declare @str nvarchar(10)declare @tablename varchar(50)declare @colname varchar(50)declare @counts intdeclare @sql nvarchar(2000)--以上定义变量declare cur1 cursor for select a.name tablename,B.name colname from sys.objects a,
把所有表的所有字段都转换成字符型的然后连起来,变成一个单一字段的数据集.然后用like查呗.
select id ,name ,len(remark)-len(replace(remark,'1','')) as remark_1 ,len(remark)-len(replace(remark,'2','')) as remark_2 ,len(remark)-len(replace(remark,'3','')) as remark_3 ,len(remark)-len(replace(remark,'x','')) as remark_xfrom nameremark
select count(uid)??as num from 表 group by uid order by num desc limit 0,50;
先给你说下思路,,,5个数字的是特定的,,,,,3,4的都涉及到排列组合declare @s varchar(200)set @s='245';WITH CTE AS(SELECT ROW_NUMBER() OVER (ORDER BY ID) AS SEQ FROM SYSOBJECTS),CTE2 AS(SELECT
likenbsp;'25,%'nbsp;ornbsp;likenbsp;'%,25,%'nbsp;ornbsp;likenbsp;'%,25'nbsp;补充:nbsp;写成存储过程nbsp;或拼接sql字符串//自己动手,丰衣足食!阿弥陀佛!!!!
以前写过类似的,参考一下:declare @str nvarchar(20) declare @tablename varchar(50) declare @colname varchar(50) declare @counts int declare @sql nvarchar(2000)--以上定义变量 declare cur1 cursor for select a.name tablename,B.name
你可以用locate和substr配合使用达到模糊匹配的效果,然后用replace替换UPDATE `aaa` SET `name`=replace(`link`,substr(`link`, locate('http://',`link`), locate( '/',`link`,locate( 'http://',`link`)+7)-locate( 'http://',`link`)), '') 补充,看错了,你要的是删除
傻方法 select left('20-45-122-25-1',charindex('-','20-45-122-25-1')-1)自己一步一步截取.也可做个函数调用
select name from 表名where id='01'结果:name 张三 李四select name from 表名where id='02'结果: name 王五 abc abc123