how to format table.ajax.params() to same format as a normal ajax request?

gwiqu

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

allan

I’m not sure what you mean - in that URL, you have a dt parameter which contains JSON. In PHP for example you could use:

json_decode( $_GET[‘dt’] );

Not that you should use encodeURIComponent() to make the dt value URL safe there.

Or do you mean you don’t want to send params as JSON? If so, you could use jQuery’s param method.

Allan

gwiqu

@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

allan

I'm with you now thanks. Your want the jQuery params method I mentioned in that case.

Allan