我想将我的颤振应用程序中的帖子分享到社交媒体平台,如LinkedIn和脸书。我该怎么做。
您可以使用 Flutter Share 插件。
Share.share('Your content here.');
此外,您还可以使用 Firebase 动态链接将用户重定向到您的应用。
我不知道LinkedIn但对于Facebook,你可以使用social_share_plugin。除了自述文件中的说明外,我还按照示例在android/app/src/main/res/xml中添加了provider_paths.xml。您需要确保使用的是最新的 facebook SDK 实现,因此在 android/app/build.gradle 中,将以下代码添加到依赖项中:
implementation 'com.facebook.android:facebook-android-sdk:[5,6)'
您也可以使用flutter_share_me包。
String response;
String url = 'https://cardaji.com';
String msg = "Look my great score on cardaji";
String strSeparator=" n";
final FlutterShareMe flutterShareMe = FlutterShareMe();
// facebook post
response = await flutterShareMe.shareToFacebook(url: url, msg: msg);
// twitter post
response = await flutterShareMe.shareToTwitter(url: url, msg: msg);
// whatsapp post
response = await flutterShareMe.shareToTelegram(msg: msg+ strSeparator+ url);
// telegram post
response = await flutterShareMe.shareToWhatsApp(msg: msg+strSeparator+url);
非常重要
当您构建iOS项目时,您可以在SwiftFlutterShareMePlugin.swift文件的sharefacebook功能中找到以下错误。
'init((' 已在此处明确标记为不可用 (FBSDKShareKit.ShareDialog(
func sharefacebook(message:Dictionary<String,Any>, result: @escaping FlutterResult) {
let viewController = UIApplication.shared.delegate?.window??.rootViewController
let shareDialog=ShareDialog() // **this line make the error**
let shareContent = ShareLinkContent()
shareContent.contentURL = URL.init(string: message["url"] as! String)!
shareContent.quote = message["msg"] as? String
shareDialog.mode = .automatic
ShareDialog(fromViewController: viewController, content: shareContent, delegate: self).show()
result("Sucess")
}
这是因为 facebook sdk 已升级,因此您可以更新 SwiftFlutterShareMePlugin.swift flutter_share_me 包(版本 1.2.0( 文件,如下所示。
func sharefacebook(message:Dictionary<String,Any>, result: @escaping FlutterResult) {
let viewController = UIApplication.shared.delegate?.window??.rootViewController
let shareContent = ShareLinkContent()
shareContent.contentURL = URL.init(string: message["url"] as! String)!
shareContent.quote = message["msg"] as? String
ShareDialog(viewController: viewController, content: shareContent, delegate: self).show()
result("Sucess")
}