我有愚蠢的表和数据。 我需要： 1-将每行中的每个句子分成新行 根据soundex函数
，根据句子的最后部分对每行中的单词进行计数create table a (id number(9), words varchar(500));
insert into a values(1,'UK,LONDON,YEMEN,JOHN,CAIRO,OMAR ALI,EGYPT,Cairo,YEMAN,OMAR AMR ALI,LONDAN');
insert into a values(2,'UK,SUDAI,SUDAIN,AYHAM SHAHER YAFOOZ,ALI YAFOOZ');
insert into a values(3,'MALAYSIA, AHMED ALI,MALYSIAN');
exxted output
create table temp_words（id number（9），words varchar2（100），count_words number（9））;
id words count_words
1 UK 1
1 LONDON 2
1 YEMEN 2
1 CAIRO 2
1 OMAR ALI 2
1 JOHN 1
2 UK 1
2 SUDAI 2
2 AYHAM SHAHER YAFOOZ 2
3 MALAYSIA 2
3 AHMED ALI 1
问候
答案 0 :(得分：1)
根据需要拆分数据，可以使用“connect by”作为行生成器。
SQL> with src as (select id,',' || words || ',' as words,
2 length(words) - length(translate(words, '.,', '.')) + 1 no_of_words
3 from a)
4 select a.id,
5 substr(a.words,
6 instr(words, ',', 1, r) + 1,
7 instr(words, ',', 1, r + 1) - instr(words, ',', 1, r) - 1) word,
8 a.no_of_words
9 from (select level r
10 from dual
11 connect by level <= (select max(no_of_words) from src)) d
12 inner join src a
13 on d.r <= a.no_of_words
14 where a.no_of_words is not null
15 order by a.id, d.r
16 /
ID WORD NO_OF_WORDS
---------- -------------------- -----------
1 UK 11
1 LONDON 11
1 YEMEN 11
1 JOHN 11
1 CAIRO 11
1 OMAR ALI 11
1 EGYPT 11
1 Cairo 11
1 YEMAN 11
1 OMAR AMR ALI 11
1 LONDAN 11
2 UK 5
2 SUDAI 5
2 SUDAIN 5
2 AYHAM SHAHER YAFOOZ 5
2 ALI YAFOOZ 5
3 MALAYSIA 3
3 AHMED ALI 3
3 MALYSIAN 3
19 rows selected.
SQL>
答案 1 :(得分：1)
select id,words,
case when i=0 then
SUBSTR(words,
1,
case when INSTR(words,',', 1, 1)=0
then 100000
else
INSTR(words,',', 1, 1)-1
end
)
ELSE
SUBSTR(words,
INSTR(words,',', 1, i)+1,
case when INSTR(words,',', 1, i+1)=0
then 100000
else
INSTR(words,',', 1, i+1)-INSTR(words,',', 1, i)-1
end
)
END word,
i+1 COUNTWORDS
from a,
(
select * from
(
select 0 i from dual
union
select 1 i from dual
union
select 2 i from dual
union
select 3 i from dual
union
select 4 i from dual
union
select 5 i from dual
union
select 6 i from dual
union
select 7 i from dual
union
select 8 i from dual
union
select 9 i from dual
union
select 10 i from dual
union
select 11 i from dual
union
select 12 i from dual
)
)
table_i
where
case when i>0 then INSTR(words,',', 1, i)
else 100000 end <>0
order by id,i
答案 2 :(得分：1)
另一种方法（使用regexp_count和regexp_substr正则表达式函数）：
SQL> with Occurence(oc) as(
2 select level
3 from ( select max(regexp_count(words, '[^,]+')) ml
4 from a
5 ) t
6 connect by level <= t.ml
7 )
8 select id
9 , word
10 , count(word) over(partition by id, soundex(word) order by id) as count_words
11 From ( select a.id
12 , regexp_substr(words, '[^,]+', 1, o.oc) as word
13 from occurence o
14 cross join a
15 ) s
16 where s.word is not null
17 order by id
18 ;
ID WORD COUNT_WORDS
---------- -------------------- -----------
1 Cairo 2
1 CAIRO 2
1 EGYPT 1
1 JOHN 1
1 LONDAN 2
1 LONDON 2
1 OMAR ALI 1
1 OMAR AMR ALI 1
1 UK 1
1 YEMEN 2
1 YEMAN 2
2 ALI YAFOOZ 1
2 AYHAM SHAHER YAFOOZ 1
2 SUDAI 1
2 SUDAIN 1
2 UK 1
3 AHMED ALI 1
3 MALAYSIA 1
3 MALYSIAN 1
19 rows selected
答案 3 :(得分：0)
您需要将数据作为单独的记录插入。如果你愿意的话，你可以把它们作为一个串联的字符串，但这只会让你的生活变得非常困难。所以：
create table words (
id number,
w varchar2(100),
s varchar2(4)
);
create or replace trigger words_auto
before insert or update on words
for each row
begin
select trim(upper(:new.w)), soundex(:new.w)
into :new.w, :new.s
from dual;
end;
insert into words (id, w) values (1, 'UK');
insert into words (id, w) values (1, 'LONDON');
...
insert into words (id, w) values (3, ' AHMED ALI');
insert into words (id, w) values (3, 'MALYSIAN');
您可以编写一个过程来拆分连接字符串并适当填充words表。请注意，我创建了一个触发器，将输入规范化为大写，删除所有无关的空格并自动生成Soundex代码。
现在这里有一个问题：你想用Soundex代码对单词进行分组;但是，您如何确定基线？例如，'LONDON'和'LONDAN'都有代码'L535'，但你怎么知道哪个记录是'主'记录？...你不能，没有进一步的查找表！因此，您可以做的最好的事情是按Soundex代码分组。这不必存储在表中，作为一个视图更有意义：
create or replace view word_counts as
select id,
s soundex,
count(w) count_rows
from words
group by id,
s;
请注意，我调用了计数字段count_rows，因为它计算记录而不是不同的行。也就是说：“LONDON”，“LONDAN”和“LONDON”的记录显示的是3，而不是2（你可能会期待）。无论如何，对于您的数据，视图将如下所示：
id soundex count_rows
----- --------- -----------
1 U200 1
1 L535 2
... ... ...
3 M420 2
3 A534 1
正如我所说，如果没有进一步的基础设施，这真的是你能想到的最好的。