【发布时间】:2024-01-24 12:32:01
【问题描述】:
Update table
set col = replace(col, 'value1','') where col like '%value1%'
or set col = replace(col, 'value2','') where col like '%value2%'
or set col = replace(col, 'value3','') where col like'%value3%';
【问题讨论】:
标签: mysql sql sql-update where-clause sql-like