select substring(reverse(‘0->星光‘),PATINDEX(‘%[0-9]%‘,reverse(‘0->星光‘)),1)
原文地址:https://www.cnblogs.com/guangzhou11/p/9093066.html
时间: 2024-10-07 22:15:43
select substring(reverse(‘0->星光‘),PATINDEX(‘%[0-9]%‘,reverse(‘0->星光‘)),1)
原文地址:https://www.cnblogs.com/guangzhou11/p/9093066.html