兩個串列中的字符是不一樣的, 只知道存在包含關系. 串列1的順序不確定.
如果兩包都是一樣的字串, sdiff 或者資料庫的 not in 都可實作.
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with tab1 as (
select '111' id from dual union all
select '1121' id from dual union all
select '1131' id from dual
),
tab2 as (
select '111' id from dual union all
select '121' id from dual union all
select '11331' id from dual
)
select * from tab1 t1
where not exists(
select 1 from tab2 t2 where instr(t1.id, t2.id) > 0
)
;
with tab1 as (
select '111' id from dual union all
select '1121' id from dual union all
select '1131' id from dual
),
tab2 as (
select '111' id from dual union all
select '121' id from dual union all
select '11331' id from dual
)
select * from tab1 t1
where not exists(
select 1 from tab2 t2 where instr(t1.id, t2.id) > 0
)
;
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