if( Session["id"] == null )

解决方案 »

  1.   

    dim id,fl
    id=session("id")
    fl=session("fl_sf")
    application.Lock
    if fl="1" or fl="2" then
    dim i,newstr,xx

    newstr=""
    if fl="1" then
    xx=split(application("online_user_danwei"),"#")
    elseif session("fl_sf")="2" then
    xx=split(application("online_user_geren"),"#")
    end if
    num=ubound(xx)
    for i=0 to num if cstr(id)<>cstr(trim(xx(i))) then
    newstr=newstr&xx(i)
    if i<>num and num<>1 then
    newstr=newstr&"#"
    end if
    end if
    next
    if fl="1" then
    application("online_user_danwei")=newstr
    elseif fl="2" then
    application("online_user_geren")=newstr
    end if
    end ifif( Session["id"] == null ) 她是不应该执行for的!!