Question

我有几个要查询的项ID，即使它们是几个。我解释一下：

我有一张桌子id | item 我有一些要查询的ID（例如25,62,89,123,12,25,25,62）我需要显示所有这些，即使有多次相同的id，也不需要进行多次查询。如果我使用“where in”查询，它会对ID进行分组，因此只显示一次ID 25，例如......

我怎么能设法做到这一点？

我的查询：

$result = mysqli_query($link,"select p.id_product, p.price, c.link_rewrite as catrew, i.id_image, l.link_rewrite, l.name, s.reduction, s.reduction_type
    from rec_product p
    left join rec_image i on i.id_product=p.id_product and i.cover=1
    left join rec_product_lang l on l.id_product=p.id_product and l.id_lang = 1
    left join rec_category_lang c on c.id_category=p.id_category_default
    left join rec_specific_price s on s.id_product=p.id_product
    WHERE p.id_product IN ($id_prod_array)
    order by rand()");

示例：

$id_prod_array = "13,13,119";

结果：

enter image description here

预期：
第一个（id 13）的2倍和id 119的一次

[编辑]

找到解决方案：id为13,13,119的示例

select p.id_product, p.price, c.link_rewrite as catrew, i.id_image, l.link_rewrite, l.name, s.reduction, s.reduction_type
        from rec_product p
        join (select 13 as id_product
              union all select 13
              union all select 119) as T1 on p.id_product = T1.id_product
        left join rec_image i on i.id_product=p.id_product and i.cover=1
        left join rec_product_lang l on l.id_product=p.id_product and l.id_lang = 1
        left join rec_category_lang c on c.id_category=p.id_category_default and c.id_lang = 1
        left join rec_specific_price s on s.id_product=p.id_product

Answer 1

您需要FIND_IN_SET功能才能搜索$id_prod_array的值。

更改：

WHERE p.id_product IN ($id_prod_array)

致：

WHERE FIND_IN_SET( p.id_product, REPLACE( $id_prod_array, ' ', '' ) ) > 0

它应该有效。

您的查询无效的原因分析如下：

如果

 `$id_prod_array` = '25, 62, 89, 123, 12, 25, 25, 62'

然后假设当与sql IN一起使用时，它等同于：

WHERE p.id_product IN ( 25, 62, 89, 123, 12, 25, 25, 62 )

但是，它相当于：

WHERE p.id_product IN ( '25, 62, 89, 123, 12, 25, 25, 62' )

含义：IN

对两种解释都不相同

'25, 62, 89, 123, 12, 25, 25, 62' != 25, 62, 89, 123, 12, 25, 25, 62

示例：

mysql> select 89 in ( 25, 62, 89, 123, 12, 25, 25, 62 );
+-------------------------------------------+
| 89 in ( 25, 62, 89, 123, 12, 25, 25, 62 ) |
+-------------------------------------------+
|                                         1 |
+-------------------------------------------+
1 row in set (0.08 sec)

mysql> select 89 in ( '25, 62, 89, 123, 12, 25, 25, 62' );
+---------------------------------------------+
| 89 in ( '25, 62, 89, 123, 12, 25, 25, 62' ) |
+---------------------------------------------+
|                                           0 |
+---------------------------------------------+
1 row in set, 1 warning (0.04 sec)

mysql> show warnings;
+---------+------+------------------------------------------+
| Level   | Code | Message                                  |
+---------+------+------------------------------------------|
| Warning | 1292 | Truncated incorrect DOUBLE               |
|         |      | value: '25, 62, 89, 123, 12, 25, 25, 62' |
+---------+------+------------------------------------------+

请参阅 ：

MySQL: FIND_IN_SET(str,strlist)

Mysql：“在哪里”但也需要重复

1 个答案: