根据论坛llm1978的思路,取栏目图片的，方法：我写出了栏目介绍

栏目介绍语句：

select top 100 id,foldername,mid(classbasicinfo,instr(classbasicinfo,"||||")+4) as s,mid(s,1,instr(s,"||||")-1) as 栏目介绍 from ks_class where tj=1 order by folderorder

取栏目图片:

select top 100 id,foldername,mid(classbasicinfo,1,instr(classbasicinfo,"||||")-1) as 栏目图片 from ks_class where tj=1 order by folderorder

综合写就成了这样:

select top 100 id,foldername,mid(classbasicinfo,instr(classbasicinfo,"||||")+4) as s,mid(s,1,instr(s,"||||")-1) as 栏目介绍 ,mid(classbasicinfo,1,instr(classbasicinfo,"||||")-1) as 栏目图片 from ks_class where tj=1 order by folderorder

用这段代码的时候提示:'mid' 不是可以识别的 函数名。