Oracle REGEXP_SUBSTR 不会匹配点字符

Question

我正在尝试从以下字符串中提取信息：

FOO-BAR-AUDIT-DATABASE.NUPKG
FOO.BAR.DATABASE-2.0.0.NUPKG

信息如下：

'FOO.BAR.DATABASE' '2.0.0'
        |             |
   module_name     version

目前，当 module_name 部分包含 . 个字符时，我无法正确解析。请参阅下面的 table。

下面的例子展示了我是如何提取信息的。第一组正则表达式无法正常工作'(.*?)，其余组处理不同版本信息的情况。

select case module_name when expected then 'pass' else 'fail' end as test, y.* from(
select lower(regexp_substr(t.pck, g.regex, 1, 1, '', 1)) as module_name, 
       t.expected,
       to_number(regexp_substr(t.pck, g.regex, 1, 1, '', 3)) as major,
       to_number(regexp_substr(t.pck, g.regex, 1, 1, '', 5)) as minor,
       to_number(regexp_substr(t.pck, g.regex, 1, 1, '', 7)) as patch,
       (t.pck) as package_name
from   (select 'FUNKY_LOG_DATABASE-1.0.0.NUPKG' as pck, 'funky_log_database' as expected from dual
        union select 'FOO.BAR.DATABASE-2.0.0.NUPKG', 'foo.bar.database' from dual 
        union select 'FOO-BAR-AUDIT-DATABASE.NUPKG', 'foo-bar-audit-database' from dual
        union select 'funk-database-1.nupkg', 'funk-database' from dual
        union select 'funk-database-1.2.nupkg', 'funk-database' from dual
        union select 'baz-database-1.0.1.nupkg', 'baz-database' from dual) t
cross  join (select '(.*?)(-(\d+)(\.(\d+))?(\.(\d+))?)?(\..*)' as regex from dual) g
)y;

上面的查询产生以下结果 (Oracle 19c)：

test	module_name	expected	major	minor	patch	package_name
pass	foo-bar-audit-database	foo-bar-audit-database				FOO-BAR-AUDIT-DATABASE.NUPKG
fail	foo	foo.bar.database				FOO.BAR.DATABASE-2.0.0.NUPKG
pass	funky_log_database	funky_log_database	1	0	0	FUNKY_LOG_DATABASE-1.0.0.NUPKG
pass	baz-database	baz-database	1	0	1	baz-database-1.0.1.nupkg
pass	funk-database	funk-database	1	2		funk-database-1.2.nupkg
pass	funk-database	funk-database	1			funk-database-1.nupkg

我试过使用 ([[:alnum:]._-]*?) 作为第一组，但结果相同。切换到 greedy 匹配匹配太多。

有什么好的建议吗？

Answer 1

这样可以吗？它并不复杂，而是 - return 你想要的数据（至少，我认为是这样）。

第 1 - 8 行 - 样本数据
temp CTE：删除扩展名 (.nupkg)，为简单起见[=26=]
最终查询：
- 第 18 行是 module name；如果它包含数字，则获取第一个数字的子字符串。否则，删除整个 PCT 值
- 第 20 - 22 行 return version：如果没有数字，return NULL。否则，return 从第一个数字开始的子串

SQL> with
  2  test as
  3   (select 'FUNKY_LOG_DATABASE-1.0.0.NUPKG' as pck, 'funky_log_database' as expected from dual
  4          union select 'FOO.BAR.DATABASE-2.0.0.NUPKG', 'foo.bar.database' from dual
  5          union select 'FOO-BAR-AUDIT-DATABASE.NUPKG', 'foo-bar-audit-database' from dual
  6          union select 'funk-database-1.nupkg', 'funk-database' from dual
  7          union select 'funk-database-1.2.nupkg', 'funk-database' from dual
  8          union select 'baz-database-1.0.1.nupkg', 'baz-database' from dual),
  9  temp as
 10    -- remove extension
 11    (select pck pck_old, expected,
 12       replace(lower(pck), '.nupkg', '') pck
 13     from test
 14    )
 15  select pck_old,
 16         expected,
 17    --
 18    nvl(substr(pck, 1, regexp_instr(pck, '\d') - 2), pck) module_name,
 19    --
 20    case when regexp_instr(pck, '\d') = 0 then null
 21         else substr(pck, regexp_instr(pck, '\d'))
 22    end version
 23  from temp;

PCK_OLD                        EXPECTED               MODULE_NAME             VERSION
------------------------------ ---------------------- ----------------------- --------
FOO-BAR-AUDIT-DATABASE.NUPKG   foo-bar-audit-database foo-bar-audit-database
FOO.BAR.DATABASE-2.0.0.NUPKG   foo.bar.database       foo.bar.database        2.0.0
FUNKY_LOG_DATABASE-1.0.0.NUPKG funky_log_database     funky_log_database      1.0.0
baz-database-1.0.1.nupkg       baz-database           baz-database            1.0.1
funk-database-1.2.nupkg        funk-database          funk-database           1.2
funk-database-1.nupkg          funk-database          funk-database           1

6 rows selected.

SQL>

Answer 2

可以从最后匹配得到版本，然后提取版本前的子串得到模块名：

select case module_name when expected then 'pass' else 'fail' end as test,
       y.*
from   (
  select lower(
           substr(
             t.pck,
             1,
             REGEXP_INSTR(t.pck, g.regex) - 1
           )
         ) as module_name, 
         t.expected,
         to_number(regexp_substr(t.pck, g.regex, 1, 1, '', 2)) as major,
         to_number(regexp_substr(t.pck, g.regex, 1, 1, '', 3)) as minor,
         to_number(regexp_substr(t.pck, g.regex, 1, 1, '', 4)) as patch,
         t.pck as package_name
  from   (
    select 'FUNKY_LOG_DATABASE-1.0.0.NUPKG' as pck, 'funky_log_database' as expected from dual
    union select 'FOO.BAR.DATABASE-2.0.0.NUPKG', 'foo.bar.database' from dual 
    union select 'FOO-BAR-AUDIT-DATABASE.NUPKG', 'foo-bar-audit-database' from dual
    union select 'funk-database-1.nupkg', 'funk-database' from dual
    union select 'funk-database-1.2.nupkg', 'funk-database' from dual
    union select 'baz-database-1.0.1.nupkg', 'baz-database' from dual
  ) t
  cross  join (
    select '(-(\d+)\.?(\d+)?\.?(\d+)?)?\.[^.]+$' as regex from dual
  ) g
)y;

输出：

TEST MODULE_NAME EXPECTED MAJOR MINOR PATCH PACKAGE_NAME

pass foo-bar-audit-database foo-bar-audit-database FOO-BAR-AUDIT-DATABASE.NUPKG

pass foo.bar.database foo.bar.database 2 0 0 FOO.BAR.DATABASE-2.0.0.NUPKG

pass funky_log_database funky_log_database 1 0 0 FUNKY_LOG_DATABASE-1.0.0.NUPKG

pass baz-database baz-database 1 0 1 baz-database-1.0.1.nupkg

pass funk-database funk-database 1 2 funk-database-1.2.nupkg

pass funk-database funk-database 1 funk-database-1.nupkg

db<>fiddle here

Oracle REGEXP_SUBSTR 不会匹配点字符

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