您的问题是您正在检查args中的
submit键
POST。您可以通过发送
data:{submit:true}或通过删除if语句并仅处理POST请求来伪造它$('.somebutton').click(function() { $.ajax({ url: 'controller/addBookmark', type: 'POST', data: {'submit':true}, // An object with the key 'submit' and value 'true; success: function (result) { alert("Your bookmark has been saved"); } });});
