如何使用机器人框架接受警报(How to accept an alert using robot framework)

 在你的javascript中，你正在劫持表单以便它通过ajax提交，但是你在表单上调用replaceWith，所以你被劫持的表单被删除并被一个新的，未被劫持的表单替换。 要解决这个问题，你也可以 
 
 1）只替换表单的内容 - 这应该有效，因为您只是将事件附加到表单本身而不是其子元素 
 
 2）将您的js编写为一个函数，您可以先在初始表单上调用，然后在通过ajax加载的任何新表单上调用。 
 
 更新：例如， 
 
<script type="text/javascript"> 
$(document).ready(function(){

    function hijack() {
        var form = $('form#createrecipeform');
        form.submit(function(e) {
            e.preventDefault();
            console.log('ajax form submission function called successfully.');
            //form = $(this);
            console.log(form)
            var serialized_form = form.serialize();
            $.ajax({ type: "POST", 
                url: $(this).attr('action'),
                data: serialized_form, 
                success: (function(data) { 
                    console.log('ajax success function called successfully.');
                    data = $.parseJSON(data);
                    if (data.success) {
                        console.log('success');
                    } else {        
                        console.log('failure');
                        var newForm = data.form;
                        form.replaceWith(newForm);
                        hijack();
                    }
                })
            });
            return false;
        });
    };

    hijack();

});
</script> 

In your javascript, you're hijacking the form so that it submits via ajax, but then you're calling replaceWith on the form, so your hijacked form gets obliterated and is replaced with a new, non-hijacked form. To solve this you can either
 
1) Only replace the content of the form - this should work since you're only attaching events to the form itself and not its child elements
 
2) Write your js as a function which you can call firstly on the initial form, and subsequently on any new forms loaded via ajax.
 
UPDATE: for example,
 
<script type="text/javascript"> 
$(document).ready(function(){

    function hijack() {
        var form = $('form#createrecipeform');
        form.submit(function(e) {
            e.preventDefault();
            console.log('ajax form submission function called successfully.');
            //form = $(this);
            console.log(form)
            var serialized_form = form.serialize();
            $.ajax({ type: "POST", 
                url: $(this).attr('action'),
                data: serialized_form, 
                success: (function(data) { 
                    console.log('ajax success function called successfully.');
                    data = $.parseJSON(data);
                    if (data.success) {
                        console.log('success');
                    } else {        
                        console.log('failure');
                        var newForm = data.form;
                        form.replaceWith(newForm);
                        hijack();
                    }
                })
            });
            return false;
        });
    };

    hijack();

});
</script>