Php mysql结合了两个select查询

我知道这可能会让人困惑，但请在这一点上容忍我

我有两个SELECT查询，它们之间的差别很小，返回的结果集几乎相同

SELECT products_id,options_values_id
    FROM products_attributes pa
    LEFT JOIN products_options po ON ( pa.options_id = po.products_options_id )
    WHERE products_id ='574' and pa.options_id!=6
    and pa.options_id!=3
    AND products_options_type = 6
    GROUP BY products_id,options_values_id
    ORDER BY products_id,products_options_sort_order,options_id

第二个查询在

products\u options\u type

SELECT products_id,options_values_id
    FROM products_attributes pa
    LEFT JOIN products_options po ON ( pa.options_id = po.products_options_id )
    WHERE products_id ='574' and pa.options_id!=6
    and pa.options_id!=3
    AND products_options_type = 2
    GROUP BY products_id,options_values_id
    ORDER BY products_id,products_options_sort_order,options_id

它们返回的结果是

574|193
574|204

及

我希望输出为

574|193|25
574|204|3

我尝试的是：

SELECT pa.products_id,pa.options_values_id,ord.options_values_id
        FROM products_attributes pa
        LEFT JOIN products_options po ON ( pa.options_id = po.products_options_id )


        LEFT JOIN(SELECT products_id,options_values_id
        FROM products_attributes pa
        LEFT JOIN products_options po ON ( pa.options_id = po.products_options_id )
        WHERE products_id ='574' and pa.options_id!=6
        and pa.options_id!=3
        AND products_options_type = 2
        GROUP BY products_id,options_values_id
        ORDER BY products_id,products_options_sort_order,options_id)ord ON pa.products_id=ord.products_id


        WHERE paproducts_id ='574' and pa.options_id!=6
        and pa.options_id!=3
        AND products_options_type = 2
        GROUP BY pa.products_id,pa.options_values_id
        ORDER BY pa.products_id,products_options_sort_order,options_id

然而，这又回来了

574|193|25
574|204|25

---

我对连接不太在行，所以你知道这是否以及如何做到吗？

试试下面的查询，这里我对

product\u options

进行了两次连接，1次为

po.products\u options\u type=6

，2次为

po1.products\u options\u type=2

SELECT pa.products_id,po.options_values_id,po1.options_values_id
FROM products_attributes pa
LEFT JOIN products_options po ON ( pa.options_id = po.products_options_id and po.products_options_type = 6)
LEFT JOIN products_options po1 ON ( pa.options_id = po1.products_options_id and po.products_options_type = 2)
WHERE pa.products_id ='574' and pa.options_id!=6
and pa.options_id!=3
GROUP BY products_id,po1.options_values_id,po1.options_values_id
ORDER BY products_id,products_options_sort_order,options_id

替代解决方案：

select v1,v2,v3, v4 from (SELECT products_id v1,options_values_id v2
    FROM products_attributes pa
    LEFT JOIN products_options po ON ( pa.options_id = po.products_options_id )
    WHERE products_id ='574' and pa.options_id!=6
    and pa.options_id!=3
    AND products_options_type = 6
    GROUP BY products_id,options_values_id
    ORDER BY products_id,products_options_sort_order,options_id) t1,    
(SELECT products_id v3,options_values_i v4
    FROM products_attributes pa
    LEFT JOIN products_options po ON ( pa.options_id = po.products_options_id )
    WHERE products_id ='574' and pa.options_id!=6
    and pa.options_id!=3
    AND products_options_type = 2
    GROUP BY products_id,options_values_id
    ORDER BY products_id,products_options_sort_order,options_id) t2
where t1.products_id=t2.products_id

您可以轻松地对其进行排序。

试试这个

SELECT a.products_id,a.options_values_id, b.options_values_id
FROM (SELECT @rownum:=@rownum+1 'rw', products_id, options_values_id
FROM (SELECT @rownum:=0) r, products_attributes pa
LEFT JOIN products_options po ON ( pa.options_id = po.products_options_id )
WHERE products_id ='574' and pa.options_id!=6
and pa.options_id!=3
AND products_options_type = 6
GROUP BY products_id,options_values_id
ORDER BY products_id,products_options_sort_order,options_id) a,
(SELECT @rownum:=@rownum+1 'rw', products_id,options_values_id
FROM (SELECT @rownum:=0) r, products_attributes pa
LEFT JOIN products_options po ON ( pa.options_id = po.products_options_id )
WHERE products_id ='574' and pa.options_id!=6
and pa.options_id!=3
AND products_options_type = 2
GROUP BY products_id,options_values_id
ORDER BY products_id,products_options_sort_order,options_id) b
WHERE a.products_id=b.products_id and a.rw=b.rw;

---

希望这能有所帮助。

为什么您的数据库引擎应该知道193和25与204和3一样属于一个整体？看到你的代码，我就不知道了。实际上里面没有太多的逻辑，但我正在处理一个现有的系统，我需要这部分的快捷方式（只是不要问为什么我需要在这里作弊）。我的观点不是你没有理由这样需要你的数据。只是引擎无法在不知道如何排序的情况下对数据进行排序。

SELECT a.products_id,a.options_values_id, b.options_values_id
FROM (SELECT @rownum:=@rownum+1 'rw', products_id, options_values_id
FROM (SELECT @rownum:=0) r, products_attributes pa
LEFT JOIN products_options po ON ( pa.options_id = po.products_options_id )
WHERE products_id ='574' and pa.options_id!=6
and pa.options_id!=3
AND products_options_type = 6
GROUP BY products_id,options_values_id
ORDER BY products_id,products_options_sort_order,options_id) a,
(SELECT @rownum:=@rownum+1 'rw', products_id,options_values_id
FROM (SELECT @rownum:=0) r, products_attributes pa
LEFT JOIN products_options po ON ( pa.options_id = po.products_options_id )
WHERE products_id ='574' and pa.options_id!=6
and pa.options_id!=3
AND products_options_type = 2
GROUP BY products_id,options_values_id
ORDER BY products_id,products_options_sort_order,options_id) b
WHERE a.products_id=b.products_id and a.rw=b.rw;