自从我试图完成它以来的几天,但我完全被困在这一点上。
这是我的服务工作者文件中的代码
importScripts('https://www.gstatic.com/firebasejs/6.0.2/firebase-app.js');
importScripts('https://www.gstatic.com/firebasejs/6.0.2/firebase-messaging.js');
firebase.initializeApp({
messagingSenderId: "xxxxxxxxxxxx"
});
var messaging = firebase.messaging();
messaging.setBackgroundMessageHandler(function(payload) {
console.log('[firebase-messaging-sw.js] Received background message ', payload);
// Customize notification here
var notificationTitle = payload.data.title; //or payload.notification or whatever your payload is
var notificationOptions = {
body: payload.data.body,
icon: payload.data.icon,
image: payload.data.image,
data: { url:payload.data.openURL }, //the url which we gonna use later
actions: [{action: "open_url", title: "View"}]
};
return event.waitUntil(self.registration.showNotification(notificationTitle,
notificationOptions));
});
self.addEventListener('notificationclick', function(event) {
console.log('event = ',event);
event.notification.close();
event.waitUntil(clients.openWindow(event.notification.data.url));
switch(event.action){
case 'open_url':
window.open(event.notification.data.url);
break;
case …Run Code Online (Sandbox Code Playgroud) javascript push-notification firebase service-worker firebase-cloud-messaging
我正在自定义 wordpress 主题,为什么在尝试在搜索框的输入字段中键入时出现此错误,我的 html 代码是
<div class="mysearchdiv">
<input type="" name="" class="mysearch">
<ul class="skills-list" id="skills_list"></ul>
</div>
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和backbone.js 代码是
jQuery(document).ready(function(){
var mymodel= Backbone.Model.extend({
defaults:function(){
return{
name:''
}
}
});
var mycollection=Backbone.Collection.extend({
model:mymodel
});
var mynewcollection= new mycollection();
var myview=Backbone.View.extend({
model:mynewcollection,
el:'.mysearchdiv',
events:{
'keypress .mysearch':'search'
},
initialize:function(){
console.log("init");
var view=this;
this.skill_list=this.$('ul.skill-list');
view.skills={};
this.$el.find('input.mysearch').typeahead({
minLength:0,
items:10,
source: function(query, process) {
console.log("Skill_Control typeahead source");
console.log(query);
console.log(process);
console.log(ae_globals.ajaxURL);
if (view.skills.length > 0) return view.skills;
//getting error at this point
//where ae_globals.ajaxURL is ajax url in wp …Run Code Online (Sandbox Code Playgroud)