Question

Table: entity
╔════╦════════════╦═════════════╦════════════╗
║ id ║ city_id    ║ province_id ║ country_id ║
╠════╬════════════╬═════════════╬════════════╣
║  1 ║ NULL       ║ 654         ║ 12         ║
║  2 ║ 17782      ║ 654         ║ 12         ║
║  3 ║ 17782      ║ 654         ║ 12         ║
║  4 ║ 16762      ║ NULL        ║ 32         ║
╚════╩════════════╩═════════════╩════════════╝

我有三个值来查询entity表：

city_id = 17782 province_id = 654和country_id = 12

现在我想按顺序提供的值对结果进行排序 city_id＆gt; province_id＆gt; COUNTRY_ID 即可。

如果city_id不为null且city_id = 17782，那么它应该在结果集中排在第一位
如果city_id为null，则它将按省_id = 654排序假设province_id不为null
如果province_id为null，则它将回退到country_id = 12

我有这个查询无法按预期工作

select * from entity 
order by city_id is null, city_id <> 17782, 
province_id is null, province_id <> 654, country_id <> 12

提供以下结果集

╔════╦════════════╦═════════════╦════════════╗
║ id ║ city_id    ║ province_id ║ country_id ║
╠════╬════════════╬═════════════╬════════════╣
║  2 ║ 17782      ║ 654         ║ 12         ║
║  3 ║ 17782      ║ 654         ║ 12         ║
║  4 ║ 16762      ║ NULL        ║ 34         ║
║  1 ║ NULL       ║ 654         ║ 12         ║
╚════╩════════════╩═════════════╩════════════╝

注意带有null值的province_id是第一个，而值654应该是第一个，这是因为第一个条件city_id为null

期望的结果

╔════╦════════════╦═════════════╦════════════╗
║ id ║ city_id    ║ province_id ║ country_id ║
╠════╬════════════╬═════════════╬════════════╣
║  2 ║ 17782      ║ 654         ║ 12         ║
║  3 ║ 17782      ║ 654         ║ 12         ║
║  1 ║ NULL       ║ 654         ║ 12         ║
║  4 ║ 16762      ║ NULL        ║ 34         ║
╚════╩════════════╩═════════════╩════════════╝

Answer 1

试试这个：

ORDER BY

<=>的第一部分将所有所需的值放在结果的前面，然后其余部分按优先级城市，省份，国家/地区排序。

我使用null-safe equal operator city_id = NULL，因此province_id不会强制排除void insertBefore(list *l, void *p){ node* newNode = malloc(sizeof(node) + l->SizeOfData); memcpy(newNode->item, p, l->SizeofData); l->curent->previous->next = newNode; l->current->previous = newNode; } void insertAfter(list *l, void *p){ node* newNode = malloc(sizeof(node) + l->SizeOfData); memcpy(newNode->item, p, l->SizeofData); l->curent->next->previous = newNode; l->current->previous->next = newNode; l->current = newNode; }排序。

Answer 2

试试以下内容并告诉我们。

select * from entity 
order by 
   case when city_id = 17782 and province_id is not null then 1 
        when city_id is null and province_id = 654 then 2 
        when province_id is null and country_id = 12 then 3 
   else 4 end

按城市，省和国家的Mysql顺序

2 个答案: