GROUP by inside INNER join 使用 php 在 html 下拉列表中显示

Question

我有以下表格：

Workers (id, total_pay, date_of_pay, projects_id)
Payments (id, payment, date, projects_id)
building_materials(id, pay_of_materials, date, projects_id)
additional(id, add_pay, date, projects_id)

其中 projects_id 是所有项目的外键。 我尝试进行查询以从 projects_id = projects_id 的所有表中选择所有日期，我现在有这个语句：

SELECT workers.date_of_pay, payments.date, building_materials.date, additional.date FROM
workers 
INNER JOIN 
payments ON workers.projects_id = payments.projects_id
INNER JOIN 
building_materials ON payments.projects_id = building_materials.projects_id
INNER JOIN 
additional ON building_materials.projects_id = additional.projects_id;

我到底想要什么，但无法弄清楚，是如何使用 GROUP BY 对每个字段进行分组，以及如何在一个下拉列表框中将所有列显示为一个？ 我有这个列表框代码：

<select name="Date" required class="form-control" id="Date">
                    <option value="">اختر التاريخ</option>
                    <?php $sql="    SELECT workers.date_of_pay, payments.date, building_materials.date, additional.date FROM
    workers 
    INNER JOIN 
    payments ON workers.projects_id = payments.projects_id
    INNER JOIN 
    building_materials ON payments.projects_id = building_materials.projects_id
    INNER JOIN 
    additional 

ON 
building_materials.projects_id = additional.projects_id AND additional.projects_id = ".$id;
                $result = mysqli_query($con, $sql) or die($sql."<br/><br/>".mysql_error());

                    while($rows=mysqli_fetch_array($result)){?>
                        <option value="<?php echo $rows['date'] ?>"><?php echo $rows['date'] ?></option>
                    <?php } ?>
                    </select>

所以，稍后，我可以从下拉列表中选择一个日期并获取信息。 那么我如何分组，并将结果回显到下拉列表中。

编辑 使用 concat() 后：

SELECT concat(workers.date_of_pay, payments.date, building_materials.date, additional.date) AS date FROM
    workers 
    INNER JOIN 
    payments ON workers.projects_id = payments.projects_id
    INNER JOIN 
    building_materials ON payments.projects_id = building_materials.projects_id
    INNER JOIN 
    additional ON building_materials.projects_id = additional.projects_id

我有这个：

但我希望每一行只有一个日期。

Answer 1

使用 UNION ALL 和 GROUP BY，如下所示。使用可用的代码，我可以编写这样的解决方案。您可以减少连接取决于表架构。并尝试使用 mysqli 而不是 mysql。

$sql = "SELECT * FROM (
       (SELECT workers.date_of_pay AS real_date FROM
            workers 
            INNER JOIN 
            payments ON workers.projects_id = payments.projects_id
            INNER JOIN 
            building_materials ON payments.projects_id = building_materials.projects_id
            INNER JOIN 
            additional 
            ON building_materials.projects_id = additional.projects_id AND additional.projects_id = $id)"
        . " UNION ALL"
        ."(SELECT payments.date AS real_date FROM
            workers 
            INNER JOIN 
            payments ON workers.projects_id = payments.projects_id
            INNER JOIN 
            building_materials ON payments.projects_id = building_materials.projects_id
            INNER JOIN 
            additional 
            ON building_materials.projects_id = additional.projects_id AND additional.projects_id = $id)"
        . " UNION ALL"
        . "(SELECT building_materials.date AS real_date FROM
            workers 
            INNER JOIN 
            payments ON workers.projects_id = payments.projects_id
            INNER JOIN 
            building_materials ON payments.projects_id = building_materials.projects_id
            INNER JOIN 
            additional 
            ON building_materials.projects_id = additional.projects_id AND additional.projects_id = $id)"
        . " UNION ALL"
        . "(SELECT additional.date AS real_date FROM
            workers 
            INNER JOIN 
            payments ON workers.projects_id = payments.projects_id
            INNER JOIN 
            building_materials ON payments.projects_id = building_materials.projects_id
            INNER JOIN 
            additional 
            ON building_materials.projects_id = additional.projects_id AND additional.projects_id = $id)"
        . ") AS tab GROUP BY real_date";