In order to get the MIME type of a file that doesn't have an extension, you'll need to use a different approach. One way to do this is to read the first few bytes of the file and determine the MIME type based on the file format signature (also known as "magic numbers").
import java.io.FileInputStream
import java.nio.ByteBuffer
import java.nio.ByteOrder
fun getMimeTypeWithoutExtension(filePath: String): String {
val magicNumbers = mapOf(
0x89.toByte() to "image/png",
0xff.toByte() to "image/jpeg",
0x47.toByte() to "image/gif",
0x49.toByte() to "image/tiff",
0x4d.toByte() to "image/tiff",
0x25.toByte() to "application/pdf",
0x50.toByte() to "application/vnd.ms-powerpoint",
0xD0.toByte() to "application/vnd.ms-word",
0x43.toByte() to "application/vnd.ms-word",
0x53.toByte() to "application/vnd.ms-word"
)
var mimeType = "application/octet-stream"
FileInputStream(filePath).use { inputStream ->
val buffer = ByteArray(1024)
inputStream.read(buffer, 0, buffer.size)
val magicNumber = ByteBuffer.wrap(buffer).order(ByteOrder.BIG_ENDIAN).get().toInt() and 0xff
mimeType = magicNumbers[magicNumber.toByte()] ?: mimeType
}
return mimeType
}
This code uses the FileInputStream class to read the first byte of the file into a ByteArray. The byte is then extracted as an integer and used to look up the corresponding MIME type in a map of magic numbers and MIME types. If the MIME type cannot be determined, the function returns "application/octet-stream" as a default.
Note that this code is only checking the first byte of the file, so it may not always accurately determine the MIME type of a file without an extension. For more accurate results, you may need to check additional bytes of the file or use a library that provides more comprehensive MIME type detection.