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

// 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);
}

其他回答

我尝试使用标准方法来确定mime类型,但我不能使用MimeTypeMap.getFileExtensionFromUrl(uri.getPath())保留文件扩展名。这个方法返回一个空字符串。所以我做了一个重要的解决方案来保留文件扩展名。

下面是返回文件扩展名的方法:

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

并且保留了文件扩展名,可以获得如下所示的mime类型:

public String getMimeType(File file) {
    String mimeType = "";
    String extension = getExtension(file.getName());
    if (MimeTypeMap.getSingleton().hasExtension(extension)) {
        mimeType = MimeTypeMap.getSingleton().getMimeTypeFromExtension(extension);
    }
    return mimeType;
}

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

// new processing the mime type out of Uri which may return null in some cases
String mimeType = getContentResolver().getType(uri);
// old processing the mime type out of path using the extension part if new way returned null
if (mimeType == null){mimeType URLConnection.guessContentTypeFromName(path);}