SQL，Oracle11g的：找到匹配的特定组ID的

Question

的请参考SQL Oracle : how to find records maching a particular id in the column

现在我的查询是在类似的路线记录： 我的表中有如下数据：

现在我想找到记录将asso_entity_id作为3个值的组合。 例如： - 如果我从jsp选择asso_entities为30000,30001和80002（以任何顺序），我应该得到上表的第一条记录。

Answer 1

请检查查询...位长...会尽量缩短它。

with test as (
    select * from YOURTABLE 
) 
SELECT distinct DATASETNAME FROM(
    select x.*, COUNT(*) OVER (partition by DATASETNAME ORDER BY DATASETNAME) CNT From(
    select DATASETNAME, regexp_substr (ASSO_ENTITY_ID, '[^|]+', 1, row_number() OVER (partition by DATASETNAME ORDER BY DATASETNAME)) split 
    from test 
    connect by level <= length (regexp_replace (ASSO_ENTITY_ID, '[^|]+')) + 1 
)x where SPLIT IS NOT NULL 
)xx 
    WHERE SPLIT IN ('300000', '300001', '800002') AND 
    CNT =3; 

Answer 2

从我对你刚才的问题的答案继，并继续使用，作为一个例子监守我可以看到做这listagg基础查询......你可以匹配多个值，看看有多少你匹配，并应用进一步的过滤器，以确保您匹配所有的人。喜欢的东西：

select distinct role_id, role_name, active, companyName, permission_id, 
    permission_name, rn, total_rows, roleCreated 
from (
    select t.*, 
     count(raw_permission_id) over (partition by role_id) as cnt 
    from (
     select r.role_id, 
      r.role_name, 
      r.active, 
      decode(r.entity_type_id, 1000, m.name, 3000, cour.name, 
       4000, 'Ensenda') companyName, 
      p.permission_id as raw_permission_id, 
      listagg(p.permission_id, ' | ') 
       within group (order by p.permission_id) 
        over (partition by r.role_id) permission_id, 
      listagg(p.permission_name, ' | ') 
       within group (order by p.permission_id) 
        over (partition by r.role_id) permission_name, 
      dense_rank() over (order by r.created_ts desc) as rn, 
      count(distinct r.role_id) over() as total_rows, 
      r.created_ts roleCreated 
     from t_role r 
     left join t_role_permission rp ON r.role_id = rp.role_id 
     left join t_permission p ON rp.permission_id = p.permission_id 
     left join merchant m on r.entity_id = m.merchantkey 
     left join courier cour on r.entity_id = cour.courierkey 
    ) t 
    where raw_permission_id in (301446, 301445) 
) 
where cnt = 2 
and rn between 1 and 100 
order by roleCreated desc; 

从以前的答案，唯一的区别是查询外两层：

select ... 
from (
    select t.*, 
     count(raw_permission_id) over (partition by role_id) as cnt 
    from (
    ... -- no changes here 
    ) t 
    where raw_permission_id in (301446, 301445) 
) 
where cnt = 2 
... 

所以它现在找两个可能的权限和计数 - 分析，因此每个角色 - 有多少人被匹配。外部查询检查匹配的数字是2，这显然必须根据您尝试匹配的选项数量来调整。

对于这个问题，它会添加类似：

 count(raw_asso_entity_id) over (partition by <some_id>) as cnt 
    ... 
    where raw_asso_id in (30000, 30001 and 80002) 
... 
where cnt = 3 

Answer 3

techdo的例子有点简化版本：

这是测试表结构：

ID DATASET_NAME DATASET_VAL 
------------------------------------------ 
1 DATASET1  3000 | 30001 | 80002 
2 DATASET1  3000 | 80002 
3 DATASET1  3000 | 80002 


SELECT LISTAGG(str, ' | ') WITHIN GROUP (ORDER BY str) asso_ety_id 
    FROM 
    (
    SELECT DISTINCT id, dataset_name 
     , TRIM(REGEXP_SUBSTR (dataset_val, '[^|]+', 1, LEVEL)) str 
    FROM your_tab 
    CONNECT BY LEVEL <= LENGTH(REGEXP_REPLACE(dataset_val, '[^|]+')) + 1 
) 
    WHERE str IN ('80002', '30001', '3000') -- in any order -- 
    -- AND id = 1 -- optional -- 
/

输出：

3000 | 30001 | 80002