Question

我有一个查询，从同一个数据库中的3个SQL表中随机选择6个图像，一旦有人点击该图像，我想根据从中选择图像的表将它们发送到单独的页面。要通过查询执行此操作，我想找出选择该随机图像的表名。

我的SQL查询是

$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   }
    ?>

Answer 1

创建一个新字段，指示SQL中的源表：

$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
";

然后，您的PHP可以将此列显示为：

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

Answer 2

在查询中包含表名：

SELECT 'salehotel' as which, sh.*
FROM   salehotel sh
UNION ALL
SELECT 'saleland' as which, sl.*
FROM   saleland sl
UNION ALL
SELECT 'salehouse' as which, sho.*
FROM   salehouse sho
ORDER BY RAND()
LIMIT 6

然后适当地更改php代码以处理额外的列。

Answer 3

select *, 'table_name'如何作为字段，然后使用该列将它们发送到适当的位置。

如何查找从中随机选择的图像的表名

3 个答案: