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)

而从资产/文件(注意，在MimeTypeMap中缺少一些情况)。

private String getMimeType(String path) {
    if (null == path) return "*/*";

    String extension = path;
    int lastDot = extension.lastIndexOf('.');
    if (lastDot != -1) {
        extension = extension.substring(lastDot + 1);
    }

    // Convert the URI string to lower case to ensure compatibility with MimeTypeMap (see CB-2185).
    extension = extension.toLowerCase(Locale.getDefault());
    if (extension.equals("3ga")) {
        return "audio/3gpp";
    } else if (extension.equals("js")) {
        return "text/javascript";
    } else if (extension.equals("woff")) {
        return "application/x-font-woff";
    } else {
        // TODO
        // anyting missing from the map (http://www.sitepoint.com/web-foundations/mime-types-complete-list/)
        // reference: http://grepcode.com/file/repo1.maven.org/maven2/com.google.okhttp/okhttp/20120626/libcore/net/MimeUtils.java#MimeUtils
    }

    return MimeTypeMap.getSingleton().getMimeTypeFromExtension(extension);
}

而使用ContentResolver

contentResolver.getType(uri)

当http/https请求

    try {
        HttpURLConnection conn = httpClient.open(new URL(uri.toString()));
        conn.setDoInput(false);
        conn.setRequestMethod("HEAD");
        return conn.getHeaderField("Content-Type");
    } catch (IOException e) {
    }

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;
}

请密切关注上面umerk44的解决方案。getMimeTypeFromExtension调用guessMimeTypeTypeFromExtension并且区分大小写。我花了一个下午的时间，然后仔细观察- getMimeTypeFromExtension将返回NULL，如果你传递它“JPG”，而它将返回“image/jpeg”，如果你传递它“JPG”。

是否也返回空值 在我的案例中，路径是

/storage/ emululated /0/Music/01 - Ghost on the Dance Floor.mp3

在工作中使用

瓦尔·奥尼。replace(“”、“”)

方法是这样的

@JvmStatic
    fun getMimeType(inUrl: String?): String {
        if (inUrl == null) return ""

        val url = inUrl.replace(" ","")
        var type: String? = null

        val extension = MimeTypeMap.getFileExtensionFromUrl(url)
        if (extension != null) {
            type = MimeTypeMap.getSingleton().getMimeTypeFromExtension(extension.toLowerCase())
        }

        if(type ==null){
            val cR = WifiTalkie.getApplicationContext().contentResolver
            type = cR.getType(Uri.parse(url))
        }

        if (type == null) {
            type = "*/*" // fallback method_type. You might set it to */*
        }
        return type
    }

结果返回成功结果:

音频/ mpeg

希望能对大家有所帮助

File file = new File(path, name);

    MimeTypeMap mime = MimeTypeMap.getSingleton();
    int index = file.getName().lastIndexOf('.')+1;
    String ext = file.getName().substring(index).toLowerCase();
    String type = mime.getMimeTypeFromExtension(ext);

    intent.setDataAndType(Uri.fromFile(file), type);
    try
    {
      context.startActivity(intent);
    }
    catch(ActivityNotFoundException ex)
    {
        ex.printStackTrace();

    }