Question

我的字符串就像＆＃34; Vinoth ^ Vinoth Karthick Vinoth ^ Vinoth ^ Vinoth＆＃34;由＆＃34; ^＆＃34;分隔。我想用XXX代替Vinoth。

I/P String : Vinoth^Vinoth Karthick Vinoth^Vinoth^Vinoth
Expected output : XXX^Vinoth Karthick Vinoth^XXX^XXX

请使用Regexp_replace或ORACLE SQL语句中的任何其他函数建议如何执行此操作。

Answer 1

将分隔符let data = new FormData() data.append('name', 'hey') fetch('http://homestead.test/api/customers', { method: 'POST', headers: { 'Content-type': 'multipart/form-data' }, body: data }) .then((response) => { // now you can access response here console.log(response) })加倍并将字符串换成分隔符^字符，以便每个元素都有自己独特的前导和尾随分隔符，然后您只需用{{^替换^Vinoth^ 1}}并反转分隔符的加倍并修剪前导和尾随分隔符：

SQL Fiddle

Oracle 11g R2架构设置：

^XXX^

查询1 ：

SELECT 1 FROM DUAL;

<强> Results ：

SELECT TRIM(
         '^' FROM
         REPLACE(
           REPLACE(
             '^' ||
             REPLACE(
               'Vinoth^Vinoth Karthick Vinoth^Vinoth^Vinoth',
               '^',
               '^^'
             )
             || '^',
             '^Vinoth^',
             '^XXX^'
           ),
           '^^',
           '^'
         )
       ) AS replaced
FROM   DUAL

Answer 2

又一个选择：

SQL> with test (col) as
  2    (select 'Vinoth^Vinoth Karthick Vinoth^Vinoth^Vinoth' from dual),
  3  inter as
  4    (select regexp_substr(replace(col, '^', ':'), '[^:]+', 1, level) col,
  5            level lvl
  6     from test
  7     connect by level <= regexp_count(col, '\^') + 1
  8    )
  9  select listagg(regexp_replace(col, '^Vinoth$', 'XXX'), '^')
 10    within group (order by lvl) result
 11  from inter;

RESULT
-----------------------------------------------------------------------------

XXX^Vinoth Karthick Vinoth^XXX^XXX

SQL>

关于Oracle Regexp_Replace

2 个答案: