如何使用 Intent 进行 Amazon App Store 搜索并按开发者名称过滤? [英] How can I do an Amazon App Store search using an Intent and filter it by developer name?

问题描述

有没有办法在 Kindle Fire 上启动 Intent 以导致 AppStore 应用打开并显示某个开发者的所有应用?例如，在安装了 Android Market 的手机/平板电脑上，我可以这样做:

Is there a way to start an Intent on the Kindle Fire that will cause the AppStore app to open and display all the apps for a certain developer? For instance, on a phone/tablet with the Android Market installed, I can do this:

           Intent otherApps = new Intent(Intent.ACTION_VIEW,Uri.parse("market://search?q=pub:"" + developerName + """));
           activity.startActivity(otherApps);

并在 Android 电子市场中显示我的所有应用.我可以通过亚马逊应用商店做到这一点吗?如果是这样，怎么做?我已经用其他看似有效的名称(例如ZeptoLab")尝试了该 Intent，但没有得到任何过滤.它只是让我进入完全未经过滤的 App Store.使用market://details?id=package.name"查找特定应用似乎确实有效.

And show all my apps in the Android Market. Can I do that with the Amazon App Store? If so, how? I've tried that Intent with other seemingly valid names (such as "ZeptoLab") and I don't get any filtering. It just drops me in the full unfiltered App Store. Looking up a specific app with "market://details?id=package.name" does seem to work.

推荐答案

来自 https:///developer.amazon.com/help/faq.html#Marketing:

为了营销目的指向您的应用，请使用 URL http://www.amazon.com/gp/mas/dl/android?p=packagename(其中 packagename 是您的应用程序包名称).

To point to your app for marketing purposes use the URL http://www.amazon.com/gp/mas/dl/android?p=packagename (where packagename is your app package name).

如果您想链接到 Amazon Appstore 上所有应用程序的列表，请使用 URL http://www.amazon.com/gp/mas/dl/android?p=packagename&showAll=1.

If you want to link to the list of all your applications on the Amazon Appstore use the URL http://www.amazon.com/gp/mas/dl/android?p=packagename&showAll=1.

例如http://www.amazon.com/gp/mas/dl/android?p=com.rovio.angrybirds&showAll=1

所有这些都可以在这里看到:https:///developer.amazon.com/sdk/in-app-purchasing/sample-code/deeplink.html

All this can be seen here: https://developer.amazon.com/sdk/in-app-purchasing/sample-code/deeplink.html

amzn://apps/android?p=

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