How to check a uploaded file whether it is an image or other file?

Question

I'm assuming that you're running this in a servlet context. If it's affordable to check the content type based on just the file extension, then use ServletContext#getMimeType() to get the mime type (content type). Just check if it starts with image/.

String fileName = uploadedFile.getFileName();
String mimeType = getServletContext().getMimeType(fileName);
if (mimeType.startsWith("image/")) {
    // It's an image.
}

The default mime types are definied in the web.xml of the servletcontainer in question. In for example Tomcat, it's located in /conf/web.xml. You can extend/override it in the /WEB-INF/web.xml of your webapp as follows:

<mime-mapping>
    <extension>svg</extension>
    <mime-type>image/svg+xml</mime-type>
</mime-mapping>

But this doesn't prevent you from users who are fooling you by changing the file extension. If you'd like to cover this as well, then you can also determine the mime type based on the actual file content. If it's affordable to check for only BMP, GIF, JPG or PNG types (but not TIF, PSD, SVG, etc), then you can just feed it directly to ImageIO#read() and check if it doesn't throw an exception.

try (InputStream input = uploadedFile.getInputStream()) {
    try {
        ImageIO.read(input).toString();
        // It's an image (only BMP, GIF, JPG and PNG are recognized).
    } catch (Exception e) {
        // It's not an image.
    }
}

But if you'd like to cover more image types as well, then consider using a 3rd party library which does all the work by sniffing the file headers. For example JMimeMagic or Apache Tika which support both BMP, GIF, JPG, PNG, TIF and PSD (but not SVG). Apache Batik supports SVG. Below example uses JMimeMagic:

try (InputStream input = uploadedFile.getInputStream()) {
    String mimeType = Magic.getMagicMatch(input, false).getMimeType();
    if (mimeType.startsWith("image/")) {
        // It's an image.
    } else {
        // It's not an image.
    }
}

You could if necessary use combinations and outweigh the one and other.

That said, you don't necessarily need ImageIO#write() to save the uploaded image to disk. Just writing the obtained InputStream directly to a Path or any OutputStream like FileOutputStream the usual Java IO way is more than sufficient (see also Recommended way to save uploaded files in a servlet application):

try (InputStream input = uploadedFile.getInputStream()) {
    Files.copy(input, new File(uploadFolder, fileName).toPath());
}

Unless you'd like to gather some image information like its dimensions and/or want to manipulate it (crop/resize/rotate/convert/etc) of course.

I used org.apache.commons.imaging.Imaging in my case. Below is a sample piece of code to check if an image is a jpeg image or not. It throws ImageReadException if uploaded file is not an image.

    try {
        //image is InputStream
        byte[] byteArray = IOUtils.toByteArray(image);
        ImageFormat mimeType = Imaging.guessFormat(byteArray);
        if (mimeType == ImageFormats.JPEG) {
            return;
        } else {
            // handle image of different format. Ex: PNG
        }
    } catch (ImageReadException e) {
        //not an image
    }

This is built into the JDK and simply requires a stream with support for

byte[] data = ;
InputStream is = new BufferedInputStream(new ByteArrayInputStream(data));
String mimeType = URLConnection.guessContentTypeFromStream(is);
//...close stream

Try using multipart file instead of BufferedImage

import org.apache.http.entity.ContentType;
...

    public void processImage(MultipartFile file) {

       if(!Arrays.asList(ContentType.IMAGE_JPEG.getMimeType(), ContentType.IMAGE_PNG.getMimeType(), ContentType.IMAGE_GIF.getMimeType()).contains(file.getContentType())) {
            throw new IllegalStateException("File must be an Image");
                 }
      }