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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