。

我希望这是你所期望的。

来源

2016-07-27 15:04:48

感谢阿肖克纪... – user2851129

选择expenses.date，随着收入从支出 expenses.cost， '0'，其中在（选择收入incomes.date）expenses.date不 UNION ALL 选择不同的（费用.date），expenses.cost，从费用incomes.cost 内从收入其中收入加入收入上expenses.date = incomes.date UNION ALL 选择incomes.date， '0' 作为成本，incomes.cost .date不在（select expenses.date from expenses） esma distinct date haruko sum kasari nikalneho – user2851129

select expenses.date，expenses.cost，'0'作为收入来自费用其中expenses.date不在（select incomesdate from incomes） union all select distinct（expenses.date），sum（expenses.cost，），总和（收入。从费用内成本）加入收入上expenses.date = incomes.date UNION ALL 选择incomes.date， '0' 作为成本，incomes.cost 从收入其中incomes.date未在（选择expenses.date来自费用） esari sum garna khojda问题aauncha – user2851129

合并表使用日期字段，如果共同，并添加新行，如果日期字段不常见

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