Play framework 1.2 file upload with play.db.jpa.Blob

26 April 2011

Peter Hilton

by Peter Hilton

This article describes how to use the Play framework’s built-in features for handling HTTP file uploads and saving the data in files on the server’s file system, in Play 1.2. This is frequently-used in web applications for things like allowing users to upload photos of themselves.

Source code for this article is available at

Architectural considerations

There are generally two approaches to persisting binary data: either save the data as a file on the server’s file system, or store the data directly in a database table. There are pros and cons to each approach. Using the file system may be easier to implement, but might not scale. Using the database allows you to support transactions, but might not scale.

Confusingly, play.db.jpa.Blob data is stored in a file outside the database, and does not use the java.sql.Blob type with a BLOB in the database. On the server, Play stores the uploaded image in a file in the attachments/ folder, inside the application folder. The file name (a UUID) and MIME type are stored in a database attribute whose SQL type is VARCHAR.

To store the data in a database column, you would annotate a model property with @javax.persistence.Lob which results in the data being stored in a column whose SQL type is BLOB or CLOB. How to implement this approach and how it compares to using the file system are beyond the scope of this article.

Upload a file and store it on the server

The basic use case of uploading, storing and serving a file is extremely easy in Play. This is because the binding framework automatically binds a file uploaded from an HTML form to your JPA model, and because Play provides convenience methods that make serving binary data as easy as serving plain text.

To store file uploads in your model, add a property of type play.db.jpa.Blob

\{% highlight java %} package models; import play.db.jpa.Blob; import play.db.jpa.Model; import javax.persistence.Entity; @Entity public class User extends Model \{ public String name; public Blob photo; } \{% endhighlight %}

To upload files, add a form to your view template:

\{% highlight groovy %} #\{form @addUser(), enctype:'multipart/form-data'} #\{/form} \{% endhighlight %}

Then, in the controller, add an action that saves the upload in a new model object:

\{% highlight java %} public static void addUser(User user) \{; index(); } \{% endhighlight %}

This code does not appear to do anything other than save the JPA entity, because the file upload is handled automatically by Play. First, before the start of the action method, the uploaded file is saved in a sub-folder of tmp/uploads/. Next, when the entity is saved, the file is copied to the attachments/ folder, with a UUID as the file name. Finally, when the action is complete, the temporary file is deleted.

To save attachments in a different folder, specify a different path in the application.conf file. This can be an absolute path, or a relative path to a folder inside the Play application folder:

\{% highlight java %} attachments.path=photos \{% endhighlight %}

To display the uploaded images, add image tags to a view:

\{% highlight groovy %} #\{list items:models.User.findAll(), as:'user'} image #\{/list} \{% endhighlight %}

Finally, add a controller method to load the model object and render the image:

\{% highlight java %} public static void userPhoto(long id) \{ final User user = User.findById(id); notFoundIfNull(user); response.setContentTypeIfNotSet(; renderBinary(; } \{% endhighlight %}

Update the upload

If you provide a form (request) parameter, you can update an entry the same way you save one:

\{% highlight java %} public static void updateUser(User user) \{; index(); } \{% endhighlight %}

If a file upload is included, this will be saved as a new file, whose name is a new UUID. This means that the original file will now be orphaned. If you do not have unlimited disk space then you will have to implement your own scheme for cleaning up. There are several possible approaches.

The brute force approach is to implement an asynchronous job that periodically queries the JPA model for all currently-used file references, scans the attachments/ folder, and deletes unused files. This might scale badly.

Depending on your application, it might make sense to maintain references to all previous versions of attachments in your model, so that you can display them in the user interface, as a wiki generally does for previous versions of each page. Then the clean-up could either be manual, triggered when uploading a new file or from an asynchronous job. The clean up policy might be to keep a certain number of versions, or delete previous versions beyond a certain age. The @PreUpdate and @PreRemove JPA interceptors are useful for doing this kind of thing.

A more hard-core solution would be to modify play.db.jpa.Blob to create an alternative BinaryField field implementation that handles transaction-aware file upload updates.

Delete the upload

If you delete an object with a play.db.jpa.Blob property, the file in the attachments/ folder is not deleted automatically. You can delete the file manually via a reference to a property, as in the following action method.

\{% highlight java %} public static void deleteUser(long id) \{ final User user = User.findById(id);; user.delete(); index(); } \{% endhighlight %}

Alternatively, you could encapsulate the file deletion in the model, by overriding the delete method in to delete the file after the database entity has been successfully deleted.

\{% highlight java %} @Override public void _delete() \{ super._delete(); photo.getFile().delete(); } \{% endhighlight %}

Upload a file and save the file name

Sometimes you want to store the name of the originally uploaded file, so that you can map the file extension to a MIME type on the server, or so that you can serve the file as an attachment with the original file name.

To get at the file name, you need to bind the form control to a action method parameter. This means that you need a new action method in your controller that constructs the model object from separate form parameters, instead of binding the whole model object as in the first example.

Add the new action method to the controller, to instantiate a new model instance and its Blog property:

\{% highlight java %} public static void addUserWithFileName(File photo) throws FileNotFoundException \{ final User user = new User(); user.photoFileName = photo.getName(); = new Blob(); FileInputStream(photo), MimeTypes.getContentType(photo.getName()));; index(); } \{% endhighlight %}

To make this work, first add the new photoFileName property to the model:

\{% highlight java %} @Entity public class User extends Model \{ public String photoFileName; public Blob photo; } \{% endhighlight %}

Next, in the template, display the saved file name with the image, and change the form to use the new controller, and so that its file upload control’s name is just photo instead of

\{% highlight groovy %} #\{list items:models.User.findAll(), as:'user'} image #\{/list} #\{form @addUserWithFileName(), enctype:'multipart/form-data'} #\{/form} \{% endhighlight %}

Download the file as an attachment

When you serve a binary file to a web browser, the browser will normally display the data in the browser window, if possible. For example, the images in the example above are shown in-line in the browser window if you access their URLs directly.

However, you can set an HTTP header to instruct the web browser to treat the file as an ‘attachment’, which generally results in the web browser downloading the file to the user’s computer.

First, add a new action for the download. Assuming that the file name is always set, the only difference to the userPhoto action is that we pass the file name as a parameter to the renderBinary method, which causes Play to set the Content-Disposition response header, providing a file name.

\{% highlight java %} public static void downloadUserPhoto(long id) \{ final User user = User.findById(id); notFoundIfNull(user); response.setContentTypeIfNotSet(; renderBinary(, user.photoFileName); } \{% endhighlight %}

Now update the list of photos in the view template to include a link to the download URL.

\{% highlight groovy %} #\{list items:models.User.findAll(), as:'user'} image #\{/list} \{% endhighlight %}

Support custom content types

The response’s content type is set in the controller’s userPhoto action method, using the type stored in the Blob.

The play.libs.MimeTypes looks up the MIME type for the given file name’s extension, using the list in $PLAY_HOME/framework/src/play/libs/

Since Play 1.2 you can add your own types to the conf/application.conf file. For example, to add a MIME type for GIMP images with the .xcf extension, add the line:

\{% highlight java %} mimetype.xcf=application/x-gimp-image \{% endhighlight %}

Note that with the code examples above, this only works if you use the addUserWithFileName action above, which explicitly looks up the MIME type based on the original file name. The earlier addUser example uses the MIME type sent in the file upload HTTP request. My web browser (Safari 5.0.4) sets the request content type to image/png for PNG images, but does not recognise an .xcf file and sets its content type to application/octet-stream.


The Play framework greatly simplifies the task of handling and storing file uploads in a web application. However, this is for storing files in the server’s file system, which might not be the ideal solution for your application. To store the binary data in a database table, you would need to work out how to use JPA’s @Lob annotation.

Thanks to Niko Schmuck for additional review comments on this article.