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Firebase应用索引Android搜索结果

[英]Firebase App Indexing Android Search Result

I have implemented all the steps that are mentioned on this link: Firebase App Indexing Start . 我已经实现了此链接上提到的所有步骤: Firebase App Indexing Start All the test your implementation steps mentioned in above are working fine. 上面提到的所有实施步骤测试都可以正常工作。 My app is also searchable in Google App ScreenShot . 我的应用程序也可以在Google App ScreenShot中搜索。 I have cross checked back button violation , content mismatch as mentioned in google documentation. 我已经交叉检查了后退按钮违规,如Google文档中所述,内容不匹配。 We have a live website with properly indexed pages,my app is in play store.My website can be found with google search result, if you tap the link it all ask to open via app,if opened in app it will redirect to that specific content. 我们有一个带有正确索引页面的实时网站,我的应用程序在Play商店中。可以在Google搜索结果中找到我的网站,如果您点击所有链接都要求通过应用程序打开,如果在应用程序中打开,它将重定向到该特定网站内容。 Yet the app icons which generally appear with indexed URLs along with web links are not showing in search results. 但是,通常与索引URL以及Web链接一起出现的应用程序图标并未显示在搜索结果中。 I am unable to identify the problem. 我无法确定问题所在。 Can some body explain where I am going wrong. 有人可以解释我哪里出问题了。

The Indexing in way you implemented is local. 您实施的索引编制方式是本地的。 it means they are available only on a phone and not accessible through Internet. 这意味着它们仅在电话上可用,而不能通过Internet访问。 first you need to have Website. 首先,您需要拥有网站。 then when search result from you website shown on Google and user click on your website link, Android System will ask user how to open it. 然后,当您的网站上的搜索结果显示在Google上并且用户单击您的网站链接时,Android系统会询问用户如何打开它。 inside target app or in Browser. 在目标应用程序内部或在浏览器中。

<intent-filter android:label="@string/app_name" android:autoVerify="true">
        <action android:name="android.intent.action.VIEW" />
        <category android:name="android.intent.category.DEFAULT" />
        <category android:name="android.intent.category.BROWSABLE" />
        <!-- Accepts URIs that begin with "http://recipe-app.com/recipe" -->
        <data android:scheme="http"
            android:host="recipe-app.com"
            android:pathPrefix="/recipe" />
        <!-- Accepts URIs that begin with "https://recipe-app.com/recipe" -->
        <data android:scheme="https"
            android:host="recipe-app.com"
            android:pathPrefix="/recipe" />
    </intent-filter>

here website is www.recipe-app.com so if user see some result from this website on google search result then he can open it inside your app. 这里的网站是www.recipe-app.com,因此,如果用户在google搜索结果上看到该网站的某些结果,则可以在您的应用程序中打开它。 and app should be published on google play store. 和应用程序应在Google Play商店上发布。 hope it helps! 希望能帮助到你!

also i recommend to read this Codelab http://search-codelabs.appspot.com/codelabs/android-deep-linking#1 我也建议阅读此Codelab http://search-codelabs.appspot.com/codelabs/android-deep-linking#1

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