PHP - 删除具有相同引用的重复值

Question

我有以下名为 store

的 MySQL 表

id ref item_no supplier
1  10    x1      usa
2  10    x1      usa
3  11    x1      china
4  12    x2      uk
5  12    x3      uk
6  13    x3      uk
7  13    x3      uk

现在我要输出的内容如下：

id ref item_no supplier
1  10    x1      usa
3  11    x1      china
4  12    x2      uk
5  12    x3      uk
6  13    x3      uk

如您所见item_no x1 和 x3 具有相同的参考和供应商来源，所以我想要删除重复记录以保留一个仅限item_no！

我已将此 PHP 代码创建为仅SELECT 结果：

$query1 = "SELECT 
                DISTINCT(item_no) AS field, 
                COUNT(item_no) AS fieldCount, 
                COUNT(ref) AS refcount 
            FROM 
                store 
            GROUP BY item_no HAVING fieldCount > 1";

$result1 = mysql_query($query1);

if(mysql_num_rows($result1)>0){
    while ($row1=mysql_fetch_assoc($result1)) {
        echo $row1['field']."<br /><br />";
    }
} else {
    //IGNORE
}

如何将查询告诉SELECT 在创建DELETE 查询之前根据我的需要正确复制记录。

谢谢大家

Answer 1

您可以使用以下查询来生成所需的结果集：

SELECT t1.*
FROM store AS t1
JOIN (
   SELECT MIN(id) AS id, ref, item_no
   FROM store
   GROUP BY ref, item_no
) AS t2 ON t1.id > t2.id AND t1.ref = t2.ref AND t1.item_no = t2.item_no 

Demo here

您可以使用DELETE：

DELETE t1
FROM store AS t1
JOIN (
   SELECT MIN(id) AS id, ref, item_no
   FROM store
   GROUP BY ref, item_no
) AS t2 ON t1.id > t2.id AND t1.ref = t2.ref AND t1.item_no = t2.item_no 

Answer 2

如果要删除行的选择，请使用

select *  from  store 
where id not in (
    select max(id)  from  store 
    group by  distinct ref, item_no, supplier);

或者你可以直接使用命令直接删除使用

delete from  store 
where id not in (
    select max(id)  from  store 
    group by  distinct ref, item_no, supplier);

Answer 3

要仅查找重复记录，您可以使用

 SELECT * FROM store WHERE id NOT IN 
 (SELECT id FROM store AS outerStore WHERE id = 
 (SELECT MAX(id) FROM store AS innerStore 
 WHERE outerStore.ref = innerStore.ref AND 
 outerStore.supplier = innerStore.supplier AND outerStore.item_no = innerStore.item_no))

可能很长，但应该可以。