我有一个react原生应用程序，登录后在web视图中加载url。用户在web视图中移动，单击支持的链接。现在，我想通过单击web视图中的按钮从web视图中打开共享选项。当我们点击按钮时，我们有一个url，但我无法检测到带有“onnavigationstatechange”的url。那么，如果没有url，我们如何处理这个问题呢？我尝试了很多链接，但都没有得到结果。从未到达“开放共享控制器”，也未到达“导航类型单击”。以下是我的示例代码：

<WebView scalesPageToFit
    style={{ flex: 1 }} source={{ uri: url  }} onLoadStart={() => (setLoading(true))} onLoadEnd={() => (setLoading(false))}
    javaScriptEnabled={true}
    domStorageEnabled={true}
    onNavigationStateChange = {handleNavigationStateChange}

    onLoadProgress = {({ nativeEvent }) => {

      console.log(nativeEvent.url)     
  }}
    />

const handleNavigationStateChange = navState => {
console.log(navState.url);
console.log(navState.navigationType)
const { url } = navState.url;
if (!url) return;

if (url.includes('Logout')) {
  Alert.alert(
    "Logout",
    "Are you sure you want to logout?",
    [
      {
        text: "No",
        onPress: () => console.log("Cancel Pressed"),
        style: "cancel"
      },
      { text: "Yes", onPress: () => navigation.pop(1) }
    ]
  );
}
if (url.includes('Register')) {
  console.log('OPEN SHARE CONTROLLER')
}
if (navState.navigationType === 'click') {
  // User clicked something
  console.log('User clicked something')

} };