如何在 Oracle 的文本中保留相同的大小写？

Question

我有一个包含单词和句子列的表格。这个想法是如果在单词列中找到单词，则用链接（包括单词本身）替换句子中的单词。下面的查询完美替换，但由于链接是从 temp.word 列构造的，因此句子中单词的大小写更改为 word 列中单词的大小写。有没有办法在句子本身中保留相同的大小写？

Create table temp(
  id       NUMBER,
  word     VARCHAR2(1000),
  sentence VARCHAR2(2000)
);

insert into temp
SELECT 1,'automation testing', 'automtestingation Testing is popular kind of testing' FROM DUAL UNION ALL
SELECT 2,'testing','manual testing' FROM DUAL UNION ALL
SELECT 3,'manual testing','this is an old method of testing' FROM DUAL UNION ALL
SELECT 4,'punctuation','automation testing,manual testing,punctuation,automanual testing-testing' FROM DUAL UNION ALL
SELECT 5,'B-number analysis','B-number analysis table' FROM DUAL UNION ALL
SELECT 6,'B-number analysis table','testing B-number analysis' FROM DUAL UNION ALL
SELECT 7,'Not Matched','testing testing testing' FROM DUAL;

with words(id, word, word_length, search1, replace1, search2, replace2) as (
  select id, word, length(word),
  '(^|\W)' || REGEXP_REPLACE(word, '([][)(}{|^$\.*+?])', '\\\1') || '($|\W)',
  '\1{'|| id ||'}\2',
  '{'|| id ||'}',
  'http://localhost/' || id || '/<u>' || word || '</u>'
  FROM temp
)
, joined_data as (
  select w.search1, w.replace1, w.search2, w.replace2,
    s.rowid s_rid, s.sentence,
    row_number() over(partition by s.rowid order by word_length desc) rn
  from words w
  join temp s
  on instr(UPPER(s.sentence), UPPER(w.word)) > 0
  and regexp_like(s.sentence, w.search1)
)
, unpivoted_data as (
  select S_RID, SENTENCE, PHASE, SEARCH_STRING, REPLACE_STRING,
    row_number() over(partition by s_rid order by phase, rn) rn,
    case when row_number() over(partition by s_rid order by phase, rn)
      = count(*) over(partition by s_rid)
      then 1
      else 0
    end is_last
  from joined_data
  unpivot(
    (search_string, replace_string) 
    for phase in ( (search1, replace1) as 1, (search2, replace2) as 2 ))
)
, replaced_data(S_RID, RN, is_last, SENTENCE) as (
  select S_RID, RN, is_last,
    regexp_replace(SENTENCE, search_string, replace_string,1,0,'i')
  from unpivoted_data
  where rn = 1
  union all
  select n.S_RID, n.RN, n.is_last,
    case when n.phase = 1
      then regexp_replace(o.SENTENCE, n.search_string, n.replace_string,1,0,'i')
      else replace(o.SENTENCE, n.search_string, n.replace_string)
    end
  from unpivoted_data n
  join replaced_data o
    on o.s_rid = n.s_rid and n.rn = o.rn + 1  
)
select s_rid, sentence from replaced_data
where is_last = 1
order by s_rid;

For example, for id = 1, the sentence is automtestingation Testing is popular kind of testing
After replacement it will be automtestingation http://localhost/2/<u>testing</u> is popular kind of http://localhost/2/<u>testing</u>.

The word Testing is replaced with testing(from the temp.word column).

预期的结果是

automtestingation http://localhost/2/<u>Testing</u> is popular kind of http://localhost/2/<u>testing</u>

Answer 1

Oracle 设置：

Create table temp(
  id       NUMBER,
  word     VARCHAR2(1000),
  Sentence VARCHAR2(2000)
);

insert into temp
SELECT 1,'automation testing', 'automtestingation TeStInG TEST is popular kind of testing' FROM DUAL UNION ALL
SELECT 2,'testing','manual testing' FROM DUAL UNION ALL
select 2,'test', 'test' FROM DUAL UNION ALL
SELECT 3,'manual testing','this is an old method of testing' FROM DUAL UNION ALL
SELECT 4,'punctuation','automation testing,manual testing,punctuation,automanual testing-testing' FROM DUAL UNION ALL
SELECT 5,'B-number analysis','B-number analysis table' FROM DUAL UNION ALL
SELECT 6,'B-number analysis table','testing B-number analysis' FROM DUAL UNION ALL
SELECT 7,'Not Matched','testing testing testing' FROM DUAL UNION ALL
SELECT 8,'^[($','testing characters ^[($ that need escaping in a regular expression' FROM DUAL;

SQL 类型：

CREATE TYPE stringlist IS TABLE OF VARCHAR2(4000);
/
CREATE TYPE intlist IS TABLE OF NUMBER(20,0);
/

PL/SQL 函数：

CREATE FUNCTION replace_words(
  word_list IN  stringlist,
  id_list   IN  intlist,
  sentence  IN  temp.sentence%TYPE
) RETURN temp.sentence%TYPE
IS
  p_sentence       temp.sentence%TYPE := UPPER( sentence );
  p_pos            PLS_INTEGER := 1;
  p_min_word_index PLS_INTEGER;
  p_word_index     PLS_INTEGER;
  p_start          PLS_INTEGER;
  p_index          PLS_INTEGER;
  o_sentence       temp.sentence%TYPE;
BEGIN
  LOOP
    p_min_word_index := NULL;
    p_index          := NULL;
    FOR i IN 1 .. word_list.COUNT LOOP
      p_word_index := p_pos;
      LOOP
        p_word_index := INSTR( p_sentence, word_list(i), p_word_index );
        EXIT WHEN p_word_index = 0;
        IF (   p_word_index  > 1
           AND REGEXP_LIKE( SUBSTR( p_sentence, p_word_index - 1, 1 ), '\w' )
           )
           OR  REGEXP_LIKE( SUBSTR( p_sentence, p_word_index + LENGTH( word_list(i) ), 1 ), '\w' )
        THEN
           p_word_index := p_word_index + 1;
           CONTINUE;
        END IF;
        IF p_min_word_index IS NULL OR p_word_index < p_min_word_index THEN
          p_min_word_index := p_word_index;
          p_index := i;
        END IF;
        EXIT;
      END LOOP;
    END LOOP;
    IF p_index IS NULL THEN
      o_sentence := o_sentence || SUBSTR( sentence, p_pos );
      EXIT;
    ELSE
      o_sentence := o_sentence
                    || SUBSTR( sentence, p_pos, p_min_word_index - p_pos )
                    || 'http://localhost/'
                    || id_list(p_index)
                    || '/<u>'
                    || SUBSTR( sentence, p_min_word_index, LENGTH( word_list( p_index ) ) )
                    || '</u>';
      p_pos := p_min_word_index + LENGTH( word_list( p_index ) );
    END IF;
  END LOOP;
  RETURN o_sentence;
END;
/

合并：

MERGE INTO temp dst
USING (
  WITH lists ( word_list, id_list ) AS (
    SELECT CAST(
             COLLECT(
               UPPER( word )
               ORDER BY LENGTH( word ) DESC, UPPER( word ) ASC, ROWNUM
             )
             AS stringlist
           ),
           CAST(
             COLLECT(
               id
               ORDER BY LENGTH( word ) DESC, UPPER( word ) ASC, ROWNUM
             )
             AS intlist
           )
    FROM   temp
  )
  SELECT t.ROWID rid,
         replace_words(
           word_list,
           id_list,
           sentence
         ) AS replaced_sentence
  FROM   temp t
         CROSS JOIN lists
) src
ON ( dst.ROWID = src.RID )
WHEN MATCHED THEN
  UPDATE SET sentence = src.replaced_sentence;

输出：

SELECT * FROM temp;

身份证 |字 |句子 -: | :------------------------ | :------------------------------------------------ -------------------------------------------------- -------------------------------------------------- -------------------------------------------------- -- 1 |自动化测试|自动化测试 http://localhost/2/TestInG http://localhost/2/TEST 是一种流行的 http://localhost/2/测试 2 |测试 | http://localhost/3/手动测试 2 |测试 | http://localhost/2/测试 3 |手动测试 |这是 http://localhost/2/testing 的旧方法 4 |标点符号 | http://localhost/1/自动化测试,http://localhost/3/手动测试,http://localhost/4/标点符​​号 u>,automanual http://localhost/2/testing-http://localhost/2/testing 5 | B数分析| http://localhost/6/B数分析表 6 | B数分析表| http://localhost/2/测试 http://localhost/5/B数分析 7 |不匹配 | http://localhost/2/测试 http://localhost/2/测试 http://localhost/2/测试 8 | ^[($ | http://localhost/2/testing 字符 http://localhost/8/^[($ 需要在正则表达式中转义

db小提琴here

Answer 2

虽然肯定有一种方法可以在单个 SQL 语句中执行此操作，但我认为使用单独的函数可以更好地解决此问题：

create or replace function replace_words(p_word varchar2, p_sentence varchar2) return varchar2 is
    v_match_position number;
    v_match_count    number := 0;
    v_new_sentence   varchar2(4000) := p_sentence;
begin
    --Find all matches.
    loop
        --Find Nth case-insensitive match
        v_match_count := v_match_count + 1;
        v_match_position := regexp_instr(
            srcstr     => v_new_sentence,
            pattern    => p_word,
            occurrence => v_match_count,
            modifier   => 'i');

        exit when v_match_position = 0;

        --Insert text, instead of replacing, to keep the original case.
        v_new_sentence :=
            substr(v_new_sentence, 1, v_match_position - 1) || 'http://localhost/2/<u>'||
            substr(v_new_sentence, v_match_position, length(p_word)) || '</u>'||
            substr(v_new_sentence, v_match_position + length(p_word));

    end loop;

    return v_new_sentence;
end;
/

那么SQL查询是这样的：

select id, word, sentence, replace_words(word, sentence) from temp;