链接回亚马逊应用程序商店进行评级

Question

我目前在谷歌播放中使用以下代码为我的Android应用程序请求审查和评级我的应用程序.

如何链接回亚马逊应用商店或亚马逊市场实现同样的事情？

Intent goToMarket = new Intent(Intent.ACTION_VIEW, Uri.parse("market://details?id=com.yapp.blah"));
goToMarket.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
context.startActivity(goToMarket);

Answer 1

从这个SO答案找到:

Intent goToAppstore = new Intent(Intent.ACTION_VIEW,Uri.parse("http://www.amazon.com/gp/mas/dl/andro" +"id?p=com.yapp.blah"));
goToAppstore.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
context.startActivity(goToAppstore);

希望有帮助:)

Answer 2

你可以做以下事情:

1. 检查设备是否基于其制造商.例如:https://developer.amazon.com/sdk/fire/specifications.html

2. 要撰写评论并打开Amazon App Store,请使用以下意图

   amzn://apps/android?p=package_name
   

   其中p=Link,以特定包名的详细页面.

参考:亚马逊开发者链接.

https://developer.amazon.com/sdk/in-app-purchasing/sample-code/deeplink.html

Answer 3

只需使用以下代码.它首先尝试通过URI打开市场应用程序,但如果没有找到打开的Web链接.

public static final String MARKET_AMAZON_URL = "amzn://apps/android?p=";
public static final String WEB_AMAZON_URL = "http://www.amazon.com/gp/mas/dl/android?p=";

    private static void openOnAmazonMarket(Context context, String packageName) {

    try {
        Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(MARKET_AMAZON_URL + packageName));
        intent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
        context.startActivity(intent);
    } catch (android.content.ActivityNotFoundException anfe) {
        Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(WEB_AMAZON_URL + packageName));
        intent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
        context.startActivity(intent);
    }

}

对于Google Play和Samsung Galaxy Apps,链接如下:

public static final String MARKET_GOOGLE_URL = "market://details?id=";
public static final String WEB_GOOGLE_URL = "http://play.google.com/store/apps/details?id=";

public static final String MARKET_SAMSUNG_URL = "samsungapps://ProductDetail/";
public static final String WEB_SAMSUNG_URL = "http://www.samsungapps.com/appquery/appDetail.as?appId=";