我在网上找到了一个字符串标记生成器查询,并将其打包到下面的函数中,该函数返回动态标记集。该函数编译成功,但不知何故我收到错误“ORA-00933:SQL 命令未正确结束”。有人可以帮我调试这个吗?谢谢。

CREATE OR REPLACE TYPE KEY_VALUE_TYPE is object (k varchar2(4000), v varchar2(4000));
CREATE OR REPLACE TYPE KEY_VALUE_TABLE is table of key_value_type;
CREATE OR REPLACE FUNCTION StrTokenizer
        (string IN VARCHAR2, delimiter IN VARCHAR2)
RETURN key_value_table AS
    v_ret key_value_table;
BEGIN
    SELECT
        CAST( 
            multiset(
            SELECT
                LEVEL  k, 
                SUBSTR(STRING_TO_TOKENIZE, 
                    DECODE(LEVEL, 1, 1, INSTR(STRING_TO_TOKENIZE, DELIMITER, 1, LEVEL-1)+1), 
                    INSTR(STRING_TO_TOKENIZE, DELIMITER, 1, LEVEL) 
                        - DECODE( LEVEL, 1, 1, INSTR(STRING_TO_TOKENIZE, DELIMITER, 1, LEVEL-1)+1)) v
            FROM
                (
                    SELECT
                        ':string'||':delimiter' AS STRING_TO_TOKENIZE , ':delimiter' AS DELIMITER
                    FROM
                        DUAL
                )
            CONNECT BY INSTR(STRING_TO_TOKENIZE, DELIMITER, 1, LEVEL)>0 
            ORDER BY level ASC) 
        As key_value_table)
    INTO
        v_ret
    FROM dual;
    return v_ret;
END;

select * from strtokenizer('a,b,c',',')
ORA-00933: SQL command not properly ended

编辑:

  1. select * from table(strtokenizer('a,b,c',',')) 给出“ORA-30732:表不包含用户可见的列”。
  2. ORA-30732 当我以函数所有者身份登录时,该函数消失了,但该函数无法替换 ':string'||':d elimiter' 替换为传入的值。我该如何纠正这个问题?当我没有以函数所有者身份登录时,如何运行该函数?谢谢。
有帮助吗?

解决方案

假设该函数已编译,请尝试:

SELECT * 
  FROM TABLE(strtokenizer('a,b,c',','));

参考:

该功能需要更正 - 使用:

CREATE OR REPLACE FUNCTION StrTokenizer (string IN VARCHAR2, 
                                         delimiter IN VARCHAR2)
RETURN key_value_table AS v_ret key_value_table
BEGIN
  SELECT CAST(MULTISET(SELECT LEVEL k, 
                              SUBSTR(STRING_TO_TOKENIZE, DECODE(LEVEL, 1, 1, INSTR(STRING_TO_TOKENIZE, DELIMITER, 1, LEVEL-1)+1), 
                              INSTR(STRING_TO_TOKENIZE, DELIMITER, 1, LEVEL) - DECODE( LEVEL, 1, 1, INSTR(STRING_TO_TOKENIZE, DELIMITER, 1, LEVEL-1)+1)) v
                         FROM (SELECT string || delimiter AS STRING_TO_TOKENIZE , 
                                      delimiter AS DELIMITER
                                 FROM DUAL)
                   CONNECT BY INSTR(STRING_TO_TOKENIZE, DELIMITER, 1, LEVEL)>0 
                     ORDER BY level ASC) AS key_value_table)
    INTO v_ret
    FROM DUAL;

  RETURN v_ret;

END;
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