我有一个像“1-3,4,9,11-15”这样的字符串，我正在寻找一种简单的方法来将它转换为一个逗号分隔的字符串，其中包含单个数字，比如“1,2,3,4 ,9,11,12,13,14,15' 带有 PL/SQL 中的函数。
感谢您的帮助！
【问题讨论】：
我试了一下，可以达到类似的效果，
CREATE OR REPLACE FUNCTION get_series_of_numbers(p_input_string IN VARCHAR2)
RETURN VARCHAR2 IS
lo_start NUMBER;
lo_end NUMBER;
lo_final_string VARCHAR2(4000);
--a convinient method to get the series for a strig like '1-3' or 9-15'
FUNCTION get_series(p_series_string VARCHAR2) RETURN VARCHAR2 IS
lo_string_to_return VARCHAR2(4000);
BEGIN
IF instr(p_series_string
,'-') > 0
THEN
lo_start := to_number(substr(p_series_string
,1
,instr(p_series_string
,'-') - 1));
lo_end := to_number(substr(p_series_string
,instr(p_series_string
,'-') + 1));
--query to generate a series of numbers between a start and end point and then concatenate all with ','
SELECT listagg(actual_numbers
,',') within GROUP(ORDER BY actual_numbers)
INTO lo_string_to_return
FROM (SELECT LEVEL actual_numbers FROM dual WHERE LEVEL >= lo_start CONNECT BY LEVEL <= lo_end);
ELSE
lo_string_to_return := p_series_string;
END IF;
RETURN lo_string_to_return;
END;
BEGIN
--this loop is to get all the elements in the string separated by ',' as column
--so that we can loop over all
FOR i IN (SELECT regexp_substr(str
,'[^,]+'
,1
,rownum) split
FROM (SELECT p_input_string str FROM dual)
CONNECT BY LEVEL <= length(regexp_replace(str
,'[^,]+')) + 1)
LOOP
IF lo_final_string IS NOT NULL
THEN
lo_final_string := lo_final_string || ',' || get_series(i.split);
ELSE
lo_final_string := get_series(i.split);
END IF;
END LOOP;
RETURN lo_final_string;
END get_series_of_numbers;
一些测试结果：
DECLARE
input_string VARCHAR2(4000) := '1,2,8-3,4';
result_string VARCHAR2(4000);
begin
dbms_output.put_line('Input string is: '||input_string);
result_string := get_series_of_numbers(p_input_string => input_string);
dbms_output.put_line('Output string is: '||result_string);
end;
/*
Input string is: 1-3,4,9,11-15
Output string is: 1,2,3,4,9,11,12,13,14,15
Input string is: 1-3,4-6,9-10,11-15
Output string is: 1,2,3,4,5,6,9,10,11,12,13,14,15
Input string is: 1,2,3,4
Output string is: 1,2,3,4
Input string is: 1,2,3-8,4
Output string is: 1,2,3,4,5,6,7,8,4
--a negative case
Input string is: 1,2,8-3,4
Output string is: 1,2,,4
*/
希望它能提供有关可以进一步优化以处理极端情况或尚未涵盖的任何内容的要求的想法。干杯！！
【讨论】：
考虑这种使用 CTE 分解步骤的不同方法。见里面的 cmets。可以将其中的一些结合起来，但保持步骤分开会使其更简单。它也可以并且应该被制成一个可重用的过程或函数。
这也处理 NULL 列表元素，并且数字排序之外的数字将在输出中按数字排序。尝试使用'1-3,17-20,4,9,,11-15' 之类的数据。总是期待意外！
-- tbl_orig only creates a source for the original data
WITH tbl_orig(orig_str) AS (
SELECT '1-3,4,9,11-15' FROM dual
),
-- tbl_rows then contains that data split into rows on
-- the comma
tbl_rows(str_element) AS (
SELECT REGEXP_SUBSTR(orig_str, '(.*?)(,|$)', 1, LEVEL, NULL, 1)
FROM tbl_orig
CONNECT BY LEVEL <= REGEXP_COUNT(orig_str, ',')+1
),
-- Next look at those rows and if does not contain a hyphen just keep it,
-- else use an inline view to expand the range using listagg and regex's to
-- get the start and end of the range
tbl_expanded(str_expanded) AS (
SELECT
CASE INSTR(str_element, '-', 1)
WHEN 0 THEN str_element
ELSE (SELECT LISTAGG(n, ',') WITHIN GROUP (ORDER BY n)
FROM (SELECT ROWNUM n FROM dual CONNECT BY LEVEL <= REGEXP_SUBSTR(str_element, '\d+-(\d+)', 1, 1, NULL, 1))
WHERE n >= REGEXP_SUBSTR(str_element, '(\d+)-\d+', 1, 1, NULL, 1)
)
END AS str_expanded
FROM tbl_rows
)
-- Lastly put it all back together, but order by the numeric value of the first part of
-- the character strings.
SELECT LISTAGG(str_expanded, ',') WITHIN GROUP (ORDER BY TO_NUMBER(REGEXP_REPLACE(str_expanded, '(\d+).*', '\1')))
as fullrange
FROM tbl_expanded;
FULLRANGE
--------------------------------------------------------------------------------
1,2,3,4,9,11,12,13,14,15
【讨论】：