how to format table.ajax.params() to same format as a normal ajax request?
how to format table.ajax.params() to same format as a normal ajax request?
gwiqu
Posts: 8Questions: 4Answers: 1
var params = table.ajax.params();
window.location.href = './ServerSide.php?ExportToExcel=Yes&dt='+JSON.stringify( params );
im following this code, but the code that gets sent to my server does not follow the format of a normal AJAX request, is there anyway i can convert "params" to the same format?
im using window.location.href as i am creating a button to download the filtered data table
Answers
I’m not sure what you mean - in that URL, you have a
dtparameter which contains JSON. In PHP for example you could use:Not that you should use
encodeURIComponent()to make thedtvalue URL safe there.Or do you mean you don’t want to send
paramsas JSON? If so, you could use jQuery’s param method.Allan
@allan im currently using python as my backend with flask and if i want to get the search value in a normal ajax request for the data table generation, i could use
search = request.args.get('search[value]')
but for
window.location.href = './ServerSide.php?ExportToExcel=Yes&dt='+JSON.stringify( params );
i need to use
json.loads(request.args.get('dt'))['search']['value']
instead, i was wondering if i could change the format to match so that i can combine my code into a single function instead of having what is essentially the same code in 2 places
I'm with you now thanks. Your want the jQuery params method I mentioned in that case.
Allan