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

// 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);}

其他回答

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;
}

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;
    }
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();

    }

Mime从本地文件:

String url = file.getAbsolutePath();
FileNameMap fileNameMap = URLConnection.getFileNameMap();
String mime = fileNameMap.getContentTypeFor("file://"+url);