forget load in this case.
<div id="target">
</div>
<script>
var counter = 0;
var minutes = 2;
var max_calls = 10;
var url = "{{=URL('call')}}";
var f = function() {
counter = counter + 1;
jQuery.get(url).done(function(data) {
jQuery('#target').html(data);
if(counter<max_calls)
setTimeout(f, minutes*60*1000);
});
}
f();
</script>
On Tuesday, 12 July 2016 08:47:44 UTC-5, Chetan Jain wrote:
>
> URL : https://pqr.pythonanywhere.com/ajax/default/index
>
> Best,
> Chetan Jain
>
> On Tue, Jul 12, 2016 at 6:07 PM, Massimo Di Pierro <
> massimo.dipie...@gmail.com> wrote:
>
>> It is not clear. Where is your app?
>>
>>
>> On Tuesday, 12 July 2016 06:32:29 UTC-5, Chetan Jain wrote:
>>>
>>> Hi Anthony,
>>>
>>> i've created one 24 hours web2py application at pythonanywhere.com,
>>> please let me know when you have 10-15 minutes of time, so that i can
>>> rectify my problem.
>>> Thanks in advance.
>>>
>>>
>>>
>>>
>>> On Tue, Jul 12, 2016 at 7:43 AM, Anthony <abasta...@gmail.com> wrote:
>>>
>>>> If your view includes an {{=i}} variable, then your function must
>>>> return a dictionary with "i" as one of its keys. Hard to say what's wrong
>>>> without seeing the current version of this function.
>>>
>>>
>>>
>>>
>>> Best,
>>> Chetan Jain
>>>
>> --
>> Resources:
>> - http://web2py.com
>> - http://web2py.com/book (Documentation)
>> - http://github.com/web2py/web2py (Source code)
>> - https://code.google.com/p/web2py/issues/list (Report Issues)
>> ---
>> You received this message because you are subscribed to the Google Groups
>> "web2py-users" group.
>> To unsubscribe from this group and stop receiving emails from it, send an
>> email to web2py+unsubscr...@googlegroups.com.
>> For more options, visit https://groups.google.com/d/optout.
>>
>
>
--
Resources:
- http://web2py.com
- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
- https://code.google.com/p/web2py/issues/list (Report Issues)
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