http://127.0.0.1:8000/appname/default/form <http://127.0.0.1:8000/admin/default/design/child> On Tuesday, 2 July 2013 12:01:54 UTC+5:30, Anthony wrote: > > Your code includes: > > web2py_component("{{=URL('default', 'form.load')}}" + "/" + > jQuery(this).val(), target='form') > > That will issue an ajax request. I was asking what URL ends up getting > requested by the browser. > > Anthony > > On Thursday, June 27, 2013 1:12:00 AM UTC-4, Rohitraj Sharma wrote: >> >> >> Hi Anthony >> >> I am not able to get your question. I have one form controller with one >> view. http://127.0.0.1:8000/tracker/default/form when i run this url its >> show 404 error. >> Actually what is my requirement, I have one csv file. which i have to >> upload. Before saving the file in database, i want to edit in browser. >> After editing i want to save in database. Is there any method to do. >> Thanks >> >> On Wednesday, 26 June 2013 19:09:00 UTC+5:30, Anthony wrote: >>> >>> Using the browser developer tools, what URL gets requested when the ajax >>> call is made? >>> >>> On Wednesday, June 26, 2013 2:44:18 AM UTC-4, Rohitraj Sharma wrote: >>>> >>>> I want to select some particular table from database in a drop dowan. >>>> Then import csv file. edit that file and update. . Can Any one can help me >>>> plz >>>> >>>> i am using the fallowing code for that >>>> >>>> >>>> View >>>> >>>> <script> >>>> jQuery(function() { >>>> jQuery('#table').change(function() { >>>> web2py_component("{{=URL('default', 'form.load')}}" + "/" + >>>> jQuery(this).val(), target='form') >>>> })})</script> >>>> {{=SELECT('Select a table', *db.tables, _id='table')}}<div id="form"></div> >>>> >>>> >>>> Controller >>>> >>>> def form(): >>>> if request.args(0) in db.tables: >>>> response.generic_patterns = ['load'] >>>> return dict(form=SQLFORM(db[request.args(0)]).process()) >>>> else: >>>> raise HTTP(404) >>>> >>>> >>>> every time it toll else part. >>>> >>>> >>>> can any one help me. >>>> >>>>
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