Posts by Felipe Soupinski • 1 point

• 0

  votes

  3

  answers

  852

  views

  A: Multiple parameters in xmlhttp.send();

  I was having difficulties with mine too, so I put the suggested code: xmlreq.open("POST","inserir.php",true); xmlreq.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); var dados…

  javascript ajax form

  answered 6 years, 1 month ago Felipe Soupinski 1