CODEWITHSUNDEEP
reactjs
handsome
handsome

Reputation: 2422

dynamically load images with react

I created a component that renders images

this is the component.

import React, { lazy, Suspense } from "react";

const Icon = (props) => {
    const { src } = props;

    return (
        <img src={src}  />
    );
};

export default Icon;

then I use it like this

import ExampleIcon from "./../images/icons/example.png";
...
<Icon src={ExampleIcon} />

is there a more efficient way to load the icons? and then just "load" example.png and use it as a source? tried to change it to:

const Icon = (props) => {
    const src = lazy(() => import("./../images/icons/" + props.src + ".png"));
    return (
        <Suspense fallback={<p>loading...</p>}><img src={src} /></Suspense>
    );
};

looks like it doesn´t work that way. any other ideas? thanks!

Upvotes: 5

Views: 6713

Answers (2)

Inspiraller
Inspiraller

Reputation: 3806

You could apply this approach: Preloading images with JavaScript

const img=new Image();
img.src=url;

And how to do it with a hook with an online example:

https://www.selbekk.io/blog/2019/05/how-to-write-a-progressive-image-loading-hook/

Another approach just using hooks:

https://codesandbox.io/s/magical-pine-419kz?file=/src/App.tsx

import React, { useEffect } from "react";
import "./styles.css";

const loadImg = (src: string): Promise<string> =>
  new Promise((resolve, reject) => {
    const img = new Image();
    img.src = src;
    img.onload = () => resolve(src);
    img.onerror = () => reject(new Error("could not load image"));
  });

export default function App() {
  const [src, setSrc] = React.useState("preloadimg");
  useEffect(() => {
    const load = async () => {
      await loadImg(
        "https://upload.wikimedia.org/wikipedia/commons/thumb/8/8c/Chess_Large.JPG/800px-Chess_Large.JPG"
      ).then((src) => {
        setSrc(src);
      });
    }; // Execute the created function directly
    load();
  }, [src, setSrc]);

  return (
    <div className="App">
      <h1>Hello CodeSandbox</h1>
      <h2>Start editing to see some magic happen!</h2>
      <img src={src} alt="example" />
    </div>
  );
}

Upvotes: 1

Mordechai
Mordechai

Reputation: 16294

No, you can't do this, since React.lazy() must be at the top level and only return React components. To lazily load images you can do inside an effect:

function Icon = props => {
    const [image, setImage] = useState()

    useEffect(() => {
        import("./../images/icons/" + props.src + ".png").then(setImage)
    }, [props.src])


    return image ? <img src={image} /> : 'Loading...'
}

Edit: there's one little problem with this, namely, Webpack will not be able to figure out the file to code split since the string in the import function is not a literal. You could still access files in the public directory dynamically using fetch instead. And perhaps you don't need fetch at all, just provide an url to src and spare the whole hassle.

Upvotes: 2

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