对于Xamarin Android(来自@HoaLe的回答)

public String getMimeType(Uri uri) {
    String mimeType = null;
    if (uri.Scheme.Equals(ContentResolver.SchemeContent))
    {
        ContentResolver cr = Application.Context.ContentResolver;
        mimeType = cr.GetType(uri);
    }
    else
    {
        String fileExtension = MimeTypeMap.GetFileExtensionFromUrl(uri.ToString());
        mimeType = MimeTypeMap.Singleton.GetMimeTypeFromExtension(
        fileExtension.ToLower());
    }
    return mimeType;
}

I don't realize why MimeTypeMap.getFileExtensionFromUrl() has problems with spaces and some other characters, that returns "", but I just wrote this method to change the file name to an admit-able one. It's just playing with Strings. However, It kind of works. Through the method, the spaces existing in the file name is turned into a desirable character (which, here, is "x") via replaceAll(" ", "x") and other unsuitable characters are turned into a suitable one via URLEncoder. so the usage (according to the codes presented in the question and the selected answer) should be something like getMimeType(reviseUrl(url)).

private String reviseUrl(String url) {

        String revisedUrl = "";
        int fileNameBeginning = url.lastIndexOf("/");
        int fileNameEnding = url.lastIndexOf(".");

        String cutFileNameFromUrl = url.substring(fileNameBeginning + 1, fileNameEnding).replaceAll(" ", "x");

        revisedUrl = url.
                substring(0, fileNameBeginning + 1) +
                java.net.URLEncoder.encode(cutFileNameFromUrl) +
                url.substring(fileNameEnding, url.length());

        return revisedUrl;
    }

在kotlin中，这要简单得多。

解决方案1:

fun getMimeType(file: File): String? = 
    MimeTypeMap.getSingleton().getMimeTypeFromExtension(file.extension)

方案二:(文件扩展名函数)

fun File.mimeType(): String? = 
    MimeTypeMap.getSingleton().getMimeTypeFromExtension(this.extension)

它适用于我和灵活的内容和文件

public static String getMimeType(Context context, Uri uri) {
    String extension;

    //Check uri format to avoid null
    if (uri.getScheme().equals(ContentResolver.SCHEME_CONTENT)) {
        //If scheme is a content
        final MimeTypeMap mime = MimeTypeMap.getSingleton();
        extension = mime.getExtensionFromMimeType(context.getContentResolver().getType(uri));
    } else {
        //If scheme is a File
        //This will replace white spaces with %20 and also other special characters. This will avoid returning null values on file name with spaces and special characters.
        extension = MimeTypeMap.getFileExtensionFromUrl(Uri.fromFile(new File(uri.getPath())).toString());

    }

    return extension;
}

这里没有一个答案是完美的。以下是一个结合了所有热门答案的最佳元素的答案:

public final class FileUtil {

    // By default, Android doesn't provide support for JSON
    public static final String MIME_TYPE_JSON = "application/json";

    @Nullable
    public static String getMimeType(@NonNull Context context, @NonNull Uri uri) {

        String mimeType = null;
        if (uri.getScheme().equals(ContentResolver.SCHEME_CONTENT)) {
            ContentResolver cr = context.getContentResolver();
            mimeType = cr.getType(uri);
        } else {
            String fileExtension = getExtension(uri.toString());

            if(fileExtension == null){
                return null;
            }

            mimeType = MimeTypeMap.getSingleton().getMimeTypeFromExtension(
                    fileExtension.toLowerCase());

            if(mimeType == null){
                // Handle the misc file extensions
                return handleMiscFileExtensions(fileExtension);
            }
        }
        return mimeType;
    }

    @Nullable
    private static String getExtension(@Nullable String fileName){

        if(fileName == null || TextUtils.isEmpty(fileName)){
            return null;
        }

        char[] arrayOfFilename = fileName.toCharArray();
        for(int i = arrayOfFilename.length-1; i > 0; i--){
            if(arrayOfFilename[i] == '.'){
                return fileName.substring(i+1, fileName.length());
            }
        }
        return null;
    }

    @Nullable
    private static String handleMiscFileExtensions(@NonNull String extension){

        if(extension.equals("json")){
            return MIME_TYPE_JSON;
        }
        else{
            return null;
        }
    }
}

EDIT

我为此创建了一个小型库。 但是底层代码几乎是一样的。

它在GitHub上可用

MimeMagic-Android

2020年9月

使用芬兰湾的科特林

fun File.getMimeType(context: Context): String? {
    if (this.isDirectory) {
        return null
    }

    fun fallbackMimeType(uri: Uri): String? {
        return if (uri.scheme == ContentResolver.SCHEME_CONTENT) {
            context.contentResolver.getType(uri)
        } else {
            val extension = MimeTypeMap.getFileExtensionFromUrl(uri.toString())
            MimeTypeMap.getSingleton().getMimeTypeFromExtension(extension.toLowerCase(Locale.getDefault()))
        }
    }

    fun catchUrlMimeType(): String? {
        val uri = Uri.fromFile(this)

        return if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.O) {
            val path = Paths.get(uri.toString())
            try {
                Files.probeContentType(path) ?: fallbackMimeType(uri)
            } catch (ignored: IOException) {
                fallbackMimeType(uri)
            }
        } else {
            fallbackMimeType(uri)
        }
    }

    val stream = this.inputStream()
    return try {
        URLConnection.guessContentTypeFromStream(stream) ?: catchUrlMimeType()
    } catch (ignored: IOException) {
        catchUrlMimeType()
    } finally {
        stream.close()
    }
}

这似乎是最好的选择，因为它结合了前面的答案。

首先，它尝试使用URLConnection获取类型。guessContentTypeFromStream，但如果这个失败或返回null，它会尝试在Android O和以上使用mimetype

java.nio.file.Files
java.nio.file.Paths

否则，如果Android版本低于O或方法失败，它将使用ContentResolver和MimeTypeMap返回类型