Service Worker 返回 javascript 文件的脱机 html 页面

Question

我是服务工作者和离线功能的新手。我创建了一个简单的 service worker 来处理网络请求并在离线时返回一个离线 html 页面。这是根据 Google 的 PWA 指南创建的。

问题是服务工作者在请求javascript文件（未缓存）时返回offline.html。相反，它应该返回网络错误或其他东西。代码如下：

const cacheName = 'offline-v1900'; //increment version to update cache
// cache these files needed for offline use
const appShellFiles = [
    './offline.html',
    './css/bootstrap.min.css',
    './img/logo/logo.png',
    './js/jquery-3.5.1.min.js',
    './js/bootstrap.min.js',
];

self.addEventListener("fetch", (e) => {
  // We only want to call e.respondWith() if this is a navigation request
  // for an HTML page.
  // console.log(e.request.url);
  e.respondWith(
    (async () => {
      try {
        // First, try to use the navigation preload response if it's supported.
        const preloadResponse = await e.preloadResponse;
        if (preloadResponse) {
            // console.log('returning preload response');
          return preloadResponse;
        }

        const cachedResponse = await caches.match(e.request);
            if (cachedResponse) {
                // console.log(`[Service Worker] Fetching cached resource: ${e.request.url}`);
                return cachedResponse;
            }

        // Always try the network first.
        const networkResponse = await fetch(e.request);
        return networkResponse;
      } catch (error) {
        // catch is only triggered if an exception is thrown, which is likely
        // due to a network error.
        // If fetch() returns a valid HTTP response with a response code in
        // the 4xx or 5xx range, the catch() will NOT be called.

        // console.log("Fetch failed; returning offline page instead.", error);
        const cachedResponse = await caches.match('offline.html');
        return cachedResponse;
      }
    })()
  );

离线时，我在我的网站上打开一个 url，它会从缓存中加载页面，但并非所有资产都被缓存以供离线使用。因此，当发出网络请求时，比如说https://www.gstatic.com/firebasejs/9.1.3/firebase-app.js，我得到的响应是offline.html 页面的html。由于 javascript 错误，这会破坏页面。

它应该返回一个网络错误或其他东西。

Answer 1

我认为相关的示例代码来自https://googlechrome.github.io/samples/service-worker/custom-offline-page/

self.addEventListener('fetch', (event) => {
  // We only want to call event.respondWith() if this is a navigation request
  // for an HTML page.
  if (event.request.mode === 'navigate') {
    event.respondWith((async () => {
      try {
        // First, try to use the navigation preload response if it's supported.
        const preloadResponse = await event.preloadResponse;
        if (preloadResponse) {
          return preloadResponse;
        }

        const networkResponse = await fetch(event.request);
        return networkResponse;
      } catch (error) {
        // catch is only triggered if an exception is thrown, which is likely
        // due to a network error.
        // If fetch() returns a valid HTTP response with a response code in
        // the 4xx or 5xx range, the catch() will NOT be called.
        console.log('Fetch failed; returning offline page instead.', error);

        const cache = await caches.open(CACHE_NAME);
        const cachedResponse = await cache.match(OFFLINE_URL);
        return cachedResponse;
      }
    })());
  }

  // If our if() condition is false, then this fetch handler won't intercept the
  // request. If there are any other fetch handlers registered, they will get a
  // chance to call event.respondWith(). If no fetch handlers call
  // event.respondWith(), the request will be handled by the browser as if there
  // were no service worker involvement.
});

具体来说，fetch 处理程序会检查是否event.request.mode === 'navigate'，如果是这种情况，则仅在脱机时返回 HTML。这就是确保您最终不会为其他类型的资源返回离线 HTML 所必需的。