javascript - How to find device type whether is it a computer or mobile in vb.net with SQL Server -

i have project in need know device type (computer or mobile). have searched everywhere cannot find answer.

i'm using below function check mobile browser or not? otherwise it's computer browser.

public shared function ismobilebrowser() boolean     'gets current user context     dim context httpcontext = httpcontext.current      'first try built in asp.nt check     if context.request.browser.ismobiledevice         return true     end if     'then try checking http_x_wap_profile header     if context.request.servervariables("http_x_wap_profile") isnot nothing         return true     end if     'then try checking http_accept exists , contains wap     if context.request.servervariables("http_accept") isnot nothing andalso context.request.servervariables("http_accept").tolower().contains("wap")         return true     end if     'and check http_user_agent      'header variable 1 of following     if context.request.servervariables("http_user_agent") isnot nothing         'create list of mobile types         dim mobiles string() = {"midp", "j2me", "avant", "docomo", "novarra", "palmos", _         "palmsource", "240x320", "opwv", "chtml", "pda", "windows ce", _         "mmp/", "blackberry", "mib/", "symbian", "wireless", "nokia", _         "hand", "mobi", "phone", "cdm", "up.b", "audio", _         "sie-", "sec-", "samsung", "htc", "mot-", "mitsu", _         "sagem", "sony", "alcatel", "lg", "eric", "vx", _         "philips", "mmm", "xx", "panasonic", "sharp", "wap", _         "sch", "rover", "pocket", "benq", "java", "pt", _         "pg", "vox", "amoi", "bird", "compal", "kg", _         "voda", "sany", "kdd", "dbt", "sendo", "sgh", _         "gradi", "jb", "dddi", "moto", "iphone"}         ' "nec",          'loop through each item in list created above          'and check if header contains text         each s string in mobiles             if context.request.servervariables("http_user_agent").tolower().contains(s.tolower())                 return true             end if         next     end if      return false end function