更改AJAX调用中的DOM属性不起作用

时间:2020-05-31 00:48:11

标签: javascript ajax dom

我有一个函数可以进行AJAX调用,并根据答案设置输入的颜色。函数核心本身可以正常工作。但是,在更改输入的颜色时,它不起作用。

function Nueva_Red() {
  var e1 = document.getElementById("rrss_tipo");
  var e2 = document.getElementById("rrss_usuario");
  var e3 = document.getElementById("redes");
  var e4 = document.getElementById("alta_nueva_red");
  var id_red = e1.options[e1.selectedIndex].value;
  var red = e1.options[e1.selectedIndex].text;
  var usuario = e2.value;
  e2.style.backgroundColor = "none";
  e2.style.borderColor = "black";
  if ((id_red > 0) && (usuario != "")) {
    var datastr = 'seccion=rrss_check&dato=' + id_red + "|" + usuario;
    $.ajax({
      type: 'GET',
      url: 'validacion.php',
      data: datastr,
      success: function(html) {
        if (html == 1) {
          document.getElementById("nuevas_rrss").style.visibility = "visible";
          document.getElementById("nuevas_rrss").style.display = "inline-block";
          var opcion = document.createElement("option");
          opcion.value = id_red + "-" + usuario;
          opcion.innerHTML = red + " -> " + usuario;
          e3.appendChild(opcion);
          e1.value = 0;
          e2.value = "";
          e2.title = "";
          e2.style.backgroundColor = "none";
          e2.style.borderColor = "black";
        } else {
          e2.title = html;
          e2.style.backgroundColor = error_error;
          e2.style.borderColor = error_error;
        }
      },
      error: function(html) {
        e2.style.borderColor = error_app;
        e2.style.backgroundColor = error_app;
      }
    });
  }
}

谢谢!

0 个答案:

没有答案