Php 我对如何从另一个表中获取列fr有问题

Php 我对如何从另一个表中获取列fr有问题,php,mysql,sql,Php,Mysql,Sql,这是我的表格和每个表格的数据 餐桌预订: 表格组合: 我想从组合表到预订表中获取组合名称 应该是这样的 +-------------+-----------+---------+------------+ | rid | r_date | r_time |combo_name | +-------------+-----------+---------+-----+------+ | 1 | 2019-12-20| 10:00:00| Package

这是我的表格和每个表格的数据

餐桌预订:

表格组合:

我想从组合表到预订表中获取组合名称 应该是这样的

+-------------+-----------+---------+------------+
|  rid        | r_date    | r_time  |combo_name  |
+-------------+-----------+---------+-----+------+
|   1         | 2019-12-20| 10:00:00| Package 1  |
|   2         | 2019-12-20| 10:00:00| Package 2  |
|   3         | 2019-12-20| 10:00:00| Package 3  |
+-------------+-----------+---------+-----+------+
这是我的代码,请帮忙

$query = mysqli_query($con, "select * 
                             from reservation 
                             where  r_status='Approved' 
                               and r_date>='$today' 
                             order by r_date"
         ) or die(mysqli_error($con))`;
您可以使用联接查询:


与联接一起使用,您将获得:

$query = mysqli_query($con, "select r.rid ,r.r_date,r. r_time ,c.combo_name 
                         from reservation r
                         join combo c on c.combo_id = r.rid
                         where  r_status='Approved' 
                           and r_date>='$today' 
                         order by r_date"
     ) or die(mysqli_error($con))`;

请尝试下面的查询,让我知道它不适合你

$query = mysqli_query($con, "SELECT r.rid
    ,r.r_date
    ,r_time
    ,c.Combo_Name
FROM reservation AS r
LEFT JOIN combo AS c ON r.rid = c.combo_id
WHERE r.r_status = 'Approved'
    AND r.r_date >= CONVERT(VARCHAR, GETDATE(), 23)
ORDER BY r_date"
         ) or die(mysqli_error($con))`;

不管怎样,我的问题用这段代码解决了,多亏了这些建议

Select reservation.rid, reservation.r_time, reservation.r_last, combo.combo_name, reservation.r_first, reservation.r_date from reservation INNER JOIN combo on reservation.combo_id=combo.combo_id where r_status='Approved'

这两个表之间没有明显的关系?使用diemysqli_error$conn是一个非常糟糕的主意;因为它可能会泄露敏感信息。更多解释请参见本文:使用diemysqli_error$conn是一个非常糟糕的主意;因为它可能会泄露敏感信息。更多解释请参见本文:使用diemysqli_error$conn是一个非常糟糕的主意;因为它可能会泄露敏感信息。更多解释请参见本文:使用diemysqli_error$conn是一个非常糟糕的主意;因为它可能会泄露敏感信息。有关更多说明,请参阅本文:
$query = mysqli_query($con, "select r.rid ,r.r_date,r. r_time ,c.combo_name 
                         from reservation r
                         join combo c on c.combo_id = r.rid
                         where  r_status='Approved' 
                           and r_date>='$today' 
                         order by r_date"
     ) or die(mysqli_error($con))`;
$query = mysqli_query($con, "SELECT r.rid
    ,r.r_date
    ,r_time
    ,c.Combo_Name
FROM reservation AS r
LEFT JOIN combo AS c ON r.rid = c.combo_id
WHERE r.r_status = 'Approved'
    AND r.r_date >= CONVERT(VARCHAR, GETDATE(), 23)
ORDER BY r_date"
         ) or die(mysqli_error($con))`;
Select reservation.rid, reservation.r_time, reservation.r_last, combo.combo_name, reservation.r_first, reservation.r_date from reservation INNER JOIN combo on reservation.combo_id=combo.combo_id where r_status='Approved'