I know my question could simply have a "This cannot be done, this defines the purpose of SPA". But...
I navigate to mydomain in my REACT web app. This page loads data from the backend and populates elaborate grids. It takes it about 2 seconds to load and render.
Now I click a link on that elaborate page and navigate to mydomain/otherPage. When I click the browser's BACK button to return to mydomain, it's blank, and has to be rebuilt from scratch as SPA dictates the DOM must be erased and re-built with every page change (at least the page-specific dynamic parts of it, as routes can be inside a fixed layout of header/footer etc). I get that...
Other than migrating to nextJS and using SSR....
Is there any magic solution in REACT to somehow 'retain' the DOM for a page when navigating out of it, so that when you browser-back into it, that page is instantly shown and not rendered from scratch?
I know my question could simply have a "This cannot be done, this defines the purpose of SPA". But...
I navigate to mydomain.com in my REACT web app. This page loads data from the backend and populates elaborate grids. It takes it about 2 seconds to load and render.
Now I click a link on that elaborate page and navigate to mydomain.com/otherPage. When I click the browser's BACK button to return to mydomain.com, it's blank, and has to be rebuilt from scratch as SPA dictates the DOM must be erased and re-built with every page change (at least the page-specific dynamic parts of it, as routes can be inside a fixed layout of header/footer etc). I get that...
Other than migrating to nextJS and using SSR....
Is there any magic solution in REACT to somehow 'retain' the DOM for a page when navigating out of it, so that when you browser-back into it, that page is instantly shown and not rendered from scratch?
Yes, it is very much possible to switch routes while keeping the DOM rendered, but hidden! If you are building a SPA, it would be a good idea to use client side routing. This makes your task easy:
For hiding, while keeping components in the DOM, use either of the following css:
.hidden { visibility: hidden } only hides the unused component/route, but still keeps its layout.
.no-display { display: none } hides the unused component/route, including its layout.
For routing, using react-router-dom, you can use the function children prop on a Route component:
children: func
Sometimes you need to render whether the path matches the location or not. In these cases, you can use the function children prop. It works exactly like render except that it gets called whether there is a match or not.The children render prop receives all the same route props as the component and render methods, except when a route fails to match the URL, then match is null. This allows you to dynamically adjust your UI based on whether or not the route matches.
Here in our case, I'm adding the hiding css classes if the route doesn't match:
App.tsx:
export default function App() {
return (
<div className="App">
<Router>
<HiddenRoutes hiddenClass="hidden" />
<HiddenRoutes hiddenClass="no-display" />
</Router>
</div>
);
}
const HiddenRoutes: FC<{ hiddenClass: string }> = ({ hiddenClass }) => {
return (
<div>
<nav>
<NavLink to="/1">to 1</NavLink>
<NavLink to="/2">to 2</NavLink>
<NavLink to="/3">to 3</NavLink>
</nav>
<ol>
<Route
path="/1"
children={({ match }) => (
<li className={!!match ? "" : hiddenClass}>item 1</li>
)}
/>
<Route
path="/2"
children={({ match }) => (
<li className={!!match ? "" : hiddenClass}>item 2</li>
)}
/>
<Route
path="/3"
children={({ match }) => (
<li className={!!match ? "" : hiddenClass}>item 3</li>
)}
/>
</ol>
</div>
);
};
styles.css:
.hidden {
visibility: hidden;
}
.no-display {
display: none;
}
Working CodeSandbox: https://codesandbox.io/s/hidden-routes-4mp6c?file=/src/App.tsx
Compare the different behaviours of visibility: hidden vs. display: none.
Note that in both cases, all of the components are still mounted to the DOM! You can verify with the inspect tool in the browser's dev-tools.
For a reusable solution, you can create a reusable HiddenRoute component.
In the following example, I use the hook useRouteMatch, similar to how the children Route prop works. Based on the match, I provide the hidden class to the new components children:
import "./styles.css";
import {
BrowserRouter as Router,
NavLink,
useRouteMatch,
RouteProps
} from "react-router-dom";
// Reusable components that keeps it's children in the DOM
const HiddenRoute = (props: RouteProps) => {
const match = useRouteMatch(props);
return <span className={match ? "" : "no-display"}>{props.children}</span>;
};
export default function App() {
return (
<div className="App">
<Router>
<nav>
<NavLink to="/1">to 1</NavLink>
<NavLink to="/2">to 2</NavLink>
<NavLink to="/3">to 3</NavLink>
</nav>
<ol>
<HiddenRoute path="/1">
<li>item 1</li>
</HiddenRoute>
<HiddenRoute path="/2">
<li>item 2</li>
</HiddenRoute>
<HiddenRoute path="/3">
<li>item 3</li>
</HiddenRoute>
</ol>
</Router>
</div>
);
}
Working CodeSandbox for the reusable solution: https://codesandbox.io/s/hidden-routes-2-3v22n?file=/src/App.tsx
You can simply put your generated elements that need intensive calculation in a state, in a component that never gets unmounted while changing page.
Here is an example with a Parent component holding 2 children and some JSX displayed after 5 seconds. When you click on the links you navigate to children, and when you click on browser's back button, you get back on the URL path. And when on / path again, the "intensive" calculation needing element is displayed immediately.
import React, { useEffect, useState } from "react";
import { Route, Link, BrowserRouter as Router } from "react-router-dom";
function Parent() {
const [intensiveElement, setIntensiveElement] = useState("");
useEffect(() => {
const intensiveCalculation = async () => {
await new Promise((resolve) => setTimeout(resolve, 5000));
return <p>Intensive paragraph</p>;
};
intensiveCalculation().then((element) => setIntensiveElement(element));
}, []);
return (
<Router>
<Link to="/child1">Go to child 1</Link>
<Link to="/child2">Go to child 2</Link>
<Route path="/" exact>
{intensiveElement}
</Route>
<Route path="/child1" exact>
<Child1 />
</Route>
<Route path="/child2" exact>
<Child2 />
</Route>
</Router>
);
}
function Child1() {
return <p>Child 1</p>;
}
function Child2() {
return <p>Child 2</p>;
}
My solution above works for not doing slow things twice like API calls. But following the remarks of Mordechai, I have made an example repository to compare DOM loading time of really big HTML for 4 solutions when using browser back button:
overflow: hidden; height: 0px; (more efficient than display: none; and the elements do not take any space contrary to visibility: hidden;, opacity: 0; etc. but maybe there is a better CSS way)Each exemple loads an initial page of 100 000 <span> elements, and has links to navigate to small pages, so that we can try the back button on the browser.
You can test yourself the static version of the examples on github pages here (the pages take several seconds to load on a normal computer, so maybe avoid clicking on them if on mobile or so).
I've added some CSS to make the elements small enough to see all of them on the screen, and compare how does the browser update the whole display.
On Firefox:
On Chrome:
Something important to note also: for Chrome when Next.js or plain HTML take 8 seconds, they actually load elements little by little on the page, and I have no cache with the back button.
On Firefox I don't have that little by little displaying, either there is nothing or everything is displayed (like what I have on Chrome with react state usage).
I don't really know what I can conclude with that, except maybe that testing things is useful, there are sometimes surprises...
I've misread the question initially. I'll leave the initial answer for the case when a user goes to a page on another domain.
You've wrote in the comments
I was clear enough
Well... judging by discussions here, that's not the case.
Here some points to consider, and when you'll answer them that should be the solution to your problem... whatever it is:
router? If, yes, then which one and how? Because some of then can retain the previous route in memory, so with a little bit of luck you could've never stumble on you blank page problem. And that can be the answer.mydomain.com/otherPage is not under control of the React or/and maybe it's not a true SPA we a talking about here. And the effects of going to this page is the same as going to another domain? Then my initial answer holds.In a nutshell:
Is there any magic solution in REACT to somehow 'retain' the DOM for a page when navigating out of it.
Yes, if by navigating out of it and a page you mean navigating to another route in you SPA and just rendering some other component, without executing a GET request through a "standard" <a>-click, window.location.href change or something similar which will lead to the browser initiating a new page loading.
For that just read your router's docs.
No if your are actually leaving your SPA.
For this case I would suggest serviceWorker. As to my taste, it's a much simpler and more flexible solution compared to a change of the architecture of your project with SSR.
as SPA dictates the DOM must be erased and re-built with every page change
Not at all. DOM will be erased only if the state of a component or the props are changed. But to help you with that we need to see the code.
It's not totally clear what is your question about. You are focused on the idea of preventing the DOM rebuild, but at the same time you're saying that the bottleneck is the API calls. And they're two quite different things to deal with.
And possible solution to you problem heavily depends on the architecture of you code.
If you have control over the server side, you can setup caching for your calls. If not, you can setup caching on the client side in a PWA-style manner.
If you have a centralized store, you can save its state to the localStorage on an <a> click and repopulate your page from the localStorage when the user gets back onto your page. If not you can resort to Service Worker API again to intercept API calls and to return cached responses. (Or just override fetch or whatever)
You can even "emulate" the SSR by saving HTML to the localStorage and showing it right away when the user gets back. (But the page will not be fully functional for a couple of seconds and need to be replaced at the moment you API-calls are completed)
But there is no feasible way to prevent DOM rebuild, because while theoretically possible, it's probably impractical to cache the whole React internal state. And if your main problem is indeed the DOM rebuild itself then probably your code is in need of serious optimizations.
One solution I often use, is to persist that data in location.state. And then when navigating back, the component first checks for data in location.state before attempting to fetch the data again.
This allows the page to render instantly.
const Example = (props) => {
const history = useHistory();
const location = useLocation();
const initialState = location.state;
const [state, setState] = useState(initialState);
useEffect(() => {
const persistentState = state;
history.replace({ pathname: location.pathname }, persistentState);
},[state]);
return ();
}
Using out of the box routing, I would say: it's impossible.
But who said we need to use routes?
Solution 1:
Why not using Portals?
This probably won't work if you want to 'retain' the DOM for any navigation on your page. But if you want to 'retain' it on only one specific page, then you could just open a fullscreen portal/modal/dialog (or whatever you wanna call it).
Solution 2:
If you want to 'retain' the DOM for all navigation, then you could also write a "router-component" yourself.
The logic of your component could look like this:
First you need a lookup-table. Give every url a related component that should be rendered, when the url is called.
Writing such a component shouldn't be too hard. I only see two problems with it: