方法如下:
CREATE OR REPLACE TYPE ty_str_split IS TABLE OF VARCHAR2 (4000);
CREATE OR REPLACE FUNCTION fn_split (p_str IN VARCHAR2, p_delimiter IN VARCHAR2)
RETURN ty_str_split
IS
j INT := 0;
i INT := 1;
len INT := 0;
len1 INT := 0;
str VARCHAR2 (4000);
str_split ty_str_split := ty_str_split ();
BEGIN
len := LENGTH (p_str);
len1 := LENGTH (p_delimiter);
WHILE j < len
LOOP
j := INSTR (p_str, p_delimiter, i);
IF j = 0
THEN
j := len;
str := SUBSTR (p_str, i);
str_split.EXTEND;
str_split (str_split.COUNT) := str;
IF i >= len
THEN
EXIT;
END IF;
ELSE
str := SUBSTR (p_str, i, j - i);
i := j + len1;
str_split.EXTEND;
str_split (str_split.COUNT) := str;
END IF;
END LOOP;
RETURN str_split;
END fn_split;
上面是一个拆分函数，我现在要调用来拆分一个表里面的一个字段 我是这样调用的
with emailaddress as(
SELECT variable_value from ledup_variables where variable_type='Email')
select fn_split(variable_value,',') AS ty_str_split from emailaddress
目的是拆分ledup_variables 中的variable_value 字段
但这样执行结果是返回一个数据集 ，点开数据集才是拆分后的数据列表，我怎么才能直接执行出正确的数据？请教
------解决方案--------------------
- SQL code
--如果是单行SQL> WITH emailaddress AS 2 (SELECT 'abc@sina.com,bbb@yahoo.com,ccc@21cn.com' variable_value FROM dual) 3 SELECT * 4 FROM TABLE (SELECT fn_split(variable_value, ',') AS ty_str_split 5 FROM emailaddress);COLUMN_VALUE--------------------------------------------abc@sina.combbb@yahoo.comccc@21cn.comSQL>
------解决方案--------------------
- SQL code
WITH emailaddress AS (SELECT variable_value from ledup_variables where variable_type='Email')SELECT substr(b.variable_value, instr(b.variable_value, ',', 1, column_value) + 1, decode(instr(b.variable_value, ',', 1, column_value + 1), 0, length(b.variable_value) + 1, instr(b.variable_value, ',', 1, column_value + 1)) - instr(b.variable_value, ',', 1, column_value) - 1) s FROM (select ','||variable_value variable_value from emailaddress) b, TABLE(CAST(MULTISET (SELECT rownum FROM DUAL CONNECT BY rownum <= length(variable_value) - length(REPLACE(variable_value, ','))) AS ty_str_split));