背景

我的应用程序( here )可以在整个文件系统(不仅仅是已安装的应用程序)中搜索 APK 文件，显示每个文件的信息，允许删除、共享、安装...

作为 Android Q 范围存储功能的一部分，谷歌宣布 SAF(存储访问框架)将取代正常的存储权限。这意味着即使您尝试使用存储权限，它也只会授予对特定类型文件的访问权限，以便使用或完全沙盒化的 File 和 file-path (写于 here )。

这意味着许多框架将需要依赖 SAF 而不是 File 和 file-path。

问题

其中之一是packageManager.getPackageArchiveInfo ，给定文件路径，返回 PackageInfo ，我可以获得有关以下方面的各种信息:

@TargetApi(Build.VERSION_CODES.O)
override fun onCreate(savedInstanceState: Bundle?) {
    super.onCreate(savedInstanceState)
    setContentView(R.layout.activity_main)
    setSupportActionBar(toolbar)
    packageInstaller = packageManager.packageInstaller

    val intent = Intent(Intent.ACTION_OPEN_DOCUMENT)
    intent.addCategory(Intent.CATEGORY_OPENABLE)
    intent.type = "application/vnd.android.package-archive"
    startActivityForResult(intent, 1)
}

override fun onActivityResult(requestCode: Int, resultCode: Int, resultData: Intent?) {
    super.onActivityResult(requestCode, resultCode, resultData)
    if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.LOLLIPOP && requestCode == 1 && resultCode == Activity.RESULT_OK && resultData != null) {
        val uri = resultData.data
        val isDocumentUri = DocumentFile.isDocumentUri(this, uri)
        if (!isDocumentUri)
           return
        val documentFile = DocumentFile.fromSingleUri(this, uri)
        val inputStream = contentResolver.openInputStream(uri)
        //TODO do something with what you got above, to parse it as APK file

AsyncTask.execute {
    val packageInfo = packageManager.getPackageInfo(packageName, 0)
    val apkFilePath = packageInfo.applicationInfo.publicSourceDir
    // I'm using the path only because it's easier this way, but in reality I will have a Uri or inputStream as the input, which is why I use FileInputStream to mimic it.
    val zipInputStream = ZipInputStream(FileInputStream(apkFilePath))
    while (true) {
        val zipEntry = zipInputStream.nextEntry ?: break
        if (zipEntry.name.contains("AndroidManifest.xml")) {
            Log.d("AppLog", "zipEntry:$zipEntry ${zipEntry.size}")
            val bytes = zipInputStream.readBytes()
            val xmlTranslator = XmlTranslator()
            val resourceTable = ResourceTable()
            val locale = Locale.getDefault()
            val apkTranslator = ApkMetaTranslator(resourceTable, locale)
            val xmlStreamer = CompositeXmlStreamer(xmlTranslator, apkTranslator)
            val buffer = ByteBuffer.wrap(bytes)
            val binaryXmlParser = BinaryXmlParser(buffer, resourceTable)
            binaryXmlParser.locale = locale
            binaryXmlParser.xmlStreamer = xmlStreamer
            binaryXmlParser.parse()
            val apkMeta = apkTranslator.getApkMeta();
            Log.d("AppLog", "apkMeta:$apkMeta")
            break
        }
    }
}

最佳答案

好的，我想我找到了一种使用 Android 框架的方法(reddit 上的某个人给了我这个解决方案)，使用文件路径并使用它，但它并不完美。一些注意事项:

val fileDescriptor = contentResolver.openFileDescriptor(uri, "r") ?: return
val packageArchiveInfo = packageManager.getPackageArchiveInfo("/proc/self/fd/" + fileDescriptor.fd, 0)

class MainActivity : AppCompatActivity() {
    override fun onCreate(savedInstanceState: Bundle?) {
        super.onCreate(savedInstanceState)
        setContentView(R.layout.activity_main)
        startActivityForResult(
                Intent(Intent.ACTION_OPEN_DOCUMENT).addCategory(Intent.CATEGORY_OPENABLE)
                        .setType("application/vnd.android.package-archive"), 1
        )
    }

    override fun onActivityResult(requestCode: Int, resultCode: Int, data: Intent?) {
        super.onActivityResult(requestCode, resultCode, data)
        try {
            val uri = data?.data ?: return
            val takeFlags = Intent.FLAG_GRANT_READ_URI_PERMISSION or Intent.FLAG_GRANT_WRITE_URI_PERMISSION
            contentResolver.takePersistableUriPermission(uri, takeFlags)
            val isDocumentUri = DocumentFile.isDocumentUri(this, uri)
            if (!isDocumentUri)
                return
            val documentFile = DocumentFile.fromSingleUri(this, uri) ?: return
            val fileDescriptor = contentResolver.openFileDescriptor(uri, "r") ?: return
            val packageArchiveInfo = packageManager.getPackageArchiveInfo("/proc/self/fd/" + fileDescriptor.fd, 0)
            Log.d("AppLog", "got APK info?${packageArchiveInfo != null}")
            if (packageArchiveInfo != null) {
                val appLabel = loadAppLabel(packageArchiveInfo.applicationInfo, packageManager)
                Log.d("AppLog", "appLabel:$appLabel")
            }
        } catch (e: Exception) {
            e.printStackTrace()
            Log.e("AppLog", "failed to get app info: $e")
        }
    }

    fun loadAppLabel(applicationInfo: ApplicationInfo, packageManager: PackageManager): String =
            try {
                applicationInfo.loadLabel(packageManager).toString()
            } catch (e: java.lang.Exception) {
                ""
            }
    }
}

关于android - 如何在不使用 File 或 file-path 的情况下获取文件系统(不仅仅是已安装的)中 APK 文件的信息？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/56309165/