So, let me make sure I understand this right, you want to stream the
response, from the server, to the browser. (have something in the
controller display realtime like the print command, instead of display
all of the information at once)
Or, are you talking about actually using print, and having it output
to the page? For that I redirect stdout to a stringIO var, and then
simply read the variable out to the view.
On Feb 24, 11:33 am, AsmanCom <d.as...@web.de> wrote:
> Hi Tiango,
>
> no i need an equivalent for print, like:
>
> def print_1():
> print "1"
> print "2"
> =
> 1
> 2
>
> def print_2():
> while True:
> print "1"
> =
> 1
> 1
> 1
> 1
> ...
>
> I need to execute scripts like in terminal, but output rendered in
> html. (should be very simple?)
>
> Best regards,
>
> Dieter Asman
>
> - AsmanCom -
> Germany
>
> On 24 Feb., 18:17, Tiago Almeida <tiago.b.alme...@gmail.com> wrote:
>
>
>
> > > for quick development-shots, i need to execute my untouched(may i
> > > could change "print" to "return") Python scripts(..with loops) in the
> > > Controller.
> > > The output should be renderd direct via http, without long-winded
> > > setting up the view for each function.
> > > Is that possible?
>
> > If I understand you want to return a string directly without being processed
> > by the view.
> > On a controller function, if you return a string (not a dict) it will be
> > rendered as is.
>
> > Tiago
> > --
>
> > On Wed, Feb 24, 2010 at 4:48 PM, AsmanCom <d.as...@web.de> wrote:
> > > Hi,
>
> > > for quick development-shots, i need to execute my untouched(may i
> > > could change "print" to "return") Python scripts(..with loops) in the
> > > Controller.
> > > The output should be renderd direct via http, without long-winded
> > > setting up the view for each function.
> > > Is that possible?
>
> > > To my next concern....
>
> > > On my actual Project i have simple form field with submit button on my
> > > controller:
> > > form = SQLFORM.factory(Field('update_url', default='http://.domain.org/
> > > text_file.txt <http://domain.org/%0Atext_file.txt>',
> > > requires=IS_NOT_EMPTY()))
>
> > > On Submit the text file should be downloaded to Server indicated by an
> > > Progress Bar on the Page.
> > > Next the file should be inserted to the DB line by line(for loop with
> > > regex) and again indicated by the same Progress Bar .....
>
> > > My Code so far(... relies on clienttools example specifically on
> > > jqueryUI-progressbar and an urllib2 +report_hook snippet i´ve found
> > > somewhere):
> > > _______________________________________________________________________
> > > def chunk_report(bytes_so_far, chunk_size, total_size):
> > > percent = float(bytes_so_far) / total_size
> > > percent = round(percent*100)
> > > return percent
>
> > > if bytes_so_far >= total_size:
> > > return ('100.0')
>
> > > def chunk_read(response, chunk_size=8192, report_hook=None):
> > > total_size = response.info().getheader('Content-Length').strip()
> > > total_size = int(total_size)
> > > bytes_so_far = 0
>
> > > while 1:
> > > chunk = response.read(chunk_size)
> > > bytes_so_far += len(chunk)
>
> > > if not chunk:
> > > break
>
> > > if report_hook:
> > > report_hook(bytes_so_far, chunk_size, total_size)
>
> > > percent = float(bytes_so_far) / total_size
> > > percent = round(percent*100)
> > > return percent #bytes_so_far
>
> > > def url_progress():
> > > wrapper = DIV(progress,_style="width:400px;")
> > > jqueryui_progress()
> > > form = SQLFORM.factory(Field('update_url',
> > > default='http://.domain.org/text_file.txt', requires=IS_NOT_EMPTY()))
> > > if form.accepts(request.vars, session):
> > > response.flash = "Got it!"
> > > try:
> > > import urllib2
> > > openurl = urllib2.urlopen(request.vars.update_url)
> > > chunk_read(openurl, report_hook=chunk_report)
> > > except Exception, e:
> > > session.flash = DIV(T('Unable to download from url
> > > because:'),PRE(str(e)))
> > > redirect(URL(r=request))
> > > event.listen('submit',"form","return confirm('Are you sure?');") #
> > > adds a confirmation to submit
> > > page.ready(jq("update_oui").focus()())
> > > # return dict(form=form)
> > > return dict(wrapper=wrapper, form=form)
>
> > > # jQuery progress bar
> > > progress = DIV(_id="progress")
> > > def jqueryui_progress():
> > > wrapper = DIV(progress,_style="width:400px;")
> > > page.include("http://ajax.googleapis.com/ajax/libs/jqueryui/1.7.2/
> > > jquery-ui.min.js<http://ajax.googleapis.com/ajax/libs/jqueryui/1.7.2/%0Ajquery-ui.min.js>",
> > > download=True)
> > > page.include("http://ajax.googleapis.com/ajax/libs/jqueryui/1.7.2/
> > > themes/ui-darkness/jquery-ui.css<http://ajax.googleapis.com/ajax/libs/jqueryui/1.7.2/%0Athemes/ui-dark...>",
> > > download=True)
> > > callback = js.call_function(get_progress)
> > > page.ready(jq(progress).progressbar( dict(value='0') )() )
> > > page.ready(js.timer(callback,2000))
> > > return dict(wrapper=wrapper)
>
> > > def get_progress():
> > > return jq(progress).progressbar('option', 'value',
> > > request.now.second)()#
> > > ________________________________________________________________
>
> > > But now i stuck.... can anyone put me on the right way?
> > > When the Page is loaded, both elemts show up correctly.
> > > When i do the submit:
> > > - The Progressbar freezes(Callback dont work while downloading the
> > > file...)
> > > - The Download works and shows 100% when completed(but the while 1
> > > loop, which calls the report_hook dont work at all! and i need it to
> > > update the Progress bar... I think?)
>
> > > + I need to change the Progressbar callback source when updating DB
> > > after downloading the file...
>
> > > Anyone any suggestions?
>
> > > Best regards,
>
> > > Dieter Asman
>
> > > - AsmanCom -
> > > Germany
>
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