Question

我有两张桌子可以说booking和supplier

从两个表中获取记录的日期范围如下

$query = "SELECT booking.book_id AS Id,
        booking.referance_no AS RefNo,
        booking.entry_date AS DepDate,
        booking.name AS Name,
        booking.mobile AS mobile,
        booking.comp_title AS Company 
    FROM booking WHERE active='1' $WHERE1
    GROUP BY booking.book_id
    UNION ALL
        SELECT supplier.id AS Id,
        supplier.reference_no AS RefNo,
        supplier.departure_date_time AS DepDate,
        supplier.name AS Name,
        supplier.mobile AS Mobile,
        supplier.company AS Company
        FROM supplier WHERE active='1' $WHERE2  
    ORDER BY `DepDate` DESC LIMIT 1000";

注意：我删除了许多与此问题无关的代码行，因此$WHERE1和$WHERE2只是日期范围子句。

查询后将数据（while循环）提取到HTML表格

<td><?php echo $row['RefNo'];?></td>
<td><?php echo $row['Name'];?></td>
<td><?php echo $row['Mobile'];?></td>
<td><?php echo $row['DepDate'];?></td>
<td><?php echo $row['Company'];?></td>
<td><a class="btn" href="delete.php?RefNo=<?php echo $row['RefNo'];?>">Delete</a></td>

在HTML视图中，我知道<?php echo $row['RefNo'];?>中的href属于哪个表booking或supplier，但PHP并不知道{{1}我已经调用了两个表，首先必须针对每个表检查delete.php，然后是否RefNo删除记录

delete.php

true

问题是，是否有更好的方法来删除记录，我不必调用这两个表，并首先针对每个表检查$ReferenceNo = $_GET['RefNo']; //Fetch records from both tables //Check records against `$ReferenceNo` //If true against `booking` table "Delete From booking where referance_no=$ReferenceNo" //else "Delete From supplier where reference_no=$ReferenceNo"。

编辑以使问题更清晰：

正如我上面提到的，PHP并不知道RefNo属于哪个表<?php echo $row['RefNo'];?>或booking所以我需要在任何操作之前的地方工作（删除，取消，编辑）我可以告诉PHP supplier属于此表<?php echo $row['RefNo'];?>或booking，因此在执行任何操作supplier <之前，无需针对这两个表检查<?php echo $row['RefNo'];?> / p>

Answer 1

html表中的每一行都有：

<input type="hidden" name="myIncrNNN" value="tableX">

这是不可见的，并且被提起以告知您在处理时该做什么。

编辑：使其更清晰

你的

<td><a class="btn" href="delete.php?RefNo=<?php echo $row['RefNo'];?>">Delete</a></td>

将被修改以获取隐藏的列线索。

你现在最终会致电：

http://example.com/somedir/delete.php?RefNo=7

在我想象的世界中它会成为

http://example.com/somedir/delete.php?RefNo=7&tc=1

其中tc表示来自隐藏输入字段的表格线索

不需要架构更改和添加表，很明显客户端正在告诉服务器该做什么，没有比原来面对的更多或更少，并且没有说客户端是一个任何事情的权威，如@Halcyon所暗示的那样。

编辑2 :(显示UNION chg）

$query = "SELECT booking.book_id AS Id,
    booking.referance_no AS RefNo,
    booking.entry_date AS DepDate,
    booking.name AS Name,
    booking.mobile AS mobile,
    booking.comp_title AS Company,
    'booking' as TableClue  -- <------ RIGHT THERE
FROM booking WHERE active='1' $WHERE1
GROUP BY booking.book_id
UNION ALL
    SELECT supplier.id AS Id,
    supplier.reference_no AS RefNo,
    supplier.departure_date_time AS DepDate,
    supplier.name AS Name,
    supplier.mobile AS Mobile,
    supplier.company AS Company,
    'supplier' as TableClue   -- <------ RIGHT THERE
    FROM supplier WHERE active='1' $WHERE2  
ORDER BY `DepDate` DESC LIMIT 1000";

Answer 2

您正在处理非规范化数据。实质上，您有两个包含相同数据的表（更多或更少）。要标准化，您必须添加一个表格，将reference_no映射到booking和supplier中的条目。您可以查询该表以查看要删除的记录，或者设置完整的英里foreign-key cascade。

这样的事情：

CREATE TABLE `booking` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `reference_no` int(11) NOT NULL,
  `etc` text NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `supplier` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `reference_no` int(11) NOT NULL,
  `etc` text NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `reference_no` (
  `reference_no` int(11) NOT NULL,
  `booking` int(11) DEFAULT NULL,
  `supplier` int(11) DEFAULT NULL,
  KEY `booking` (`booking`),
  KEY `supplier` (`supplier`),
  CONSTRAINT `reference_no_ibfk_4` FOREIGN KEY (`booking`) REFERENCES `booking` (`id`) ON DELETE CASCADE,
  CONSTRAINT `reference_no_ibfk_5` FOREIGN KEY (`supplier`) REFERENCES `supplier` (`id`) ON DELETE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

但在实践中，您可能不必这样做。两个表上的删除对我来说似乎没问题。

Answer 3

首先必须针对每个表检查RefNo，然后如果是，则删除记录

不，不。只需发出DELETE语句：

DELETE FROM booking WHERE referance_no=$ReferenceNo
DELETE FROM supplier WHERE reference_no=$ReferenceNo

如果执行DELETE语句时不存在匹配的记录，则不会删除任何记录。默认行为是您想要的...删除匹配记录（如果存在）。

处理两个不同的表时如何从表中删除记录

3 个答案:

编辑：使其更清晰

编辑2 :(显示UNION chg）