我正在使用bootstrap显示向导,如example所示。当我单击“提交”按钮时,数据不会发布到给定的URL /wizard_submit
。帮助
注意:有关完整代码,请访问给出的示例。
<div class="wizard" id="wizard-demo">
<div class="wizard-card" data-onValidated="setServerName" data-cardname="name">
<h1>First</h1>
<div class="control-group">
<input id ="name" type="text"
/>
</div>
</div>
<div class="wizard-card" data-onload="" data-cardname="location">
<h3>LAST</h3>
<div class="wizard-input-section">
<select data-placeholder="Monitor nodes" style="width:350px;" class="chzn-select">
<option value=""></option>
<option>Male</option>
<option>Female</option>
</select>
</div>
</div>
<div class="wizard-success">
<div class="alert alert-success">
Success!!!!
</div>
<a class="btn im-done">Close</a>
</div>
</div>
的Javascript
$(function() {
$.fn.wizard.logging = true;
var wizard = $("#wizard-demo").wizard({
showCancel : true
});
wizard.on("submit", function(wizard) {
$.ajax({
url : "/wizard_submit",
type : "POST",
data : wizard.serialize(),
success : function() {
wizard.submitSuccess();
wizard.hideButtons();
wizard.updateProgressBar(0);
},
error : function() {
wizard.submitError();
wizard.hideButtons();
}
});
});
$(".chzn-select").chosen();
wizard.el.find(".wizard-ns-select").change(function() {
wizard.el.find(".wizard-ns-detail").show();
});
wizard.el.find(".create-server-service-list").change(function() {
var noOption = $(this).find("option:selected").length == 0;
wizard.getCard(this).toggleAlert(null, noOption);
});
wizard.on("submit", function(wizard) {
var submit = {
"hostnamed" : $("#new-server-fqdn").val()
};
setTimeout(function() {
wizard.trigger("success");
wizard.hideButtons();
wizard._submitting = false;
wizard.showSubmitCard("success");
wizard.updateProgressBar(0);
}, 2000);
});
wizard.el.find(".wizard-success .im-done").click(function() {
wizard.reset().close();
});
wizard.el.find(".wizard-success .create-another-server").click(function() {
wizard.reset();
});
$(".wizard-group-list").click(function() {
alert("Disabled for demo.");
});
$("#open-wizard").click(function() {
wizard.show();
});
wizard.show();
});
答案 0 :(得分:0)
删除此行
var submit = {
"hostnamed" : $("#new-server-fqdn").val()
};
它应该可以工作(我遇到了同样的问题并删除了这条线。)
无论如何,该行是不必要的,看起来你在代码中有相同的代码,我猜你是否有一个带有该ID的元素,它会导致无关的错误。