select       *       from       tb1       
where       fa       like       '%a%'   
or   fa   like'%c%'
or  fa   like'%2%'

------解决方案--------------------
上面理解错了
select  *  from  tb1     
  where patindex('%a%',fa)>0
and patindex('%b%',fa)>0
patindex('%c%',fa)>0


------解决方案--------------------
SQL code
declare @tb table (id int,fa varchar(10))
insert into @tb select 1,'abc123'
insert into @tb select 2,'ac1234'
insert into @tb select 3,'bc2345'

select * from @tb 
where 
charindex('a',fa)>0 
and charindex('c',fa)>0 
and charindex('2',fa)>0

------解决方案--------------------
以前遇到过这个问题,是用函数实现的,现在不记得怎么写了.

------解决方案--------------------
id         fa  
1           abc123  
2           ac1234  
3           bc2345  

id 为 3 是不是你想要查的,如果是 用 or 不是 用 and


------解决方案--------------------
C# code

        string whereStr = "select * from tb1 where 1=1 ";
        string tempStr = "ac2";
        foreach (char c in tempStr)
        {
            whereStr += " and charindex('" + c + "',fa)> 0 ";
        }

------解决方案--------------------
可以试试启用sqlserver的全文索引

sp_fulltext_database   'enable'  

execute   sp_fulltext_catalog   'ft_test',   'create'  

execute   sp_fulltext_table   'tb1','create',   'ft_test','PK_tb1'  

execute   sp_fulltext_column   'tb1',   'fa', 'add'  

execute   sp_fulltext_catalog   'ft_test',   'start_full'  

select  * from tb1
where   freetext(*,'keyword1 keyword2')

------解决方案--------------------
SQL code
create table tb (id int,fa varchar(10))
insert into tb select 1,'abc123'
insert into tb select 2,'ac1234'
insert into tb select 3,'bc2345'

declare @s varchar(50),@where varchar(8000)
select @s='ac2',@where ='where 1=1'
while(len(@s)>0)
begin
set @where =@where+' and charindex('''+left(@s,1)+''',fa)>0 '
set @s=right(@s,len(@s)-1)
end
exec('select * from tb '+@where)

------解决方案--------------------
C# code

public string SQLStr(string searchCondition)
{
  StringBuilder sql=new StringBuilder("select * form tb1 where 1=1 ");
  for(int i=0;i<searchCondition;i++)
  {
    sql.Append(" and fa like '%"+searchCondition[i]+"%'");
  }
  return sql.ToString();
}

------解决方案--------------------