假设我有一个完整的文件路径:(/sdcard/tlogo.png)。我想知道它的mime类型。

我为它创建了一个函数

public static String getMimeType(File file, Context context)    
{
    Uri uri = Uri.fromFile(file);
    ContentResolver cR = context.getContentResolver();
    MimeTypeMap mime = MimeTypeMap.getSingleton();
    String type = mime.getExtensionFromMimeType(cR.getType(uri));
    return type;
}

但当我调用它时,它返回null。

File file = new File(filePath);
String fileType=CommonFunctions.getMimeType(file, context);

当前回答

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;
}

其他回答

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) {
    }

上面的解决方案在.rar文件的情况下返回null,使用URLConnection.guessContentTypeFromName(url)在这种情况下工作。

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”。

public static String getFileType(Uri file)
{
    try
    {
        if (file.getScheme().equals(ContentResolver.SCHEME_CONTENT))
            return subStringFromLastMark(SystemMaster.getContentResolver().getType(file), "/");
        else
            return MimeTypeMap.getFileExtensionFromUrl(file.toString()).toLowerCase();
    }
    catch(Exception e)
    {
        return null;
    }
}

public static String getMimeType(Uri file)
{
    try
    {
        return MimeTypeMap.getSingleton().getMimeTypeFromExtension(getFileType(file));
    }
    catch(Exception e)
    {
        return null;
    }
}

public static String subStringFromLastMark(String str,String mark)
{
    int l = str.lastIndexOf(mark);
    int end = str.length();
    if(l == -1)
        return str;

    return str.substring(l + 1, end);
}