with t1 (no, col) as
(
-- start of test data
select 1, 'user:12345;group:56789;group:6785;' from dual union all
select 2, 'user:12345;group:56789;group:6785;' from dual
-- end of test data
)
-- the lookup table which has the substitute strings
-- nid : concatenation of name and id as in table t1 which requires the lookup
-- tname : required substitute for each nid
, t2 (id, name, type, nid, tname) as
(
select t.*, type || ':' || id, type || ':' || name from
(
select 12345 id, 'admin' name, 'user' type from dual union all
select 56789, 'testgroup', 'group' from dual
) t
)
--select * from t2;
-- cte table calculates the indexes for the substrings (eg, user:12345)
-- no : sequence no in t1
-- col : the input string in t1
-- si : starting index of each substring in the 'col' input string that needs attention later
-- ei : ending index of each substring in the 'col' input string
-- idx : the order of substring to put them together later
,cte (no, col, si, ei, idx) as
(
select no, col, 1, case when instr(col,';') = 0 then length(col)+1 else instr(col,';') end, 1 from t1 union all
select no, col, ei+1, case when instr(col,';', ei+1) = 0 then length(col)+1 else instr(col,';', ei+1) end, idx+1 from cte where ei + 1 <= length(col)
)
,coll(no, col, sstr, idx, newstr) as
(
select
a.no, a.col, a.sstr, a.idx,
-- when a substitute is not found in t2, use the same input substring (eg. group:6785)
case when t2.tname is null then a.sstr else t2.tname end
from
(select cte.*, substr(col, si, ei-si) as sstr from cte) a
-- we don't want to miss if there is no substitute available in t2 for a substring
left outer join
t2
on (a.sstr = t2.nid)
)
select no, col, listagg(newstr, ';') within group (order by no, col, idx) from coll
group by no, col;
爲什麼PLSQL?不是普通的SQL查詢很好嗎?另外,你到目前爲止嘗試過什麼? – Aleksej
我的數據庫是oracle,我可以對它們進行操作 – user3728800
PLSQL是SQL的Oracle過程擴展。如果您需要爲Oracle工作的SQL查詢,您最好標記Oracle和SQL,而不是PLSQL。另外,請發佈您迄今爲止嘗試/搜索的內容 – Aleksej