Question

我正在处理一个包含显示carid(foreign key from tbl_vehicle) reg_num(registration number or plate number of a car, from tbl_vehicle)和amount(from tbl_fuel)的项目，我成功显示了carid，reg_num和amount的值使用此SQL语句以HTML格式从表tbl_fuel和tbl_vehicle中提取数据。

$Withdraw = query("SELECT tbl_fuel.carid,tbl_vehicle.reg_num,sum(trim(replace(amount, '$', '')) + 0.0) as amount
                   FROM tbl_fuel
                    LEFT JOIN tbl_vehicle
                    on tbl_fuel.carid=tbl_vehicle.carid
                    GROUP BY carid");

但是我忘记了另一个名为tbl_maintenance的表，它具有与tbl_fuel相同的属性，分别为carid(foreign key from tbl_vehicle)，amount。我需要以单个HTML格式显示tbl_fuel和tbl_maintenance中的此属性的值。

这是我的html表单

<div class="panel-body">
    <h3 align="center">Withdrawal Per Vehicle</h3>
    <table class="table table-striped table-bordered">
        <thead>
            <tr>
                <th>ID</th>
                <th>Plate Number</th>
                <th>Amount</th>
            </tr>
        </thead>
        <tbody>
            <?php foreach($Withdraw as $w): ?>
                <?= '<tr>' ?>
                    <?= '<td>' . $w["carid"] . '</td>' ?>
                    <?= '<td>' . $w["reg_num"] . '</td>' ?>
                    <?= '<td>' . $w["amount"] . '</td>' ?>
                <?= '</tr>' ?>
            <?php endforeach; ?>
        </tbody>
    </table>
</div>

请注意，上述代码可以正常运行，我的问题是，应在我当前的SQL语句中添加哪一行SQL行，以包含tbl_maintenance中的属性值并将其显示在我的HTML表单中，并将tbl_maintenance和tbl_fuel的龋齿分组并求和来自tbl_maintenance和tbl_fuel的金额？

Answer 1

切换一下，将tbl_vehicle设为主表，并对总和进行子查询：

SELECT v.carid,
       v.reg_num,
       IFNULL(f.sum_amount,0) + IFNULL(m.sum_amount,0) AS amount
FROM tbl_vehicle v
LEFT JOIN
  (SELECT carid,
          sum(trim(replace(amount, '$', ''))+0) sum_amount
   FROM tbl_fuel
   GROUP BY cardid) f ON f.carid = v.carid
LEFT JOIN
  (SELECT carid,
          sum(trim(replace(amount, '$', ''))+0) sum_amount
   FROM tbl_maintenance
   GROUP BY cardid) m ON m.carid = v.carid
WHERE f.sum_amount IS NOT NULL OR m.sum_amount IS NOT NULL

Answer 2

SELECT tbl_fuel.carid,tbl_vehicle.reg_num,sum(trim(replace(tbl_fuel.amount, '$', '')) + 0.0) as FuelAmount , sum(trim(replace(tbl_maintenance.amount, '$', '')) + 0.0) as MaintenanceAmount
FROM tbl_fuel 
LEFT JOIN tbl_vehicle
on tbl_fuel.carid=tbl_vehicle.carid
LEFT JOIN tbl_maintenance 
on tbl_maintenance.carid=tbl_vehicle.carid
GROUP BY carid

如何将两个具有相同属性的不同表的值合并在一起

2 个答案: