Question

<head runat="server">
    <script type="text/javascript">

  $.fn.serializeObject = function () {
      var o = {};
      var a = this.serializeArray();
      $.each(a, function () {
          if (o[this.name] !== undefined) {
              if (!o[this.name].push) {
                  o[this.name] = [o[this.name]];
              }
              o[this.name].push(this.value || '');
          } else {
              o[this.name] = this.value || '';
          }
      });
      return o;
  };

  // we will add our javascript code here
         $(document).ready(function () {
             $("#myForm").submit(function () {


                  var str = $("#myForm").serialize();
                         var str = JSON.stringify($('#myForm').serializeObject());
                         alert(str);
                         $.ajax({
                             type: 'POST',
                             contentType: 'application/json; charset=utf-16',
                             data: str,,
                             dataType: 'json',
                             url: "testAjaxPost.aspx/SaveCustomerEmail",

                             success: function (msg) {
                                 alert('done');
                                 return true;

                             },
                             error: function (msg) { alert(JSON.stringify(msg)) }

                         });

                         return form.submit();
                 return false;
             });
         });

我希望将表单数据发布到两个页面，为此我使用表单操作方法和jQuery表单提交方法。但只有jQuery表单提交事件有效，但表单方法发布事件不起作用。我怎么能这样做？

  function ValidateForm() {

      var emailID = $(".email").val();


      if (!validate_email_address(emailID)) {
          alert("Please provide a valid email address.");
          $(".email").focus();
          return false;
      }


  }

  function validate_email_address(email) {
      var regex = /^([a-zA-Z0-9_\.\-\+])+\@(([a-zA-Z0-9\-])+\.)+([a-zA-Z0-9]{2,4})+$/;
      return regex.test(email);
  }
    </script>
</head>
<body>
        <form id="myForm" action="page1.aspx" method="post" onsubmit="return ValidateForm()">
        <label>
            Email*</label>
        <input type="text" name="Email" class="email" />
        <input type="submit" title="Save" />
        </form>
    </body>

Answer 1

function ValidateForm() {     

    //Data Post to FirstForm
    var str = $("#FirstForm").serialize();

    $.ajax({
        type: "POST",
        dataType: "jsonp",
        contentType: "application/json; charset=utf-8",
        url: "https://xyz.com",
        data: str,
        success: function (msg) {console.log("Success data post ---->" + JSON.stringify(msg)) },
        error: function (msg) { console.log("Error data post ---->" + JSON.stringify(msg)) }

    });
    //End Data Post to First Form

    //Call Click event of Second form .Use Form post and action method .

       //Copy First Form text box  value into second Form text box 
      $("#btnSecondForm").trigger('click'); 
   //End Click event


  return false;
}



//Form in Body

<form  id="FirstForm" onsubmit="return ValidateForm(this)">

 //text box HTML 

</form>


 <form id="SecondForm" style='display:none' action="https://abc.com" method="post">

 <input type="submit" value="Submit" id='btnSecondForm' />

 </form>

如何使用jQuery和表单提交将表单数据发布到2个不同的URL

1 个答案: