Sql 我正在向API发送一个参数列表（比如名称），参数列表将检查数据库并返回值

我正在向API发送一个参数列表（比如名称），参数列表将检查数据库并返回值

例：2

Med_疾病表

中到疾病表

我的查询是使用like命令仅拾取第一行，而不是其他行，我希望所有第一、第二和第三行都由A、B和C组成。我使用的是R和SQLoracle。 P.N:之前不知道该列表中可以包含多少令牌。

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请提供帮助。

您好，这里有一个关于您试图实现的目标的查询

with your_data as (
  select 'A,B,C,D' medicine_name, 'D1' disease from dual union
  select 'A,C,B', 'D2' from dual union
  select 'D,C,A,B', 'D3' from dual union
  select 'D,A', 'D4' from dual 
),
split as (select distinct  regexp_substr(medicine_name,'[^,]+', 1, level) as medicine_name, disease from your_data
connect by regexp_substr(medicine_name,'[^,]+', 1, level) is not null),
aggr as (
select disease, listagg(medicine_name,',') within group (order by medicine_name) medicine_name from split
group by disease),
input as (select  distinct regexp_substr('A,B,C','[^,]+', 1, level) as medicine_name from your_data
connect by regexp_substr('A,B,C','[^,]+', 1, level) is not null),
aggr_input as (
select  listagg(medicine_name,',') within group (order by medicine_name) medicine_name from input)
select a.* from aggr a
join aggr_input ai on ai.medicine_name = substr(a.medicine_name,1,length(ai.medicine_name))

只需将您的数据替换为“从医学疾病中选择医学名称、疾病” 以及带有输入参数的硬编码“A、B、C”

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让我知道它是如何运行的：-

@ted感谢您的回复，但正如我之前提到的，我的数据集是动态的，它可以由3个令牌、4个令牌或任意数量的令牌组成，因此在这种情况下该怎么办…@KoushikRoy是的，代码支持任意数量的逗号分隔符tokens@KoushikRoy很高兴能帮上忙。请别忘了把答案标为正确。

with your_data as (
  select 'A,B,C,D' medicine_name, 'D1' disease from dual union
  select 'A,C,B', 'D2' from dual union
  select 'D,C,A,B', 'D3' from dual union
  select 'D,A', 'D4' from dual 
),
split as (select distinct  regexp_substr(medicine_name,'[^,]+', 1, level) as medicine_name, disease from your_data
connect by regexp_substr(medicine_name,'[^,]+', 1, level) is not null),
aggr as (
select disease, listagg(medicine_name,',') within group (order by medicine_name) medicine_name from split
group by disease),
input as (select  distinct regexp_substr('A,B,C','[^,]+', 1, level) as medicine_name from your_data
connect by regexp_substr('A,B,C','[^,]+', 1, level) is not null),
aggr_input as (
select  listagg(medicine_name,',') within group (order by medicine_name) medicine_name from input)
select a.* from aggr a
join aggr_input ai on ai.medicine_name = substr(a.medicine_name,1,length(ai.medicine_name))