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I'm very new to Grails so there's probably a very simple answer to this question. I'm trying to display a dynamically created image in a gsp. The image is NOT stored in a database, it's created on the fly in the controller.
What I essentially have is one gsp that has a form which takes in a set of user inputs (requestGraph.gsp). Upon submitting the form, the parameters are sent to the a displayGraph action in the controller which queries information from a database completely outside of Grails and creates a chart using the JFreeChart library. I would like to display this image within a displayGraph.gsp or something like that.
So basically within requestGraph.gsp I have a snippet similar to:
<g:form action="displayGraph">
<!-- ... bunch of labels and boxes -->
<g:submitButton name="displayGraph" value="Display Graph" />
</g:form>
Within the controller I have something like:
def requestGraph = {}
def displayGraph = {
//... code that uses params to make an image byte array and assigns to var img
return [image : img]
}
Within displayGraph.gsp:
<body>
<h1>Graph Title</h1>
<!-- ??? How to dislpay image? -->
</body>
I tried piping the image directly to the output stream in the displayGraph action. This works, but then I lose control of all page formatting in displayGraph.gsp.
Most tutorials I've found create a dedicated action to pipe the image to an output steam then call that action using a tag. The problem is that my image isn't stored in a database and I see no way of passing the image byte array (or even the form parameters) to create/render the image. Can anybody help me with this? Thanks.
367
If you write the bytes to the output stream, you can treat the controller/action as the source of the image in your GSP. Here's a quick, untested example:
// controller action
def displayGraph = {
def img // byte array
//...
response.setHeader('Content-length', img.length)
response.contentType = 'image/png' // or the appropriate image content type
response.outputStream << img
response.outputStream.flush()
}
You could then access your image in the src of an <img> tag like this:
<img src="${createLink(controller: 'myController', action: 'displayGraph')}"/>
Update:
After reading your question again, this may or may not work - it looks like you might be displaying the graph as the result of a form submission. This will only work if you're storing the state on the server somewhere (instead of just getting it from the one request where the form is submitted). If you do store enough state on the server to generate the graph, then you'd have to provide some additional parameters to your controller to get the correct image, e.g. src="${g.link(controller: 'myController', action: 'displayGraph', params: ['id': 1234])}", where id is how you retrieve the graph state.
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