摘要: select (@i:=@i+1) i,a.* from ads_building_street a,(select @i:=0) it; 结果: set @num =0 ; SELECT @num:=@num+1 as id,a.name FROM ads_building_street a; 阅读全文
posted @ 2020-03-26 17:19 代码伊甸园 阅读(548) 评论(0) 推荐(0) 编辑