Hello, I am constantly testing, but every time I log in again by clearing the browser and open the same page again.when I do this maybe 200 times a day until the evening, it is a big waste of time.when I update, how can I update the changed css and script files in the browser cache without refreshing the page? Even if I run it again in the console after updating an existing function, it sees the old codes. There are more than 300 script files in my project, I should not run all of the scripts because unnecessary scripts will run, so I'm trying to do something like this, the scripts or css that are currently loaded in the browser's memory should only be updated in the cache, I searched a lot but I couldn't find a solution, thanks
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Adarsh NigamPosted Jul 12, 2024, 3:22 AM
1. Cache Busting
Ensure that your browser gets the latest version of your files by appending a version number or a unique hash to the file URLs. This forces the browser to fetch the new files instead of using the cached versions.
For CSS and JavaScript Files
In your HTML or layout file, you can add a query string with a version number or a hash:
2. Using Service Workers
Service workers can intercept network requests and dynamically control the caching behavior. You can programmatically cache specific files and update them when changes are detected.
Example Service Worker
Create a
service-worker.jsfile:const CACHE_NAME = 'my-cache-v1';
const urlsToCache = [
'/css/styles.css',
'/js/scripts.js',
// Add other files you want to cache
];
self.addEventListener('install', event => {
event.waitUntil(
caches.open(CACHE_NAME)
.then(cache => cache.addAll(urlsToCache))
);
});
self.addEventListener('fetch', event => {
event.respondWith(
caches.match(event.request)
.then(response => {
// Cache hit - return response
if (response) {
return response;
}
// Clone the request for fetch and cache
const fetchRequest = event.request.clone();
return fetch(fetchRequest).then(
response => {
// Check if we received a valid response
if (!response || response.status !== 200 || response.type !== 'basic') {
return response;
}
// Clone the response for cache
const responseToCache = response.clone();
caches.open(CACHE_NAME)
.then(cache => {
cache.put(event.request, responseToCache);
});
return response;
}
);
})
);
});
Try with these 2 approaches maybe u'lll be able to avoid this situation. Thanks!!