matplotlib passing parameters from the view

[Paolo Amboni]

I'm trying to build a single function (in a controller) that generate different matplotlib graph..

So the function that generate the stream works fine:

def grafico(title='title',xlab='x',ylab='y' data=.... ):

     fig=Figure()

     ....

     return stream.getvalue()

the function that call the privious works fine but not as i want:

def graf_comp_gior():

    test_id=1 # I need to set the id on the fly

    print response.args # Always none why?

    print response.vars #Always none why?

    rows = db(db.tat.id_test==test_id).select()

    tat=pd.DataFrame(rows.as_list())

    .... #manupulate the data as i need

    graf=grafico(........)

    

    return graf

This is the view with the embedded graph works fine:

{{{extend 'layout.html'}}

<h1>TAT completi per l'analisi {{=variabili['anal']}}</h1>

{{=IMG(_src=URL('graf_comp_gior',args=[argomenti[0],vars=dict('id'=argomenti[0])]))}} How can i pass some data to the previous function?

Do i need to store them in a session?

Thanks!!

[Paolo Amboni]

I stored the id in a session variable and it works fine!!

But why the URL helper didn't send the args to the function?

[Niphlod]

there's a problem in your syntax.

it's usually URL('blabla', args=[], vars=dict())

while your code reports

args=[..., dict=()]

[Paolo Amboni]

Same problem!

the args and var are not passed to the receiving function in this specific case.

(I tried both args and vars alone first).

With session all works fine.

I find very complex the way web2py handle matplotlib plots.

Is there other plotting library that better integrate in web2py framework? 

[Niphlod]

if your generated url is /a/c/f/graf_comp_gior?var1=a&var2=b I assure you that those will be passed.

as for the graphing library, I highly suggest to switch to a javascript based solution rather than a python one for any web site.