Oracle - 需要在给定字符串之间提取文本

Question

示例 - 需要提取“Begin begin”和“End end”之间的所有内容。我试过这样：

with phrases as (
  select 'stackoverflow is awesome. Begin beginHello, World!End end It has everything!' as phrase
    from dual
         )
select regexp_replace(phrase
     , '([[:print:]]+Begin begin)([[:print:]]+)(End end[[:print:]]+)', '\2')
  from phrases
       ;

结果：你好，世界！

但是，如果我的文本包含换行符，它会失败。任何提示如何解决此问题以允许提取包含新行的文本？

[编辑]它是如何失败的：

with phrases as (
  select 'stackoverflow is awesome. Begin beginHello, 
  World!End end It has everything!' as phrase
    from dual
         )
select regexp_replace(phrase
     , '([[:print:]]+Begin begin)([[:print:]]+)(End end[[:print:]]+)', '\2')
  from phrases
       ;

结果：

stackoverflow 很棒。开始 beginHello, World!End end 它有 一切！

应该是：

你好，
世界！

[编辑]

另一个问题。让我们看看这个示例：

WITH phrases AS (
  SELECT 'stackoverflow is awesome. Begin beginHello,
 World!End end It has everything!End endTESTESTESTES' AS phrase
    FROM dual
)
SELECT REGEXP_REPLACE(phrase, '.+Begin begin(.+)End end.+', '\1', 1, 1, 'n')
  FROM phrases;

结果：

你好，
World!End end 它拥有一切！

所以它匹配最后出现的结束字符串，这不是我想要的。子字符串应该被提取到我的标签的第一次出现，所以结果应该是：

你好，
世界！

第一次出现标签字符串之后的所有内容都应该被忽略。有什么想法吗？

Answer 1

我对 POSIX [[:print:]] 字符类不太熟悉，但我使用通配符 . 使您的查询正常工作。您需要在REGEXP_REPLACE() 中指定n 匹配参数，以便. 可以匹配换行符：

WITH phrases AS (
  SELECT 'stackoverflow is awesome. Begin beginHello,
 World!End end It has everything!' AS phrase
    FROM dual
)
SELECT REGEXP_REPLACE(phrase, '.+Begin begin(.+)End end.+', '\1', 1, 1, 'n')
  FROM phrases;

我使用了\1 反向引用，因为我没有看到需要从正则表达式中捕获其他组。 如果分隔符前后没有任何内容，使用* 量词（而不是+）也是一个好主意。如果您想捕获所有组，那么您可以使用以下内容：

WITH phrases AS (
  SELECT 'stackoverflow is awesome. Begin beginHello,
 World!End end It has everything!' AS phrase
    FROM dual
)
SELECT REGEXP_REPLACE(phrase, '(.+Begin begin)(.+)(End end.+)', '\2', 1, 1, 'n')
  FROM phrases;

更新 - 仅供参考，我使用 [[:print:]] 进行了测试，但它不起作用。这并不奇怪，因为 [[:print:]] 应该匹配 printable 字符。它不匹配任何低于 32（空格）的 ASCII 值。您需要使用.。

更新 #2 - 每次更新问题 - 我认为正则表达式不会按您希望的方式工作。将惰性量词添加到 (.+) 没有任何效果，Oracle 正则表达式没有前瞻功能。您可能会做几件事，其中之一是使用INSTR() 和SUBSTR()：

WITH phrases AS (
  SELECT 'stackoverflow is awesome. Begin beginHello,
 World!End end It has everything!End endTESTTESTTEST' AS phrase
    FROM dual
)
SELECT SUBSTR(phrase, str_start, str_end - str_start) FROM (
    SELECT INSTR(phrase, 'Begin begin') + LENGTH('Begin begin') AS str_start
         , INSTR(phrase, 'End end') AS str_end, phrase
      FROM phrases
);

另一种是将INSTR()和SUBSTR()与正则表达式结合起来：

WITH phrases AS (
  SELECT 'stackoverflow is awesome. Begin beginHello,
 World!End end It has everything!End endTESTTESTTEST' AS phrase
    FROM dual
)
SELECT REGEXP_REPLACE(SUBSTR(phrase, 1, INSTR(phrase, 'End end') + LENGTH('End end')), '.+Begin begin(.+)End end.+', '\1', 1, 1, 'n')
  FROM phrases;

Answer 2

试试这个正则表达式：

([[:print:]]+Begin begin)(.+?)(End end[[:print:]]+)

示例用法：

SELECT regexp_replace(
         phrase ,
         '([[:print:]]+Begin begin)(.+?)(End end[[:print:]]+)',
         '\2',
         1,  -- Start at the beginning of the phrase
         0,  -- Replace ALL occurences
         'n' -- Let dot meta character matches new line character
)
FROM
  (SELECT 'stackoverflow is awesome. Begin beginHello, '
    || chr(10)
    || ' World!End end It has everything!' AS phrase
  FROM DUAL
  )

点元字符 (.) 匹配数据库字符集中的任何字符和换行符。但是，当调用 regexp_replace 时，match_parameter 必须包含 n 开关，以便 dot 匹配新行。

Answer 3

为了让您的第二个选项起作用，您需要添加[[:space:][:print:]]*，如下所示：

with phrases as (
  select 'stackoverflow is awesome. Begin beginHello, 
  World!End end It has everything!' as phrase
    from dual
         )
select regexp_replace(phrase
     , '([[:print:]]+Begin begin)([[:print:]]+[[:space:][:print:]]*)(End end[[:print:]]+)', '\2')
  from phrases
       ;    

但如果你有更多\n，它仍然会中断，例如它不会工作

with phrases as (
  select 'stackoverflow is awesome. Begin beginHello, 
  World!End end 
  It has everything!' as phrase
    from dual
         )
select regexp_replace(phrase
     , '([[:print:]]+Begin begin)([[:print:]]+[[:space:][:print:]]*)(End end[[:print:]]+)', '\2')
  from phrases
       ;       

那你需要添加

with phrases as (
  select 'stackoverflow is awesome. Begin beginHello, 
  World!End end 
  It has everything!' as phrase
    from dual
         )
select regexp_replace(phrase
     , '([[:print:]]+Begin begin)([[:print:]]+[[:space:][:print:]]*)(End end[[:print:]]+[[:space:][:print:]]*)', '\2')
  from phrases
       ;   

正则表达式的问题在于，您可能必须确定变体的范围并创建一个匹配所有变体的规则。如果某些内容超出了您的范围，则必须访问正则表达式并添加新的异常。

您可以找到更多信息here。

Answer 4

 Description.........: This is a function similar to the one that was available from PRIME Computers
                       back in the late 80/90's.  This function will parse out a segment of a string
                       based on a supplied delimiter.  The delimiters can be anything.
Usage:
     Field(i_string     =>'This.is.a.cool.function'
          ,i_deliiter   => '.'
          ,i_start_pos  => 2
          ,i_occurrence => 2)

     Return value = is.a

---

FUNCTION field(i_string           VARCHAR2
              ,i_delimiter        VARCHAR2
              ,i_occurance        NUMBER DEFAULT 1
              ,i_return_instances NUMBER DEFAULT 1) RETURN VARCHAR2 IS
  --
  v_delimiter      VARCHAR2(1);
  n_end_pos        NUMBER;
  n_start_pos      NUMBER := 1;
  n_delimiter_pos  NUMBER;
  n_seek_pos       NUMBER := 1;
  n_tbl_index      PLS_INTEGER := 0;
  n_return_counter NUMBER := 0;
  v_return_string  VARCHAR2(32767);
  TYPE tbl_type IS TABLE OF VARCHAR2(4000) INDEX BY PLS_INTEGER;
  tbl tbl_type;
  e_no_delimiters EXCEPTION;
  v_string VARCHAR2(32767) := i_string || i_delimiter;
BEGIN
  BEGIN
    LOOP
      ----------------------------------------
      -- Search for the delimiter in the
      -- string
      ----------------------------------------
      n_delimiter_pos := instr(v_string, i_delimiter, n_seek_pos);
      --
      IF n_delimiter_pos = length(v_string) AND n_tbl_index = 0 THEN
        ------------------------------------------
        -- The delimiter you are looking for is
        -- not in this string.
        ------------------------------------------
        RAISE e_no_delimiters;
      END IF;
      --
      EXIT WHEN n_delimiter_pos = 0;
      n_start_pos := n_seek_pos;
      n_end_pos   := n_delimiter_pos - n_seek_pos;
      n_seek_pos  := n_delimiter_pos + 1;
      --
      n_tbl_index := n_tbl_index + 1;
      -----------------------------------------------
      -- Store the segments of the string in a tbl
      -----------------------------------------------
      tbl(n_tbl_index) := substr(i_string, n_start_pos, n_end_pos);
    END LOOP;
    ----------------------------------------------
    -- Prepare the results for return voyage
    ----------------------------------------------
    v_delimiter := NULL;
    FOR a IN tbl.first .. tbl.last LOOP
      IF a >= i_occurance AND n_return_counter < i_return_instances THEN
        v_return_string  := v_return_string || v_delimiter || tbl(a);
        v_delimiter      := i_delimiter;
        n_return_counter := n_return_counter + 1;
      END IF;
    END LOOP;
    --
  EXCEPTION
    WHEN e_no_delimiters THEN
      v_return_string := i_string;
  END;
  RETURN TRIM(v_return_string);
END;