选择 2927 id 列表中出现的所有行

Question

我有一个 2927 id 的列表。我想获取该列表中所有 id 的行。我怎样才能做到这一点？它是一个逗号分隔的 id 列表。由于限制为 1000，因此 in 语句不起作用。我尝试过类似 Loop through pre defined values 的解决方案，但它没有达到我的预期。

我正在使用 toad，我希望查看数据网格中的行（多行、多列）。

提前致谢！

id 列表的外观如下：

67,122,173,256,284,285,288,289,291,294,296,298,301,320,346,359,366,425,428,454,528,573,576,584,593,654,654,694,722,838,1833,1976,1979,1979,2002,2004,2005,2045,2083,2109,2114,2126,2126,2157,2204,2204,2211,2212,2332,2576，...

当未达到 1000 的限制时，语句将如何：

Select * from tablename where tablename.id in (67,122,173,256,284,285,288,289,291,294,296,298,301,320,346,359,366,425,428,454,528,573,576,584,593,654,654,694,722,838,1833,1976,1979,1979,2002,2004,2005,2045,2083,2109,2114,2126,2126,2157,2204,2204,2211,2212,2332,2576);

Answer 1

这是另一种处理方法，即使用公用表表达式 (CTE) 将 ID 转换为逻辑表，然后像往常一样加入。这样想可能更容易理解它：

-- Build the list of IDs.
with data(str) as (
    select '67,122,173,256,284,285,288,289,291,294,296,298,301,320,346,359
     ,366,425,428,454,528,573,576,584,593,654,654,694,722,838,1833,1976,1979,1979,2002
     ,2004,2005,2045,2083,2109,2114,2126,2126,2157,2204,2204,2211,2212,2332,2576' 
    from dual
),
-- Turn the list into a table using the comma as the delimiter. Think of it
-- like a temp table in memory.
id_list(id) as (
  select regexp_substr(str, '(.*?)(,|$)', 1, level, NULL, 1)
  from data
  connect by level <= regexp_count(str, ',') + 1
)
-- Select data from the main table, joining to the id_list "temp" table where
-- the ID matches.
select tablename.*
from tablename, id_list
where tablename.id = id_list.id;

Answer 2

• 缓慢、丑陋但有效的解决方案：将字符串拆分为 id 并使用游标获取行 或
• 根据字符串生成一个动态选择查询：'select * from tbl where id = 1 or id = 2 or ...' 并执行它，不会再快但会工作。

Answer 3

不用于生产：

with list as (
    select to_number(regexp_substr('67,122,173,256,284,285,288,289,291,294,296,298,301,320,346,359
     ,366,425,428,454,528,573,576,584,593,654,654,694,722,838,1833,1976,1979,1979,2002
     ,2004,2005,2045,2083,2109,2114,2126,2126,2157,2204,2204,2211,2212,2332,2576'
     , '\d+', 1, level)) id from dual connect by level < 5000
)
select * from tablename where tablename.id in (select id from list where id is not null);

Answer 4

你可以这样使用：

... 
where ',' || id_list || ',' like '%,' || to_char(id) || ',%'

您确实需要在 to_char(id) 周围添加逗号（否则 3854 将匹配 38），并且您需要在 id_list 周围添加逗号以便第一个和最后一个值匹配（仅需要 id_list 周围的逗号因为你需要 to_char(id) 周围的逗号）。

这假定id 的数据类型为NUMBER。如果它已经是 VARCHAR2 或其他字符数据类型，则只需要 id 而不是 to_char(id)。 （严格来说，即使id 是 NUMBER，也不需要to_char()，但作为一种最佳实践，不应依赖隐式转换。）

例子：

with
     mytable(id, val) as (
       select 100, 'abc' from dual union all
       select 173, 'z'   from dual union all
       select 250, 'dvd' from dual union all
       select  30, 'vv'  from dual union all
       select 359, 'ghi' from dual
     ),
     test_data (id_list) as (
       select '67,122,173,256,284,285,288,289,291,294,296,298,301,320,346,359'
         from dual
     )
select mytable.* from mytable, test_data
where ',' || id_list || ',' like '%,' || to_char(id) || ',%'
;

        ID VAL
---------- ---
       173 z
       359 ghi