No, now you're trying to put the whole image in the URL query string. Just
let the URL point to the render_image function, and create the image in
that function. You cannot create the image in the same action that produces
the URL. First you have to return an HTML page with an image element. Then
the browser requests the src of the image element. The image must be
fetched in a separate request -- you cannot embed it in the original HTML
page that is returned.
Anthony
On Monday, July 8, 2013 3:43:22 PM UTC-4, Lamps902 wrote:
>
> Maybe I'm a bit confused on how to generate/stream the image, given the
> buffer. The following should work, right?
>
> def render_image():
> return Image.open(request.vars.image_buffer)
>
> ...
> img_buffer = io.BytesIO()
> pylab.savefig(img_buffer, format = 'png')
> img_buffer.seek(0)
>
> return dict(img = IMG(_src=URL('render_image', vars=dict(image_buffer =
> img_buffer)), _alt = 'image buffer not displaying'))
>
> Or is there something other than Image.open() that's a better solution?
>
> On Monday, July 8, 2013 1:04:38 PM UTC-5, Anthony wrote:
>>
>>
>> On Monday, July 8, 2013 12:49:19 PM UTC-5, Anthony wrote:
>>>>
>>>> The IMG() helper simply creates an HTML <img> element to be displayed
>>>> in the browser. The "src" attribute of an image element should be a URL
>>>> where the browser can request the image. You cannot pass an actual image
>>>> file object to the helper.
>>>
>>>
>> Just make the "src" URL an action that generates the image and streams
>> it.
>>
>
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