从另一个表中替换select语句中的多个标记

Question

我有两张桌子.
一个是answer_step_dtl
另一个是tag_mst,如下所示

Ans_code                      Ans_Desc
-------------------------------------------
50000000000000005770       Enter <B><APN></B> and press Ok.       
40000000000000000164       Enter <B><ACCOUNTNAME></B> in connection name.
40000000000000000165       Enter <B><ACCOUNTNAME></B> in <APN>.

等等.

我的tag_mst有这样的值

TAG_CODE                    TAG_NAME          TAG_VALUE
-------------------------------------------------------
100                          <APN>            EXAMPLE.COM
101                          <ACCOUNTNAME>    EXAMPLE

现在我的要求是这个我想选择答案表答案表,替换标签桅杆的值.一个ans_desc中可以有多个标记.

我在想这个方法

1. 循环遍历answer_dtl的结果从单个记录中获取标记

2. 循环遍历记录中的多个标记,然后替换标记

3. 存储在临时表中,然后打开新的tem表的光标.

这是方法吗？或者有任何简单的方法来做到这一点.

Answer 1

编辑：使用（递归）CTE 的额外查询，受到@Rob van Wijk的注释的启发，需要 11g（R2？）：

SQL> WITH data AS (
  2     SELECT ans_code, Ans_Desc, tag_name, tag_value,
  3            row_number() OVER (partition BY ans_code ORDER BY t.rowid) no,
  4            row_number() OVER
  5               (partition BY ans_code ORDER BY t.rowid DESC) is_last
  6        FROM answer_step_dtl a
  7        JOIN tag_mst t ON a.ans_desc LIKE '%' || t.tag_name || '%'
  8  ), n(ans_code, no, is_last, replaced) AS (
  9     SELECT ans_code, no n, is_last,
 10            replace (ans_desc, tag_name, tag_value) replaced
 11       FROM data
 12      WHERE no = 1
 13     UNION ALL
 14     SELECT d.ans_code, d.no, d.is_last,
 15            replace (n.replaced, d.tag_name, d.tag_value) replaced
 16       FROM data d
 17       JOIN n ON d.ans_code = n.ans_code
 18        AND d.no = n.no + 1
 19  )
 20  SELECT *
 21    FROM n
 22   WHERE is_last=1;

ANS_CODE             NO IS_LAST REPLACED
-------------------- -- ------- ---------------------------------------
40000000000000000164  1       1 Enter <B>EXAMPLE</B> in connection
50000000000000005770  1       1 Enter <B>EXAMPLE.COM</B> and press Ok.
40000000000000000165  2       1 Enter <B>EXAMPLE</B> in EXAMPLE.COM.

初步回答：

您可以使用 PL/SQL 函数。即使有多个标签需要替换，以下操作也将起作用：

CREATE OR REPLACE FUNCTION replacetags(p_desc VARCHAR2)
   RETURN VARCHAR2 IS
   l_result  LONG := p_desc;
   l_tag_pos INTEGER := 1;
   l_tag     tag_mst.tag_name%TYPE;
BEGIN
   LOOP
      l_tag     := regexp_substr(l_result, '<[^<]+>', l_tag_pos);
      l_tag_pos := regexp_instr(l_result, '<[^<]+>', l_tag_pos) + 1;
      EXIT WHEN l_tag IS NULL;
      BEGIN
         SELECT replace(l_result, l_tag, tag_value)
           INTO l_result
           FROM tag_mst
          WHERE tag_name = l_tag;
      EXCEPTION
         WHEN no_data_found THEN
            NULL; -- tag doesn't exist in tag_mst
      END;
   END LOOP;
   RETURN l_result;
END;

SQL> SELECT ans_code, replacetags(ans_desc)
  2    FROM answer_step_dtl;

ANS_CODE              REPLACETAGS(ANS_DESC)
--------------------- ----------------------------------------
50000000000000005770  Enter <B>EXAMPLE.COM</B> and press Ok.
40000000000000000164  Enter <B>EXAMPLE</B> in connection
40000000000000000165  Enter <B>EXAMPLE</B> in EXAMPLE.COM.