如何找到图像随机选择的表名 [英] How to find the table name which the image randomly selected from

问题描述

我有一个查询，可以从同一个数据库中的 3 个 SQL 表中随机选择 6 个图像，一旦有人点击了该图像，我想根据选择该图像的表将它们发送到单独的页面.为此，我想通过查询找出选择该随机图像的表名.

我的 SQL 查询是

$sql="SELECT *来自销售酒店联合所有选择 *来自赛尔兰联合所有选择 *从售楼处按兰德 () 排序限制 6";$result=mysql_query($sql)or die(mysql_error());?><?phpwhile($row = mysql_fetch_array($result)){?><div style="float: left; margin-left: 10px;"><a href="sale_house_detail.php?id=<?php echo $row['property_id']; ?>"><img src=<?= '"admin/uploads/'.$row['image1'].'"';?>width="172px" height="149px" style='border:5px solid #CCC'/></a><p><?php echo $row['Type'];?></p><p><?php echo $row['Location'];?></p>

<?php }?>

解决方案

在 SQL 中创建一个指示源表的新字段:

$sql="SELECT salehotel.*, 'salehotel' 作为源来自销售酒店联合所有SELECT saleland.*, 'saleland' 作为源来自赛尔兰联合所有SELECT salehouse.*, 'salehouse' 作为来源从售楼处按兰德 () 排序限制 6";

然后您的 PHP 可以将此列显示为:

<?php echo $row['source'];?>

I have a query to randomly select 6 images from 3 SQL tables in same database and once someone clicked on that image I want to send them to separate page according from what table that image is selected from. To do that by the query I want to find out the table name which that random image was selected.

My SQL query is

$sql="SELECT *
FROM   salehotel
UNION ALL
SELECT *
FROM   saleland
UNION ALL
SELECT *
FROM   salehouse
ORDER BY RAND()
LIMIT 6
";
$result=mysql_query($sql)or die(mysql_error());
?>
<?php
        while($row = mysql_fetch_array($result))
        {?>
    <div style="float: left; margin-left: 10px;">
        <a href="sale_house_detail.php?id=<?php echo $row['property_id'];  ?>">
            <img src=<?= '"admin/uploads/'.$row['image1'].'"'; ?> width="172px" height="149px" style='border:5px solid  #CCC' />
        </a>
    <p><?php  echo $row['Type']; ?></p>
    <p><?php  echo $row['Location']; ?></p>
  </div>

    <?php   }
    ?>

解决方案

Create a new field that indicates the source table in your SQL:

$sql="SELECT salehotel.*, 'salehotel' as source
FROM   salehotel
UNION ALL
SELECT saleland.*, 'saleland' as source
FROM   saleland
UNION ALL
SELECT salehouse.*, 'salehouse' as source
FROM   salehouse
ORDER BY RAND()
LIMIT 6
";

Your PHP can then show this column as:

<?php  echo $row['source']; ?>

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