使用 REGEXP_REPLACE 替换开始字符串和结束字符串之间的任何内容（包括特殊字符）

Question

update customer set cust_info= REGEXP_REPLACE(cust_info,'<start><cust_name><start><cust_name><this field has long string><end>','') where user_id=123;

ORA-12733: 正则表达式太长

所以我尝试了 REGEXP_REPLACE(cust_info,'<start>[A-Z0-9_](*.?)<end>',''); 但没有用。

我想将 <start> 字符串和 <end> 字符串之间的任何内容替换为空白。（即删除 <start> 和 <end> 之间的任何内容）。

PS:- cust_info 列包含带有 html 标签的长字符串。

Answer 1

你的正则表达式似乎不对试试表达式 <start>(.)*?<end>

WITH da AS ( 
SELECT '<start><cust_name><start><cust_name><this field has long string><end>' AS cust_info FROM dual UNION ALL
SELECT 'name_test' AS cust_info FROM dual
) SELECT REGEXP_REPLACE(cust_info,'<start>(.)*?<end>','') FROM da;

尝试

UPDATE
    customer
SET
    cust_info = REGEXP_REPLACE(cust_info, '<start>(.)*?<end>', '')
WHERE
    user_id = 123;

解释：-

<start> //Matches literal <start>
    (.) //Matches any character except linebreaks
     *  //Matches 0 or more of the preceding token of (.)
     ?  //Makes the preceding quantifier lazy, causing it to match as few characters as possible
<end>   //Matches literal <end>

如果您的字符串有换行符，请使用 match_parameter 将其纳入考虑

REGEXP_REPLACE ( cust_info, '<start>(.*?)*?<end>' , '' , 1 , 1 , 'n'   )

基于：

REGEXP_REPLACE ( source_string, search_pattern
                 [, replacement_string
                    [, star_position
                       [, nth_occurrence
                          [, match_parameter ]
                       ]
                    ]
                 ]
               )

因此：

UPDATE
    customer
SET
    cust_info = REGEXP_REPLACE ( ID_DESC, '<start>(.*?)*?<end>' , '' , 1 , 1 , 'n'   )
WHERE
    user_id = 123;

使用 REGEXP_REPLACE 替换开始字符串和结束字符串之间的任何内容（包括特殊字符）

replace anything between a start string and end string (including special char) using REGEXP_REPLACE

sql

oracle

regexp-replace