Improperly trusting ContentProvider-provided filename

OWASP category: MASVS-CODE: Code Quality

Overview

FileProvider, a subclass of ContentProvider, is intended to provide a secure method for an application ("server application") to share files with another application ("client application"). However, if the client application does not properly handle the filename provided by the server application, an attacker-controlled server application may be able to implement its own malicious FileProvider to overwrite files in the client application's app-specific storage.

Impact

If an attacker can overwrite an application's files, this can lead to malicious code execution (by overwriting the application's code), or allow otherwise modifying the application's behavior (for example, by overwriting the application's shared preferences or other configuration files).

Mitigations

Don't Trust User Input

Prefer working without user input when using file system calls by generating a unique filename when writing the received file to storage.

In other words: When the client application writes the received file to storage, it should ignore the filename provided by the server application and instead use its own internally generated unique identifier as the filename.

This example builds upon the code found at https://developer.android.com/training/secure-file-sharing/request-file:

Kotlin

// Code in
// https://developer.android.com/training/secure-file-sharing/request-file#OpenFile
// used to obtain file descriptor (fd)

try {
    val inputStream = FileInputStream(fd)
    val tempFile = File.createTempFile("temp", null, cacheDir)
    val outputStream = FileOutputStream(tempFile)
    val buf = ByteArray(1024)
    var len: Int
    len = inputStream.read(buf)
    while (len > 0) {
        if (len != -1) {
            outputStream.write(buf, 0, len)
            len = inputStream.read(buf)
        }
    }
    inputStream.close()
    outputStream.close()
} catch (e: IOException) {
    e.printStackTrace()
    Log.e("MainActivity", "File copy error.")
    return
}

Java

// Code in
// https://developer.android.com/training/secure-file-sharing/request-file#OpenFile
// used to obtain file descriptor (fd)

FileInputStream inputStream = new FileInputStream(fd);

// Create a temporary file
File tempFile = File.createTempFile("temp", null, getCacheDir());

// Copy the contents of the file to the temporary file
try {
    OutputStream outputStream = new FileOutputStream(tempFile))
    byte[] buffer = new byte[1024];
    int length;
    while ((length = inputStream.read(buffer)) > 0) {
        outputStream.write(buffer, 0, length);
    }
} catch (IOException e) {
    e.printStackTrace();
    Log.e("MainActivity", "File copy error.");
    return;
}

Sanitize Provided Filenames

Sanitize the provided filename when writing the received file to storage.

This mitigation is less desirable than the preceding mitigation because it can be challenging to handle all potential cases. Nonetheless: If generating a unique filename is not practical, the client application should sanitize the provided filename. Sanitization includes:

  • Sanitizing path traversal characters in the filename
  • Performing a canonicalization to confirm there are no path traversals

This example code builds upon the guidance on retrieving file information:

Kotlin

protected fun sanitizeFilename(displayName: String): String {
    val badCharacters = arrayOf("..", "/")
    val segments = displayName.split("/")
    var fileName = segments[segments.size - 1]
    for (suspString in badCharacters) {
        fileName = fileName.replace(suspString, "_")
    }
    return fileName
}

val displayName = returnCursor.getString(nameIndex)
val fileName = sanitizeFilename(displayName)
val filePath = File(context.filesDir, fileName).path

// saferOpenFile defined in Android developer documentation
val outputFile = saferOpenFile(filePath, context.filesDir.canonicalPath)

// fd obtained using Requesting a shared file from Android developer
// documentation

val inputStream = FileInputStream(fd)

// Copy the contents of the file to the new file
try {
    val outputStream = FileOutputStream(outputFile)
    val buffer = ByteArray(1024)
    var length: Int
    while (inputStream.read(buffer).also { length = it } > 0) {
        outputStream.write(buffer, 0, length)
    }
} catch (e: IOException) {
    // Handle exception
}

Java

protected String sanitizeFilename(String displayName) {
    String[] badCharacters = new String[] { "..", "/" };
    String[] segments = displayName.split("/");
    String fileName = segments[segments.length - 1];
    for (String suspString : badCharacters) {
        fileName = fileName.replace(suspString, "_");
    }
    return fileName;
}

String displayName = returnCursor.getString(nameIndex);
String fileName = sanitizeFilename(displayName);
String filePath = new File(context.getFilesDir(), fileName).getPath();

// saferOpenFile defined in Android developer documentation

File outputFile = saferOpenFile(filePath,
    context.getFilesDir().getCanonicalPath());

// fd obtained using Requesting a shared file from Android developer
// documentation

FileInputStream inputStream = new FileInputStream(fd);

// Copy the contents of the file to the new file
try {
    OutputStream outputStream = new FileOutputStream(outputFile))
    byte[] buffer = new byte[1024];
    int length;
    while ((length = inputStream.read(buffer)) > 0) {
        outputStream.write(buffer, 0, length);
    }
} catch (IOException e) {
    // Handle exception
}

Contributors: Dimitrios Valsamaras and Michael Peck of Microsoft Threat Intelligence

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