Question

function voip_submit()
{
  var link = encodeURIComponent('/goform/az_ifx_set_voip_sip_linebasicaccount'); 
  var loc = 'page=quickstart.asp';
  loc += '&status='+document.az_sip.status.value;
  loc += '&cvflag='+document.az_sip.cvflag.value;
  loc += '&vlstatus1='+document.az_sip.vlstatus1.value;
  loc += '&vlstatus2='+document.az_sip.vlstatus2.value;
  var retVal =  new Ajax.Request(link, {
      method: 'post',
      parameters: loc,
      encoding: 'ISO-8859-1',
      onSuccess: function(transport) {
        submit_voip1();
      },
      onFailure: function(transport) {
            Tiny.hide();
            showAlert("Apply settings failed");
      }       
  });
}

Answer 1

首先，不需要在retVal中存储新的Ajax.Request对象。 onSuccess和onFailure回调是实际处理请求“返回”的数据（或缺少）的回调。

其次，您应该使用内置的Prototype方法来编码数据

function voip_submit()
{
    new Ajax.Request('/goform/az_ifx_set_voip_sip_linebasicaccount', {
        method: 'post',
        parameters: {
            page:      'quickstart.asp',
            status:    document.az_sip.status.value,
            cvflag:    document.az_sip.cvflag.value,
            vlstatus1: document.az_sip.vlstatus1.value,
            vlstatus2: document.az_sip.vlstatus2.value
        },
        encoding: 'ISO-8859-1',
        onSuccess: function(transport) {
            submit_voip1();
        },
        onFailure: function(transport) {
            Tiny.hide();
            showAlert("Apply settings failed");
        }       
    });
}

Ajax.Request总是在IE上返回失败

1 个答案: