Read the binary image data from a URL into a ByteArrayInputStream from HttpUrlConnect::URL

Question

I'm trying to pull the image from a URL and read it directly into a ByteArrayInputStream. I found one way of doing it, but it requires an image type, and there will be various image types, so I'd like to find a simple way to just read the binary data right in.

Here is my latest attempt. I'm using a BufferedImage, which I don't think is necessary.

URL url = new URL("http://hobbylesson.com/wp-content/uploads/2015/04/Simple-Acrylic-Painting-Ideas00005.jpg");

//Read in the image
BufferedImage image = ImageIO.read(url);
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ImageIO.write(image, "png", baos);
is = new ByteArrayInputStream(baos.toByteArray());

Answer 1

Here's the solution I found to work. Thanks for the two approaches above. I'd rather avoid external libraries, but because the environment is a real pain. Similar, I should have access to Java 9 and transferTo(), but that's not working.

This answerer was also helpful: Convert InputStream(Image) to ByteArrayInputStream

URL url = new URL("http://hobbylesson.com/wp-content/uploads/2015/04/Simple-Acrylic-Painting-Ideas00005.jpg");
        
InputStream source = url.openStream();
byte[] buf = new byte[8192]; 
int bytesRead = 0;
ByteArrayOutputStream baos = new ByteArrayOutputStream();
while((bytesRead = source.read(buf)) != -1) {
     baos.write(buf, 0, bytesRead);
}
is = new ByteArrayInputStream(baos.toByteArray());

Answer 2

As an alternative to the solution proposed by @rmunge, the Apache Commons IO library provides the class IOUtils which can be vey useful in your use case.

If you are using Maven for instance, you can import the library including the following dependency in your pom.xml:

<dependency>
  <groupId>commons-io</groupId>
  <artifactId>commons-io</artifactId>
  <version>2.8.0</version>
</dependency>

Then, you can use IOUtils like this:

URL url = new URL("http://hobbylesson.com/wp-content/uploads/2015/04/Simple-Acrylic-Painting-Ideas00005.jpg");
try (
  InputStream imageInputStream = url.openStream();
  ByteArrayOutputStream bOut = new ByteArrayOutputStream()
) {
  // You can obtain a byte[] as well if required
  // Please, consider write to the actual final OutputStream instead
  // of into the intermediate byte array output stream to optimize memory
  // consumption
  IOUtils.copy(imageInputStream, bOut);

  // Create an input stream from the read bytes
  ByteArrayInputStream in = new ByteArrayInputStream(bOut.toByteArray());
  // ...
} catch (IOException ioe) {
  ioe.printStackTrace();
}

Or simply this approach:

URL url = new URL("http://hobbylesson.com/wp-content/uploads/2015/04/Simple-Acrylic-Painting-Ideas00005.jpg");
byte[] imageBytes = IOUtils.toByteArray(url);
ByteArrayInputStream in = new ByteArrayInputStream(imageBytes);

For your comments, if the problem if you are trying to avoid network latency problems, if the requirement for a ByteArrayInputStream is not strictly necessary, as you can see in the javadocs perhaps the following code may be helpful as well:

URL url = new URL("http://hobbylesson.com/wp-content/uploads/2015/04/Simple-Acrylic-Painting-Ideas00005.jpg");
try (InputStream imageInputStream = url.openStream()) {
  InputStream in = IOUtils.toBufferedInputStream(imageInputStream);
  //...
}

Of course, you can always perform the read and write "manually" using the standard Java InputStream and OutputStream mechanisms:

URL url = new URL("http://hobbylesson.com/wp-content/uploads/2015/04/Simple-Acrylic-Painting-Ideas00005.jpg");
try (
  InputStream inputStream = url.openStream();
  BufferedInputStream bufferedInputStream = new BufferedInputStream(inputStream);
  ByteArrayOutputStream outputStream = new ByteArrayOutputStream();
  BufferedOutputStream bufferedOutputStream = new BufferedOutputStream(outputStream); 
) {
  byte[] buffer = new byte[8192];
  int bytesRead;
  while ((bytesRead = bufferedInputStream.read(buffer)) != -1) {
    bufferedOutputStream.write(buffer, 0, bytesRead);
  }

  bufferedOutputStream.flush();

  // Create an input stream from the read bytes
  ByteArrayInputStream in = new ByteArrayInputStream(outputStream.toByteArray());
  // ...
} catch (IOException ioe) {
  ioe.printStackTrace();
}

If you require more control about the underlying URL connection you can use URLConnection or HttpURLConnection, or many HTTP client libraries like Apache HttpClient or OkHttp, to name some of them.

Take as example the problem pointed out by @LuisCarlos in his comment, in order to avoid possible leak connections:

URLConnection urlConn = null;
try {
  urlConn = url.openConnection();
  urlConn.setConnectTimeout(5000);
  urlConn.setReadTimeout(30000);
  InputStream inputStream = urlConn.getInputStream();
  // the rest of the code...

} catch (Exception e) {
  
}

If you need to detect the actual image type consider the use of Tika or JMimeMagic.

Answer 3

URL url = new URL("http://hobbylesson.com/wp-content/uploads/2015/04/Simple-Acrylic-Painting-Ideas00005.jpg");

ByteArrayOutputStream baos = new ByteArrayOutputStream();
url.openStream().transferTo(baos);

ByteArrayInputStream in = new ByteArrayInputStream(baos.toByteArray());

The transferTo() method exists since Java 9. If you should use an older version of Java please see here for an alternative. Main drawback of this solution is that it has to read the whole file into memory first. If you anyway plan to forward the binary data to an other process you could omit the ByteArray streams and transfer the content directly to an OutputStream.