我想替换这个:
"STORES/KOL#10/8/36#1718.00#4165570.00#119539388#PT3624496#9902001#04266#6721#PT3624496-11608091-1-55-STORES/KOL"
有了这个:
"STORES/KOL#10#8#36#1718.00#4165570.00#119539388#PT3624496#9902001#04266#6721#PT3624496-11608091-1-55-STORES/KOL"
基本上这是有条件的替换我想用/
替换#
像STORES/KOL
字符串应该是STORES/KOL
但10/8/36
字符串应该是10#8#36
这将用/
替换第2和第3个#
角色:
Oracle安装程序:
CREATE TABLE test_data ( value ) AS
SELECT '"STORES/KOL#10/8/36#1718.00#4165570.00#119539388#PT3624496#9902001#04266#6721#PT3624496-11608091-1-55-STORES/KOL"'
FROM DUAL;
查询:
SELECT REGEXP_REPLACE(
value,
'^(.*?/.*?)/(.*?)/(.*)$',
'\1#\2#\3'
) AS replacement
FROM test_data
输出:
| REPLACEMENT | | :---------------------------------------------------------------------------------------------------------------- | | "STORES/KOL#10#8#36#1718.00#4165570.00#119539388#PT3624496#9902001#04266#6721#PT3624496-11608091-1-55-STORES/KOL" |
db <>小提琴here
这是使用REGEXP_REPLACE
的一个选项。我们可以尝试定位以下正则表达式模式:
#(\d+)/(\d+)/(\d+)#
然后使用三个捕获组替换,用井号替换路径分隔符。
WITH yourTable AS (
SELECT 'STORES/KOL#10/8/36#1718.00#4165570.00#119539388#PT3624496#9902001#04266#6721#PT3624496-11608091-1-55-STORES/KOL' AS input FROM dual
)
SELECT
input,
REGEXP_REPLACE(input, '#(\d+)/(\d+)/(\d+)#', '#\1#\2#\3#') AS output
FROM yourTable;
此正则表达式替换是否足够/准确用于其余数据取决于您从未向我们展示过的数据。
with s as (select '"STORES/KOL#10/8/36#1718.00#4165570.00#119539388#PT3624496#9902001#04266#6721#PT3624496-11608091-1-55-STORES/KOL"' str from dual)
select
replace(replace(str, '/', '#'), 'STORES#KOL', 'STORES/KOL') result_str_1,
regexp_replace(str, '(\d)/', '\1#') result_str_2
from s;