3

私は2つのテーブルを持っています。

表1：

ID   SENTENCE
1    The shoes are good shoes.
2    There is a tree.
3    This is nice, nice, nice!

表 2:

ID   WORD
1    The
1    shoes
1    are
1    good
1    shoes
2    There
2    is
2    a
2    tree
3    This
3    is
3    nice
3    nice
3    nice

Table1 のすべての文で、各単語の出現回数をカウントする必要があります。いずれかの単語が複数回 (>1) 出現する場合は、それを数えます。それ以外の場合はスキップします。最終的に、結果のテーブルは次のようになります。

ID   SENTENCE                   CNT
1    The shoes are good shoes.  2
2    There is a tree.
3    This is nice, nice, nice!  3
4

4 に答える 4

3

使用できますcount() over()：

select distinct t1.id,
  t1.sentence,
  coalesce(t2.cnt, 0) cnt
from table1 t1
left join 
(
  select t1.id, 
    t1.sentence,
    t2.word,
    count(t2.word) over(partition by t1.id, t2.word) cnt
  from table1 t1
  left join table2 t2
    on t1.id = t2.id
) t2
  on t1.id = t2.id
  and t2.cnt > 1
order by t1.id

SQL Fiddle with Demoを参照してください。

または、次を使用できますcount()。

select t1.id,
  t1.sentence,
  coalesce(t2.cnt, 0) cnt
from table1 t1
left join 
(
  select t1.id, 
    t1.sentence,
    t2.word,
    count(t2.word) cnt
  from table1 t1
  left join table2 t2
    on t1.id = t2.id
  group by t1.id, t1.sentence, t2.word
  having count(t2.word) > 1
) t2
  on t1.id = t2.id
order by t1.id

デモで SQL Fiddle を参照してください

于 2013-02-27T11:09:48.847 に答える
3

SQLデモ

select t1.id, t1.sentence, 
coalesce(t2.cnt,0) as counts
from table1 t1
left join
(select id, word, count(id) cnt
from table2
group by id, word
having count(id) > 1)t2
on t1.id = t2.id
order by t1.id
;

| ID |                  SENTENCE | COUNTS |
-------------------------------------------
|  1 | The shoes are good shoes. |      2 |
|  2 |          There is a tree. |      0 |
|  3 | This is nice, nice, nice! |      3 |
于 2013-02-27T11:19:30.807 に答える
0 
SELECT table1.id, table1.sentence, COUNT(word) as cnt FROM table2 JOIN table1 ON table1.id = table2.id GROUP BY table2.word HAVING COUNT(word) > 1

私の答えはmysql用です。SQLでも機能することを確認しています

于 2013-02-27T11:02:32.500 に答える
0

結合の例はたくさんあるので、単語数の例だけを追加します。

Select REGEXP_COUNT('The shoes are good shoes.', ' ')+1 words_count
 From dual
/

WORDS_COUNT
-----------
5

SELECT id
     , LISTAGG(word, ' ') WITHIN GROUP (ORDER BY id, word) AS words
     , count(*) word_cnt
  FROM your_table2
GROUP BY id
/

ID    WORDS                    WORD_CNT
---------------------------------------
1    The are good shoes shoes    5
2    There a is tree             4
3    This is nice nice nice      5
于 2013-02-27T14:26:44.873 に答える