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Sphinx: button inside literalinclude blocks

I have a desktop application and a user guide written in Sphinx. The API is explained with the use of code blocks that are imported with literalinclude blocks.
What I would like to have is a button inside or close to each code block. The button is intended for opening the respective example in our application.
What I have achieved so far (see below) is a button above the code block, that I have to include manually every time I use a literalinclude statement.
How can I put the button inside the code block rather than above?
A particularly beautiful result would be a transparent button like Chris Holdgraf's Sphinx copybutton. But at the moment I would already be happy to have the button as it is located in the top right corner of the code block.
Minimal example of how things are right now
The button is rendered above the code block. But I want to have it inside (at the top right corner).
Folder structure:
conf.py
index.rst
source_file.py
load_source.py
Content of index.rst:
:load_example:`source_file.py`
.. literalinclude:: source_file.py
Content of source_file.py:
import numpy as np
print "hello extension"
for i in range(3):
print np.sin(i)
Content of load_souce.py extension file:
from docutils import nodes
# HTML code to generate button
button_raw = """
<script>
function openScriptFunction() {
callBackHandler.openScript("%s");
}
</script>
<button onclick="openScriptFunction()">Load %s</button>
"""
def setup(app):
app.add_role('load_example', load_example_script)
def load_example_script(name, rawtext, text, lineno, inliner, options={}, content=[]):
node = nodes.raw(rawsource = button_raw%(text, text), text = button_raw%(text, text), format="html")
return [node], []
This extension is included in the conf.py by extensions = ['load_source']

How to Serve a Video (blob) with Google App Angine

I'm trying just for learning how to serve a video with the blobstore without it takes all the screen the video, for example
here I imported Video as video_model
class ViewVideo(webapp.Reque...,blobstore_handlers.BlobstoreDownloadHandler):
def get(self):
video_id = self.request.get('video_id')
video_instance = None
if video_id:
video_instance = video_model().get_video_content(video_id)
self.response.headers['Content-Type'] = 'video/mp4'
self.send_blob(video_instance.content.key())
class Video(db.Model):
content = blobstore.BlobReferenceProperty()
title = db.StringProperty()
def get_video(self,video_id):
return Video.get_by_id(video_id)
def get_video_content(self,content):
query_str = "SELECT * FROM Video WHERE content =:content"
return db.GqlQuery(query_str,content=content).get()
Where the video_id came from a url given, but as you see I put it directly in send_blob() function and this one when I tested it takes all the screen just to see the video, I was wondering how can I serve the video from my application without happening this, I was thinking embedded HTML but I can't figure it out how the source will be
Any help will be grateful
If it lacks of content to answer the question I will edit it

Without HTML5, it's a tricky mess. With HTML5, it becomes easy & elegant. Serve to the user's browser, as part of whatever page you're serving, the following HTML (5) snippet:
<video width="320" height="240" controls>
<source src="/getmp4?video_id=whatever" type="video/mp4">
Your browser does not support the video tag: please upgrade it!
</video>
and use that ViewVideo handler to serve only the /getmp4 URL, not the URL that your user directly gets via their browser.
The 320, 240, and the choice to show controls, are all optional, of course -- as even more is the use of whatever for the video id!-)

Edit response logo in web2py

I am trying to edit the default logo to my own in my MODEL(menu.py).
The default code for the logo is:
response.logo = A(B('web',SPAN(2),'py'),XML('™ '), _class="brand", _href="http://www.web2py.com/")
How would I change this line to display my own logo? For instance, the url for the logo might be something like "http://www.web2py.com/init/static/images/logo_lb.png"

Just change the content of the anchor tag to include the image:
response.logo = A(IMG(_src=URL('static', 'images/logo_lb.png'),
_href=URL('default', 'index'))
Note, you don't really need to use response.logo at all. You can instead just put the relevant HTML directly in the layout.html view.

Generating pdf using web2py-appreport (xhtmltopdf) in Python Web2py webapp

I am from a non coding background so python, web2py is very new to me.
My app needs to export textarea content (using RTE redactor) to pdf. I get html content from textarea (redactor), can you please advice me on how to use pyfpdf to generate a pdf file on button click.
I don't know how to get the html content (images and text) on button click in view to generate pdf using appreport.
I was able to use app-report to generate a pdf (using PISA, PYPDF does not work) from an existing html file (without css) if html file has css it throws an error,
***<class 'sx.w3c.cssParser.CSSParseError'> Terminal function expression expected closing ')':: (u'Alpha(Opacity', u'=0); }\n\n\n\n.ui-state-')***
This might be due to a mistake in the controller code:
def myreport():
html = response.render('myreport.html', dict())
return plugin_appreport.REPORTPISA(html = html)
Another thing I tried was passing the html from my view to the controller using ajax post (in Javascript). Redactor is the textarea RTE I am using and alert gives me the desired html result.
View:
function getContent() {
var t= jQuery('#redactor_content').getCode();
alert(t);
jQuery.ajax({
type: "POST",
url: "http://127.0.0.1:8000/Test50/default/myreport2",
data: "{g : 'jQuery('#redactor_content').getCode()'}"
});
}
Controller:
def myreport2():
g = request.get_vars
html = response.render(g)
return plugin_appreport.REPORTPISA(html = html)
Due to my less knowledge in coding , I am not able to figure out and correct my mistake. I will be thankful if anybody can help me with this problem.
Regards,
Akash

Could it be this post request:
jQuery.ajax({
type: "POST",
url: "http://127.0.0.1:8000/Test50/default/myreport2",
data: "{g : 'jQuery('#redactor_content').getCode()'}"
});
}
I think you should have the 'data' parameter be a literal dictionary, not a string. Change this line like this (remove all but one set of quotes):
data: {g : jQuery('#redactor_content').getCode() }
This should properly send the request. The jQuery documentation says that the data parameter should be key-value pairs, not a string.

How to loading/show a picture in a Python CGI page

I have written a little survey using Python and CGI. I am trying to show a picture using the normal <img> tag, But even-though the picture is in the same directory as my cgi script, my script cannot show it. I also changed the header to this:
print "Content-type: text/html; image/jpeg"
print
print """<html>
<head>
<title>You are going to be redirected</title>
</head>
<body bgcolor = #14b585>
<br>
<p align=center><font>Helloooo</font></p>
<img src="cat.jpeg" alt="cat" width="304" height="228"/>
<form action="./sample.py" method="post">
<p align=center><input type="submit" value="YES!" /></p>
</form>
</body>
</html>
"""
Why?(it is a very small jpg file)

print "Content-type: text/html; image/jpeg"
Don't change the header to that. You can't have multiple content types for a single document (multipart documents excluded, but browsers don't support them and that isn't the right format).
You are delivering an HTML document with an reference to an image in it. The image will be a separate request and response.
print "Content-type: text/html"
Or, better:
print "Content-type: text/html; charset=utf-8"
(Assuming you are using utf-8, which you should be).
print """<html>
Your Doctype is missing. This will trigger quirks mode, which is undesirable. You also have a great deal of legacy presentational markup that should be replaced by CSS.
<img src="cat.jpeg" alt="cat" width="304" height="228"/>
The context suggests that the image is decorative, so the alternative text should probably be "" (see Alt texts in IMGS), but there is nothing wrong with the actual reference to the image.
But even-though the picture is in the same directory as my cgi script
Since the HTML seems to be OK. You need to check that the image is.
Can you reach the image by typing the URL in directly?
What do the web server logs say when you try?
It is possible that your server is configured to try to treat everything in the directory as an executable, no matter what the file extension, so it might be trying to run the image as if it were a CGI program (which will obviously fail). If so, then you could either move the image or change the configuration of the server.
And I've just noticed this comment:
I did this in my browser: localhost/cgi-bin/cat.jpg and it got an error, I checked the logs, it Exec format error: exec of '/home/hossein/public_html/cgi-bin/cat.jpg' failed
That is what is happening. Moving the image is the simplest solution.

your apache was configured to use the cgi-bin directory as an CGI scripts folder, so any request that trying to get a file from this folder apache try to execute it as an CGI script. to make your image visible move it to the www/html folder.