Question

在yii2应用中，我的Payment模型包含字段id（主键），u_id（整数，表示从User模型付款的人的ID），sum（整数）和date。例如：

+--------------------------------+
| Payment                        |
+--------------------------------+
| id | u_id | sum  | date        |
+--------------------------------+
| 1  | 1    | 400  | 2015-11-25  |
| 2  | 1    | 200  | 2015-11-25  |
| 3  | 2    | 500  | 2015-11-25  |
| 4  | 2    | 300  | 2015-11-25  |
| 5  | 1    | 100  | 2015-11-20  |
+--------------------------------+

问：我希望按日期对结果进行分组，并每天为每个sum汇总所有行的u_id个字段，并显示total行。怎么做...？也许没有total行？在yii2 或清除mysql 。结果示例：

+-------------------------------------+
| Date       | User id | Money, $     |
+-------------------------------------+
| 2015-11-25 |         | 1400 (total) |
| 2015-11-25 | 1       | 600          |
| 2015-11-25 | 2       | 800          |
| 2015-11-20 |         | 100 (total)  |
| 2015-11-20 | 1       | 100          |
+-------------------------------------+

Payment型号：

public function search($params)
{
    $query = Payment::find();

    // do we need to group and sum here?

    $dataProvider = new ActiveDataProvider([
        'query' => $query,
    ]);

    $this->load($params);

    return $dataProvider;
}

我猜ListView会更加轻松GridView，因为我们可以为每个结果进行一些计算。视图：

<table>
    <thead>
        <tr>
            <th>Date</th>
            <th>User id</th>
            <th>Money, $</th>
        </tr>
    </thead>
    <tbody>
        <?= ListView::widget([
            'dataProvider' => $dataProvider,
            'itemView' => '_item_view',
        ]) ?>
    </tbody>
</table>

_item_view：

<tr>
    <td><?= $model->date ?></td>
    <td><?= $model->u_id ?></td>
    <td><?= $model->sum ?></td>
</tr>

Answer 1

尝试这种方式：

select dt "Date", usr "User Id", 
      case when usr is null 
           then concat(money, ' (total)')
           else money
           end as "Money, $"
  from (
     select dt, null as usr, sum(vsum) as money
       from mytable
      group by dt
     union
     select dt, u_id, sum(vsum) as money
       from mytable
      group by dt, u_id
   ) a
  order by dt desc, coalesce(usr,0)

在此处查看： http://sqlfiddle.com/#!9/48102/6

您可以使用分析函数在普通MySql中获得所需的内容，但由于MySql不支持它，因此您必须模拟它。

在我的解决方案中，我已经提出了一个只按日期汇总资金的查询

select dt, null as usr, sum(vsum) as money
  from mytable
 group by dt

需要null as usr列，因此我可以将UNION运算符与第二部分一起使用。此查询将汇总所有日期和金额。

然后是第二部分

select dt, u_id, sum(vsum) as money
  from mytable
 group by dt, u_id

将由用户汇总所有日期。

第三部分是将它作为子查询，以便我可以按日期，用户订购。请记住，第一部分的用户为null，因此我将其设置为null，因为它将变为0，因此它将首先显示。

Answer 2

您可以这样使用SqlDataProvider

$count = Yii::$app->db->createCommand('
    SELECT COUNT(*) FROM payment group by `date`, u_id', )->queryScalar();

$dataProvider = new SqlDataProvider([
    'sql' => 'SELECT u_id, sum(`sum`)  as `sum` , `date`   FROM payment group by `date`, u_id',

    'totalCount' => $count,
    'sort' => [
        'attributes' => [
            'u_id',
            'sum',
            'date',
        ],
    ],
    'pagination' => [
        'pageSize' => 20,
    ],
]);

 return $this->render('index', [
        'dataProvider' => $dataProvider,
    ]);

Answer 3

一个选项是让mysql使用WITH ROLLUP计算总数。

select dt, u_id, sum(vsum)
from   payment
group by dt, u_id with rollup;

+------------+------+-----------+
| dt         | u_id | sum(vsum) |
+------------+------+-----------+
| 2015-11-20 |    1 |       100 |
| 2015-11-20 | NULL |       100 |
| 2015-11-25 |    1 |       600 |
| 2015-11-25 |    2 |       800 |
| 2015-11-25 | NULL |      1400 |
| NULL       | NULL |      1500 |
+------------+------+-----------+

如果你不想要总数，你也可以消除它。

select * from (
    select dt, u_id, sum(vsum)
    from   payment
    group by dt, u_id with rollup
) q where dt is not null;

+------------+------+-----------+
| dt         | u_id | sum(vsum) |
+------------+------+-----------+
| 2015-11-20 |    1 |       100 |
| 2015-11-20 | NULL |       100 |
| 2015-11-25 |    1 |       600 |
| 2015-11-25 |    2 |       800 |
| 2015-11-25 | NULL |      1400 |
+------------+------+-----------+

如何使sql查询按日期分组和求和？

3 个答案: