当应用程序处于后台状态时，深度链接不起作用 React native

Question

我正在创建基于 react native 的电子商务应用程序。这里我需要从 url shared 打开单个产品页面。实际上，当应用程序处于终止状态时它会工作，但如果应用程序处于后台/非活动状态，它不会工作。在后台/非活动状态下打开时，共享网址为空。我已附加我的代码。

// following code working for app killing state

componentWillMount() {

    if (Platform.OS === 'android') {
      console.log("Testing");debugger

      //Constants.OneTimeFlag == false;
          Linking.getInitialURL().then(url => {
            console.log(url);
            var str = url
            var name = str.split('/')[4]
            Constants.isLinking = true;
           this.setState({ shop_Id: name})


           if (str)
           {
            this.setState({ isFromHomeLinking:'FROM_LINK' })
            this.props.navigation.navigate('SingleProductScreen', { ListViewClickItemHolder: [this.state.shop_Id,1,this.state.isFromHomeLinking] });

           }


          });

    }

    else {
        Linking.addEventListener('url', this.handleNavigation);
      }

  }

Not working code following..



componentDidMount() {
    AppState.addEventListener('change', this._handleAppStateChange);
}

componentWillUnmount() {
    AppState.removeEventListener('change', this._handleAppStateChange);
  }

this.state.appState declared in constructor(props)

_handleAppStateChange = (nextAppState) => {
    if (this.state.appState.match(/inactive|background/) && nextAppState === 'active') {
     console.log('App has come to the foreground!');debugger
 if (Platform.OS === 'android') {
          console.log("Testing");debugger

          //Constants.OneTimeFlag == false;
              Linking.getInitialURL().then(url => {
                console.log(url);
                var str = url
                var name = str.split('/')[4]
                Constants.isLinking = true;
               this.setState({ shop_Id: name})


               if (str)
               {
                this.setState({ isFromHomeLinking:'FROM_LINK' })
                this.props.navigation.navigate('SingleProductScreen', { ListViewClickItemHolder: [this.state.shop_Id,1,this.state.isFromHomeLinking] });

               }


              });

        }

        else {
            Linking.addEventListener('url', this.handleNavigation);
          }
    }
    }

当我从 whatsapp 和应用程序打开外部链接时，后台状态 Linking.getInitialURL() 收到为 null ..

按照我在清单文件中的内容

<activity
        android:name=".MainActivity"
        android:label="@string/app_name"
        android:configChanges="keyboard|keyboardHidden|orientation|screenSize"
        android:windowSoftInputMode="adjustResize"
        android:launchMode="singleTask">
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />
            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
        <intent-filter>
 <action android:name="android.intent.action.VIEW" />
<category android:name="android.intent.category.DEFAULT" />
 <category android:name="android.intent.category.BROWSABLE" />
<data android:scheme="http"
 android:host="demo1.zgroo.com" />
</intent-filter>
      </activity>

以下是我的示例网址..

http://demo1.zgroo.com/xxxx

请告诉我任何解决方案..

提前谢谢..

Answer 1

您需要为这种情况注册链接监听器。

componentDidMount() {
  Linking.addEventListener('url', this._handleOpenURL);
},
componentWillUnmount() {
  Linking.removeEventListener('url', this._handleOpenURL);
},
_handleOpenURL(event) {
  console.log(event.url);
}

了解更多https://facebook.github.io/react-native/docs/linking

Answer 2

这是 Anurag 使用钩子回答的一个版本：

export function useDeepLinkURL() {
  const [linkedURL, setLinkedURL] = useState<string | null>(null);

  // 1. If the app is not already open, it is opened and the url is passed in as the initialURL
  // You can handle these events with Linking.getInitialURL(url) -- it returns a Promise that
  // resolves to the url, if there is one.
  useEffect(() => {
    const getUrlAsync = async () => {
      // Get the deep link used to open the app
      const initialUrl = await Linking.getInitialURL();
      setLinkedURL(decodeURI(initialUrl));
    };

    getUrlAsync();
  }, []);

  // 2. If the app is already open, the app is foregrounded and a Linking event is fired
  // You can handle these events with Linking.addEventListener(url, callback)
  useEffect(() => {
    const callback = ({url}: {url: string}) => setLinkedURL(decodeURI(url));
    Linking.addEventListener('url', callback);
    return () => {
      Linking.removeEventListener('url', callback);
    };
  }, []);

  const resetURL = () => setLinkedURL(null);

  return {linkedURL, resetURL};
}

然后您可以将其用于：

const {linkedURL, resetURL} = useDeepLinkURL();

useEffect(() => {
    // ... handle deep link
    resetURL();
}, [linkedURL, resetURL])

我添加了函数resetURL，因为如果用户与应用程序共享同一个文件两次，你会想要加载它两次。但是，因为深层链接最终会是相同的，所以 useEffect 不会再次被触发。您可以通过将linkedURL 设置为null 来再次触发它，这样下次共享文件时，您可以确定它会导致useEffect 运行。

另外，我使用decodeURI 来解码传入的 URL，因为如果您使用 react-native-fs 之类的库从指定路径加载文件，它将无法处理其中包含空格的文件名称，除非您使用 decodeURI。

Answer 3

从componentwillunmount 中删除监听器。 无需在componentwillunmount 中编写任何代码，因为链接的addListener 将始终监听，只有当您从后台（通过单击新的深层链接）进入活动状态时才会触发。