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
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;
}
其他回答
以下是我在我的Android应用程序中使用的解决方案:
public static String getMimeType(String url)
{
String extension = url.substring(url.lastIndexOf("."));
String mimeTypeMap = MimeTypeMap.getFileExtensionFromUrl(extension);
String mimeType = MimeTypeMap.getSingleton().getMimeTypeFromExtension(mimeTypeMap);
return mimeType;
}
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;
}
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。
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”。
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;
}
