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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