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I have an app built in react-native in which I need to share a post on LinkedIn with predefined content.

I used 'react-native-share' for sharing content on LinkedIn but It's not working. How can I achieve this?

Thanks in advance.

Τζιότι
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3 Answers3

4

LinkedIn supports only one parameter being passed to it, and that is the url parameter. It will look like this...

https://www.linkedin.com/sharing/share-offsite/?url={url}

Source: Official LinkedIn Sharing Documentation.

The following parameters will not work: summary, title, source, etc.. Anything besides url.

To share to LinkedIn, just make a <a href> element that points to an above formatted-URL, and make absolutely sure to do URL-encoding on your {url}.

You probably want to share title and summary, though, based on the accepted answer. You cannot do that using GET-data, BUT you can do that using og: tags.

  • <meta property='og:title' content='Title of the article"/>
  • <meta property='og:image' content='//media.example.com/ 1234567.jpg"/>
  • <meta property='og:description' content='Description that will show in the preview"/>
  • <meta property='og:url' content='//www.example.com/URL of the article" />

Source: LinkedIn Developer Docs: Making Your Website Shareable on LinkedIn.

In case you are uncertain that you've followed the LinkedIn documentation correctly to make a share URL, you can test your page's URL to see how it will look when shared here: LinkedIn Post Inspector.

HoldOffHunger
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1

You can do it with react-native-share. I have used it in one of my application.

const shareOptions = {
  title: 'Share via',
  message: `Hello,  ${description}`,
  subject: 'Subject,
  url: "data:image/png;base64," + base64Data,
  showAppsToView: false,
  filename: 'test',
};
Share.open(shareOptions).then(res => {
  console.log(res)
}).catch(e => {
  console.log(e)
});

Note: It would be greater if you can share your code.

Rajan
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1

You can use Linking from react native

import { Linking } from 'react-native';

and then you can use

Linking.openURL("https://www.linkedin.com/shareArticle?mini=true&summary=youtube&title=f1&url=https://www.youtube.com/watch?v=dQw4w9WgXcQ");

you can found more info here https://stackoverflow.com/a/10737122/6125249

  • This is the correct answer, it works for me on "react-native": "0.71.5". The only problem is that summary and title don't work. So it should be: `Linking.openURL("https://www.linkedin.com/shareArticle?mini=true&url=https://www.youtube.com/watch?v=dQw4w9WgXcQ");` – Bruno Pintos Jul 26 '23 at 13:04