替换字符串的问题,求高手指教 将strOld中的数字1替换成AstrOld=1,2,1,13,112,11,1替换后的结果:strNew=A,2,A,13,112,11,A请问:①在C#中用Regex.Replace该如何实现?②在SQL中该如何实现? 解决方案 » 免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货 SQL:UPDATE [News] SET [KeyWord]=REPLACE(]KeyWord],'1','A') (1).C#里用Linq string strOld = "1,2,1,13,112,11,1"; string strNew = strOld.Split(',').Select(x => x == "1" ? "A" : x).Aggregate((x, y) => y = x + "," + y);(2).SQL里用ReplacestrOld=Replace(','+strOld+',' , ',A,')strNew=RIGHT(LEFT(strOld,LEN(strOld)-1),LEN(strOld)-2) C# strOld="1,2,1,13,112,11,1";strOld=Regex.Replace(strOld,@"(?<=(,|\s|^))1(?=,|$","A"); string strOld = "1,2,1,13,112,11,1"; Response.Write( ("," + strOld + ",").Replace(",1,", ",A,").TrimStart(',').TrimEnd(','));给分吧 strOld=Replace(','+strOld+',' , ',1,1', ',A,') --上面少些了一个参数strNew=RIGHT(LEFT(strOld,LEN(strOld)-1),LEN(strOld)-2) SQL的少写了个参数strOld=Replace(','+strOld+',' , ',1,', ',A,') --上面少些了一个参数 string strOld = "1,2,1,13,112,11,1"; string strNew = Regex.Replace(strOld, @"(?<=,|^)1(?=,|$)", "A");declare @str varchar(100)set @str='1,2,1,13,112,11,1'select left(stuff(replace(','+@str+',',',1,',',A,'),1,1,''),len(stuff(replace(','+@str+',',',1,',',A,'),1,1,''))-1) 并非所有的代码路径都返回值 这个URL重写,挺简单的,不过放到IIS就不行了!有没有不用修改IIS的? 怪事多呀!form里面用上table,表单就跑到别的contentPlaceHoulder上去了 手动添加删除按钮如何添加确定! 如何在javascript中绑定数据中的表 javascript 如何调用SERVER端的脚本? 顶者有分!!!!!!! 如何得到URL里面的全部信息? 用客户端脚本响应asp.net文本筐控件的问题 System.NotSupportedException: 不支持给定路径的格式。 hyperlink和button的问题,急
SQL:
UPDATE [News] SET [KeyWord]=REPLACE(]KeyWord],'1','A')
string strOld = "1,2,1,13,112,11,1";
string strNew = strOld.Split(',').Select(x => x == "1" ? "A" : x).Aggregate((x, y) => y = x + "," + y);(2).SQL里用Replace
strOld=Replace(','+strOld+',' , ',A,')
strNew=RIGHT(LEFT(strOld,LEN(strOld)-1),LEN(strOld)-2)
strOld="1,2,1,13,112,11,1";
strOld=Regex.Replace(strOld,@"(?<=(,|\s|^))1(?=,|$","A");
给分吧
strNew=RIGHT(LEFT(strOld,LEN(strOld)-1),LEN(strOld)-2)
strOld=Replace(','+strOld+',' , ',1,', ',A,') --上面少些了一个参数
string strNew = Regex.Replace(strOld, @"(?<=,|^)1(?=,|$)", "A");declare @str varchar(100)
set @str='1,2,1,13,112,11,1'
select left(stuff(replace(','+@str+',',',1,',',A,'),1,1,''),len(stuff(replace(','+@str+',',',1,',',A,'),1,1,''))-1)