jQuery的形式没有提交

这是我写过的提交表单后验证并接收PHP文件中的值使用$ _POST的脚本，但我没有获得值在PHP文件。当我尝试回声一个尊重其显示空白php.Please指引，我是新来的jQueryjQuery的形式没有提交

<script> 
$("#changepassform").validate({ 


    rules: { 
    old_password: "required", 
    password: "required", 
    password2: { 
     equalTo: "#password" 
    }, 
    }, 
    messages: { 

    old_password: "Please enter old password", 

    password: "Please enter new password", 
    password2: " Enter Confirm Password Same as Password" 

    }, 
    submitHandler: function(form) { 

     var current_password = $("#current_password").val(); 
     var new_password = $("#password").val(); 
     var comfirm_password = $("#password2").val(); 
     var id = $("#id").val(); 
     var dataString = 'newpassword1=' + new_password + '&id1=' + id; 

     $.ajax({ 
     type: "POST", 
     url: "changepassword.php", 
     data: "dataString", 
     success: function(response) { 

      $("#status").html(response); 
     } 
     }); 

    } 
    //form.submit(); 

    //return false; 

}); 
// required to block normal submit since you used ajax 
//form.submit(); 
</script>

来源

2016-12-26 Hemant

更改数据：“dataString”为data：dataString， –

而不是data：“dataString”，update为dataString，并尝试解析表单。 –

谢谢我的朋友prakash，但什么是不引用dataString后面的原因...将不胜感激 – Hemant

错误在发送数据..

<script> 
    $("#changepassform").validate({ 


     rules: { 
     old_password: "required", 
     password: "required", 
     password2: { 
      equalTo: "#password" 
     }, 
     }, 
     messages: { 

     old_password: "Please enter old password", 

     password: "Please enter new password", 
     password2: " Enter Confirm Password Same as Password" 

     }, 
     submitHandler: function(form) { 

      var current_password = $("#current_password").val(); 
      var new_password = $("#password").val(); 
      var comfirm_password = $("#password2").val(); 
      var id = $("#id").val(); 
      var dataString = 'newpassword1=' + new_password + '&id1=' + id; 

      $.ajax({ 
      type: "POST", 
      url: "changepassword.php", 
      data: dataString, 
      success: function(response) { 

       $("#status").html(response); 
      } 
      }); 

     } 
     //form.submit(); 

     //return false; 

    }); 
    // required to block normal submit since you used ajax 
    //form.submit(); 
    </script>

使用serialize() method.The序列化值可以是在进行AJAX请求时在URL查询字符串中使用。

来源

2016-12-26 05:25:16

jQuery的形式没有提交

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