是的，你可以。

UPDATE table_name 
  SET column_name = 'seach_str_name'
  WHERE column_name REGEXP '[^a-zA-Z0-9()_ .\-]';

我们不用正则表达式来解决这个问题 此查询只替换精确匹配字符串。

update employee set
employee_firstname = 
trim(REPLACE(concat(" ",employee_firstname," "),' jay ',' abc '))

例子:

emp_id employee_firstname 1杰 2 jay ajay 三杰

执行查询结果后:

emp_id employee_firstname 1美国广播公司 2 ABC ajay 3 abc

我很高兴地告诉大家，既然问了这个问题，现在有了一个令人满意的答案!看看这个很棒的套餐吧:

https://github.com/mysqludf/lib_mysqludf_preg

示例SQL:

SELECT PREG_REPLACE('/(.*?)(fox)/' , 'dog' , 'the quick brown fox' ) AS demo;

我从这篇博客文章中找到了这个问题的链接。

我们可以在SELECT查询中使用IF条件，如下:

假设对于包含“ABC”，“ABC1”，“ABC2”，“ABC3”的任何东西，…，我们想用“ABC”替换，然后在SELECT查询中使用REGEXP和IF()条件，我们可以实现这一点。

语法:

SELECT IF(column_name REGEXP 'ABC[0-9]$','ABC',column_name)
FROM table1 
WHERE column_name LIKE 'ABC%';

例子:

SELECT IF('ABC1' REGEXP 'ABC[0-9]$','ABC','ABC1');

MySQL 8.0 +:

您可以使用本机REGEXP_REPLACE函数。

旧版本:

您可以使用用户定义函数(UDF)，如mysql-udf-regexp。

更新2:MySQL 8.0中提供了一组有用的regex函数，包括REGEXP_REPLACE。这将使阅读变得不必要，除非您必须使用较早的版本。

更新1:现在已经把这变成了一篇博客:http://stevettt.blogspot.co.uk/2018/02/a-mysql-regular-expression-replace.html

下面扩展了Rasika Godawatte提供的函数，但它涵盖了所有必要的子字符串，而不仅仅是测试单个字符:

-- ------------------------------------------------------------------------------------
-- USAGE
-- ------------------------------------------------------------------------------------
-- SELECT reg_replace(<subject>,
--                    <pattern>,
--                    <replacement>,
--                    <greedy>,
--                    <minMatchLen>,
--                    <maxMatchLen>);
-- where:
-- <subject> is the string to look in for doing the replacements
-- <pattern> is the regular expression to match against
-- <replacement> is the replacement string
-- <greedy> is TRUE for greedy matching or FALSE for non-greedy matching
-- <minMatchLen> specifies the minimum match length
-- <maxMatchLen> specifies the maximum match length
-- (minMatchLen and maxMatchLen are used to improve efficiency but are
--  optional and can be set to 0 or NULL if not known/required)
-- Example:
-- SELECT reg_replace(txt, '^[Tt][^ ]* ', 'a', TRUE, 2, 0) FROM tbl;
DROP FUNCTION IF EXISTS reg_replace;
DELIMITER //
CREATE FUNCTION reg_replace(subject VARCHAR(21845), pattern VARCHAR(21845),
  replacement VARCHAR(21845), greedy BOOLEAN, minMatchLen INT, maxMatchLen INT)
RETURNS VARCHAR(21845) DETERMINISTIC BEGIN 
  DECLARE result, subStr, usePattern VARCHAR(21845); 
  DECLARE startPos, prevStartPos, startInc, len, lenInc INT;
  IF subject REGEXP pattern THEN
    SET result = '';
    -- Sanitize input parameter values
    SET minMatchLen = IF(minMatchLen IS NULL OR minMatchLen < 1, 1, minMatchLen);
    SET maxMatchLen = IF(maxMatchLen IS NULL OR maxMatchLen < 1
                         OR maxMatchLen > CHAR_LENGTH(subject),
                         CHAR_LENGTH(subject), maxMatchLen);
    -- Set the pattern to use to match an entire string rather than part of a string
    SET usePattern = IF (LEFT(pattern, 1) = '^', pattern, CONCAT('^', pattern));
    SET usePattern = IF (RIGHT(pattern, 1) = '$', usePattern, CONCAT(usePattern, '$'));
    -- Set start position to 1 if pattern starts with ^ or doesn't end with $.
    IF LEFT(pattern, 1) = '^' OR RIGHT(pattern, 1) <> '$' THEN
      SET startPos = 1, startInc = 1;
    -- Otherwise (i.e. pattern ends with $ but doesn't start with ^): Set start pos
    -- to the min or max match length from the end (depending on "greedy" flag).
    ELSEIF greedy THEN
      SET startPos = CHAR_LENGTH(subject) - maxMatchLen + 1, startInc = 1;
    ELSE
      SET startPos = CHAR_LENGTH(subject) - minMatchLen + 1, startInc = -1;
    END IF;
    WHILE startPos >= 1 AND startPos <= CHAR_LENGTH(subject)
      AND startPos + minMatchLen - 1 <= CHAR_LENGTH(subject)
      AND !(LEFT(pattern, 1) = '^' AND startPos <> 1)
      AND !(RIGHT(pattern, 1) = '$'
            AND startPos + maxMatchLen - 1 < CHAR_LENGTH(subject)) DO
      -- Set start length to maximum if matching greedily or pattern ends with $.
      -- Otherwise set starting length to the minimum match length.
      IF greedy OR RIGHT(pattern, 1) = '$' THEN
        SET len = LEAST(CHAR_LENGTH(subject) - startPos + 1, maxMatchLen), lenInc = -1;
      ELSE
        SET len = minMatchLen, lenInc = 1;
      END IF;
      SET prevStartPos = startPos;
      lenLoop: WHILE len >= 1 AND len <= maxMatchLen
                 AND startPos + len - 1 <= CHAR_LENGTH(subject)
                 AND !(RIGHT(pattern, 1) = '$' 
                       AND startPos + len - 1 <> CHAR_LENGTH(subject)) DO
        SET subStr = SUBSTRING(subject, startPos, len);
        IF subStr REGEXP usePattern THEN
          SET result = IF(startInc = 1,
                          CONCAT(result, replacement), CONCAT(replacement, result));
          SET startPos = startPos + startInc * len;
          LEAVE lenLoop;
        END IF;
        SET len = len + lenInc;
      END WHILE;
      IF (startPos = prevStartPos) THEN
        SET result = IF(startInc = 1, CONCAT(result, SUBSTRING(subject, startPos, 1)),
                        CONCAT(SUBSTRING(subject, startPos, 1), result));
        SET startPos = startPos + startInc;
      END IF;
    END WHILE;
    IF startInc = 1 AND startPos <= CHAR_LENGTH(subject) THEN
      SET result = CONCAT(result, RIGHT(subject, CHAR_LENGTH(subject) + 1 - startPos));
    ELSEIF startInc = -1 AND startPos >= 1 THEN
      SET result = CONCAT(LEFT(subject, startPos), result);
    END IF;
  ELSE
    SET result = subject;
  END IF;
  RETURN result;
END//
DELIMITER ;

Demo

Rextester演示

限制

This method is of course going to take a while when the subject string is large. Update: Have now added minimum and maximum match length parameters for improved efficiency when these are known (zero = unknown/unlimited). It won't allow substitution of backreferences (e.g. \1, \2 etc.) to replace capturing groups. If this functionality is needed, please see this answer which attempts to provide a workaround by updating the function to allow a secondary find and replace within each found match (at the expense of increased complexity). If ^and/or $ is used in the pattern, they must be at the very start and very end respectively - e.g. patterns such as (^start|end$) are not supported. There is a "greedy" flag to specify whether the overall matching should be greedy or non-greedy. Combining greedy and lazy matching within a single regular expression (e.g. a.*?b.*) is not supported.

用法示例

该函数用于回答以下StackOverflow问题:

如何计数单词在MySQL /正则表达式 替代者? 如何提取第n个词和计数词出现在一个MySQL 字符串? 如何从文本字段中提取两个连续的数字 MySQL吗? 如何从一个字符串中删除所有非字母数字字符 MySQL吗? 如何替换MySQL中一个特定字符的每一个其他实例 字符串? 如何从MySQL表中的多个列中获得指定的最小长度的所有不同的单词?