我有一个服务工作者,目的是将一个名为offline.html
的页面缓存起来,如果客户端没有网络连接,则显示该页面。然而,有时它会错误地认为导航器处于离线状态,即navigator.onLine === false
,即使它实际上并不是离线状态。这意味着用户可能会在在线状态下得到offline.html
而不是实际内容,这显然是我想避免的。
这是我在main.js
中注册服务工作者的方式:
// Install service worker for offline use and caching
if ('serviceWorker' in navigator) {
navigator.serviceWorker.register('/service-worker.js', {scope: '/'});
}
我目前的service-worker.js
文件:
const OFFLINE_URL = '/mysite/offline';
const CACHE_NAME = 'mysite-static-v1';
self.addEventListener('install', (event) => {
event.waitUntil(
// Cache the offline page when installing the service worker
fetch(OFFLINE_URL, { credentials: 'include' }).then(response =>
caches.open(CACHE_NAME).then(cache => cache.put(OFFLINE_URL, response)),
),
);
});
self.addEventListener('fetch', (event) => {
const requestURL = new URL(event.request.url);
if (requestURL.origin === location.origin) {
// Load static assets from cache if network is down
if (/\.(css|js|woff|woff2|ttf|eot|svg)$/.test(requestURL.pathname)) {
event.respondWith(
caches.open(CACHE_NAME).then(cache =>
caches.match(event.request).then((result) => {
if (navigator.onLine === false) {
// We are offline so return the cached version immediately, null or not.
return result;
}
// We are online so let's run the request to make sure our content
// is up-to-date.
return fetch(event.request).then((response) => {
// Save the result to cache for later use.
cache.put(event.request, response.clone());
return response;
});
}),
),
);
return;
}
}
if (event.request.mode === 'navigate' && navigator.onLine === false) {
// Uh-oh, we navigated to a page while offline. Let's show our default page.
event.respondWith(caches.match(OFFLINE_URL));
return;
}
// Passthrough for everything else
event.respondWith(fetch(event.request));
});
我做错了什么?