如何在 Oracle SQL 中查找 col 包含字母、数字或特殊字符（连字符、撇号和空格除外）的行

Question

我必须找到像这样的名字：Robert@jr23（必须有字母，任何特殊字符或数字，除了连字符（-），撇号（'）和空格）。

我的做法如下：

select * from test where REGEXP_LIKE(trim(NAME_1), '[^- '']')

但是我没有得到正确的结果。

需要匹配：

Kevin#123
bob@jr
mike$dr

需要排除：

Alex-jr
Robert'jr
Brian jr

Answer 1

您需要将not 与要排除的值匹配的模式一起使用。否则，您将匹配包含不在排除列表中的任何字符的字符串，即所有字符。

select column_value
from   ora_mining_varchar2_nt
       ( 'Kevin#123'
       , 'bob@jr'
       , 'mike$dr'
       , 'Alex-jr'
       , 'Robert''jr'
       , 'Brian jr'
       , 'A!' )
where  1=1
and    not regexp_like(column_value,'[- '']')
and    regexp_like
       (column_value, '([A-Z0-9]+[^A-Z0-9])|([^A-Z0-9]+[A-Z0-9])', 'i') ;

编辑：添加了regex_like 条件以确保值包含字母或数字和“特殊字符”，这意味着此处的字符既不是字母、数字也不是空格，' 或 -。

Answer 2

您可以使用以下查询来包含除space, - and ' 之外的所有特殊字符

SQL> with tbl(name) as (
  2        select 'Kevin#123'          from dual union
  3        select 'bob@jr' from dual union
  4        select 'mike$dr' from dual union
  5        select 'Alex-jr'         from dual union
  6    select 'Brian jr'         from dual union
  7    select 'Brian)jr'         from dual union
  8        select 'Robert''jr'  from dual
  9      )
 10      select *
 11      from tbl
 12      where regexp_like(name, '[^[a-z]|[A-Z]|[:space:]|[:cntrl:]|/]')
 13  and  not regexp_like(name,'[- '']');

NAME
---------
Brian)jr -- see this is included
Kevin#123
bob@jr
mike$dr

SQL>

Answer 3

一种选择是用NULL 替换所有有效的东西，剩下的就是无效：

SQL> with test (col) as
  2    (select 'Robert@jr23' from dual union all
  3     select 'Kevin#123'   from dual union all
  4     select 'bob@jr'      from dual union all
  5     select 'mike$dr'     from dual union all
  6     select 'Alex-jr'     from dual union all
  7     select 'Robert''jr'  from dual union all
  8     select 'Brian jr'    from dual
  9    )
 10  select col
 11  From test
 12  where regexp_replace(col, '[[:alpha:]]|[[:digit:]]|-|''', null) is not null;

COL
-----------
Robert@jr23
Kevin#123
bob@jr
mike$dr
Brian jr

SQL>

---

我对您实际上想要得到的结果感到困惑。以下是一些显示此类正则表达式结果的示例；作为结果，您希望获得哪些 ID？

SQL> with test (id, col) as
  2    (select 1, 'Robert@jr23' from dual union all
  3     select 2, 'Kevin#123'   from dual union all
  4     select 3, 'bob@jr'      from dual union all
  5     select 4, 'mike$dr'     from dual union all
  6     select 5, 'Alex-jr'     from dual union all
  7     select 6, 'Robert''jr'  from dual union all
  8     select 7, 'Brian jr'    from dual union all
  9     select 8, 'Brian10'     from dual
 10    )
 11  select col,
 12  regexp_replace(col,
 13    '[[:alpha:]]|[[:digit:]]|[[:space:]]|-|''', null) result
 14  from test;

COL         RESULT
----------- ----------
Robert@jr23 @
Kevin#123   #
bob@jr      @
mike$dr     $
Alex-jr
Robert'jr
Brian jr
Brian10

8 rows selected.

SQL>

---

你的一个 cmets 说：

选择必须包含特殊字符或数字的客户名称（空格、连字符和撇号除外）

这意味着数字和特殊字符应该被视为“相等”。如果是这样，这有帮助吗？

SQL> with test (id, col) as
  2    (select 1, 'Robert@jr23' from dual union all
  3     select 2, 'Kevin#123'   from dual union all
  4     select 3, 'bob@jr'      from dual union all
  5     select 4, 'mike$dr'     from dual union all
  6     select 5, 'Alex-jr'     from dual union all
  7     select 6, 'Robert''jr'  from dual union all
  8     select 7, 'Brian jr'    from dual union all
  9     select 8, 'Brian10'     from dual
 10    )
 11  select id, col
 12  from test
 13  where regexp_replace(col, '[[:alpha:]]|[[:space:]]|-|''', '')
 14    is not null;

        ID COL
---------- -----------
         1 Robert@jr23
         2 Kevin#123
         3 bob@jr
         4 mike$dr
         8 Brian10

SQL>