为什么我的jquery ajax表单在第一次提交时提交一次，在第二次提交时提交两次......？

I have a simple AJAX form that works as it should when I submit it. However, if I then enter new data into the same form (without refreshing the page), then it submits the form twice. If I do this a third time, then it submits the form three time, and so on. Why is it doing this? Here is my code:

我有一个简单的AJAX表单，它在我提交时可以正常工作。但是，如果我然后将新数据输入到相同的表单中（不刷新页面），则它会提交表单两次。如果我第三次这样做，那么它会三次提交表单，依此类推。它为什么这样做？这是我的代码：

$(document).ready(function(){
    $("#myForm").submit(function() {
        var myField = $('#myID).val();
        $.ajax({
            type: "POST",
            url: 'myFile.php',
            dataType: 'html',
            data: {myData:myField},
            success: function(data){
                alert(data);
            }
        });

        return false;
    });
});

3 个解决方案

#1

Even I got stuck with the same problem while developing a AJAX login form. After an googling for couple of hours I found a solution. I hope this should help you out.

甚至在开发AJAX登录表单时我也遇到了同样的问题。经过几个小时的谷歌搜索后，我找到了解决方案。我希望这能帮到你。

Basically you have to unbind the current instance of form's submit action after your ajax request gets completed. So add this line before return false;

基本上，在ajax请求完成后，您必须取消绑定表单提交操作的当前实例。所以在返回false之前添加这一行;

$("#myform").unbind('submit');

Therefore your entire code should look like

因此，您的整个代码应该是这样的



    $(document).ready(function(){
        $("#myForm").submit(function() {
            var myField = $('#myID).val();
            $.ajax({
                type: "POST",
                url: 'myFile.php',
                dataType: 'html',
                data: {myData:myField},
                success: function(data){
                    alert(data);
                }
            });

            /* Add this line */
            $("#myform").unbind('submit');
            return false;
        });
    });

#2

I had the same problem.

我有同样的问题。

I found the solution in: jquery : submitting a form twice

我找到了解决方案：jquery：提交表单两次

Basicaly is adding .submit() to Saurabh Mishra answer:

Basicaly正在向Saurabh Mishra添加.submit（）回答：

$('#form1').unbind('submit').submit();

#3

Unbinding the form can break you Ajax from. I found that as long as I declare the form first.

解除绑定表单可能会破坏您的Ajax。我发现只要我先申报表格。

$('#my_form').ajaxForm

that all works as expected, and follows the standard protocol. I just spent a day chasing the 'missing template' error on the form submission because I unbound the form.

这一切都按预期工作，并遵循标准协议。我只花了一天时间在表单提交上追逐“缺少模板”错误，因为我解除了表单的限制。

#1