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
我认为最简单的方法是引用这个资源文件: https://android.googlesource.com/platform/libcore/+/master/luni/src/main/java/libcore/net/android.mime.types
其他回答
// 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);}
get file object....
File file = new File(filePath);
then....pass as a parameter to...
getMimeType(file);
...here is
public String getMimeType(File file) {
String mimetype = MimeTypeMap.getSingleton().getMimeTypeFromExtension(MimeTypeMap.getFileExtensionFromUrl(Uri.fromFile(file).toString()).toLowerCase());
if (mimetype == null) {
return "*/*";
}
return mimetype;///return the mimeType
}
Intent myIntent = new Intent(android.content.Intent.ACTION_VIEW);
File file = new File(filePatch);
Uri uris = Uri.fromFile(file);
String mimetype = null;
if
(uris.getScheme().equals(ContentResolver.SCHEME_CONTENT)) {
ContentResolver cr =
getApplicationContext().getContentResolver();
mimetype = cr.getType(uris);
} else {
String fileExtension =
MimeTypeMap.getFileExtensionFromUrl(uris.toString());
mimetype = MimeTypeMap.getSingleton().getMimeTypeFromExtension(fileExtension.toLowerCase());
}
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)
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;
}
