declare @Userinfo varchar(50)
set @Userinfo='admin@123'
怎么将admin截取出来
很急拜托

解决方案 »

  1.   

    select
       left(@userinfo,charindex('@',(@userinfo)-1)
      

  2.   

    declare @Userinfo varchar(50)
    set @Userinfo='admin@123'select
       left(@userinfo,charindex('@',@userinfo)-1)
         --------------------------------------------------
         /*
    admin(1 行受影响)*/
      

  3.   

    declare @Userinfo varchar(50)
    set @Userinfo='admin@123'select
       PARSENAME(replace(@Userinfo,'@','.'),2)
       
       /*--------------------------------------------------------------------------------------------------------------------------------
    admin(1 行受影响)*/