Oracle 检查表 a 中列中的字符串是否包含表 b 中列中的单词

Question

In Oracle 11.2.0.4... Given table A that has a long varchar column that has paragraphs of text, I want to check if any word in that text exists in a table that has a vertical list of singular words. I am looking for a way to do this in a single query, not with cursors.

Basically, I think I need to split the paragraph into words, then join on the table of vertical words, but am at a loss as to how to proceed.

Example:
TableA:

+----+--------------------------------------------+
| id |                  comments                  |
+----+--------------------------------------------+
|  1 | This sentence has no reference to any word |
|  2 | But this one references jungle             |
|  3 | And this one references Trees              |
+----+--------------------------------------------+

TableB:

+----+---------+
| id |  word   |
+----+---------+
|  1 | Jungle  |
|  2 | Forest  |
|  3 | Trees   |
|  4 | Animals |
|  5 | River   |
+----+---------+

Given those tables, I'd like a single SQL query that would tell me that Table A rows 2 and 3 have words that exist in Table B.

I've looked into using the xmltype feature to split a table into words and then join those together like this:

select id, 
(select count(1) from tableb f, xmltable(
  '/r/c/text()'
  passing xmltype('<r><c>'||replace((regexp_replace(comments, '[^a-z][^A-Z]')), ' ', '</c><c>')||'</c></r>')
  columns subs varchar2(4000) path '.'
) s where trim(f.word) = trim(s.subs)) words_found, comments
from tablea
where trim(comments) is not null

but that doesn't work very well. It can only process a small set of comment lines before failing.

Thanks in advance and I apologize if this has been answered. I did quite a bit of checking and didn't find quite what I am looking for.

Answer 1

Since you are only comparing the words, it's not required to split them. You may use a LIKE expression in JOIN

select a.*, b.* from TableA a  JOIN TableB b  ON 
' ' ||lower(a.comments)|| ' ' like  '% '||lower(b.word)||' %'

OR REGEXP_LIKE

select a.*, b.* from TableA a  JOIN TableB b  ON 
REGEXP_LIKE(a.comments,'(^|\W)'||b.word||'($|\W)','i');

Demo

Answer 2

How about a simple INSTR ?

SQL> with
  2  a (id, comments) as
  3    (select 1, 'This sentence has no reference to any word' from dual union all
  4     select 2, 'But this one references jungle'             from dual union all
  5     select 3, 'And this one references Trees'              from dual union all
  6     select 4, 'Jungle animals swim in a river'             from dual
  7    ),
  8  b (id, word) as
  9    (select 1, 'Jungle' from dual union all
 10     select 2, 'Forest' from dual union all
 11     select 3, 'Trees'  from dual union all
 12     select 4, 'Animals' from dual union all
 13     select 5, 'River'   from dual
 14    )
 15  select a.id, a.comments, b.word
 16  from a cross join b
 17  where instr(lower(a.comments), lower(b.word)) > 0
 18  order by a.id, b.word;

        ID COMMENTS                                   WORD
---------- ------------------------------------------ -------
         2 But this one references jungle             Jungle
         3 And this one references Trees              Trees
         4 Jungle animals swim in a river             Animals
         4 Jungle animals swim in a river             Jungle
         4 Jungle animals swim in a river             River

SQL>

Answer 3

Can anyone improve @Jeff's solution to match only string from table A that start with the word from table B?

Example:
TableA:

+----+--------------------------------------------+
| id |                  comments                  |
+----+--------------------------------------------+
|  1 | This sentence has no reference to any word |
|  2 | But this one references jungle             |
|  3 | Jungle  this one references 
+----+--------------------------------------------+

TableB:

+----+---------+
| id |  word   |
+----+---------+
|  1 | Jungle  |
|  2 | Forest  |
|  3 | Trees   |
|  4 | Animals |
|  5 | River   |
+----+---------+

Output:
3 Jungle this one references