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) {
    }

你有多种选择来获得文件扩展名: 字符串文件名= uri.getLastPathSegment();请看这个链接

2-你也可以使用这段代码

 filePath .substring(filePath.lastIndexOf(".")+1);

但这不是一个好方法。 3-如果你有文件的URI，那么使用这个代码

String[] projection = { MediaStore.MediaColumns.DATA,
MediaStore.MediaColumns.MIME_TYPE };

4-如果你有URL，那么使用下面的代码:

 public static String getMimeType(String url) {
String type = null;
String extension = MimeTypeMap.getFileExtensionFromUrl(url);
if (extension != null) { 


  type = MimeTypeMap.getSingleton().getMimeTypeFromExtension(extension);
    }

return type;
}

享受你的代码:)

首先，你应该考虑调用MimeTypeMap#getMimeTypeFromExtension()，就像这样:

// url = file path or whatever suitable URL you want.
public static String getMimeType(String url) {
    String type = null;
    String extension = MimeTypeMap.getFileExtensionFromUrl(url);
    if (extension != null) {
        type = MimeTypeMap.getSingleton().getMimeTypeFromExtension(extension);
    }
    return type;
}

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

// This will return the mimeType. 
// for eg. xyz.png it will return image/png. 
// here uri is the file that we were picked using intent from ext/internal storage.
private String getMimeType(Uri uri) {
   // This class provides applications access to the content model.  
   ContentResolver contentResolver = getContentResolver();

   // getType(Uri url)-Return the MIME type of the given content URL. 
   return contentResolver.getType(uri);
}