未捕获的SyntaxError：使用Submit Handler的意外标识符

Question

我正在创建验证脚本和ajax调用。我面临的问题是

  

警报消息在条件中不起作用

我不知道发生了什么，当我运行脚本时，在Chrome中显示：

Uncaught SyntaxError: Unexpected identifier

下面是我的代码：

  $("#myCA-Form").validate({

    rules:{
    field1:"required",
    city:"required",
    email:{
    required:true,
    email:true
    },
    phone:{
    required:true,
    number: true,
    minlength:9,
    maxlength:10
    }
    },

    messages:{
    name:"Please enter your username..!",
    email:"Please enter your email..!",
    phone:"Enter your mobile no"
    },


    submitHandler: function(form) {


    var formData =  $(form).serialize();
    var id = $('#product_id2').val();

    var name = $('#field1').val();
    $.ajax({
    url: "https://eaxmple.co.in/test.aspx?"+formData,
    type: "POST",
    cache: false,
    dataType: "jsonp",
    crossDomain: true,  
    data: {formd : formData },
    success: function(data) {
     }


    });
     if(id=='100105')
    {
    alert("sdfsd");
    document.getElementById("msg").innerHTML='submit Successfully!!!';
     }

    }

  });

Answer 1

您的代码没有问题。它的工作完美。我希望问题出在其他地方

这是代码的精确副本，带有添加的HTML代码

$("#myCA-Form").validate({

    rules:{
    field1:"required",
    city:"required",
    email:{
    required:true,
    email:true
    },
    phone:{
    required:true,
    number: true,
    minlength:9,
    maxlength:10
    }
    },

    messages:{
    name:"Please enter your username..!",
    email:"Please enter your email..!",
    phone:"Enter your mobile no"
    },


    submitHandler: function(form) {


    var formData =  $(form).serialize();
    var id = $('#product_id2').val();

    var name = $('#field1').val();
    $.ajax({
    url: "https://eaxmple.co.in/test.aspx?"+formData,
    type: "POST",
    cache: false,
    dataType: "jsonp",
    crossDomain: true,  
    data: {formd : formData },
    success: function(data) {
     }


    });
     if(id=='11')
    {
    alert("sdfsd");
    document.getElementById("msg").innerHTML='submit Successfully!!!';
     }

    }

  });

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery-validate/1.17.0/jquery.validate.min.js"></script>
<form id="myCA-Form">
	<input name="field1" type="text"/>
	<input name="city" type="text"/>
	<input name="email" type="text"/>
	<input name="phone" type="text"/>
	<input value="11" id="product_id2" type="hidden"/>
<button class="save-post">Save Post</button>
</form>

Answer 2

您不应将表单数据附加到URL，如果您想要类似的东西，则应查看GET方法而不是POST。您可能想了解有关POST和GET的更多信息。