Hmm, i've visited your links but still I don't know what to do, my knowledge on this area is very limited. As for your reply about not using cache's own webserver: I can't change that situation now. But csp (and weblink)have been installed on the iis-webserver and througt that server the cache's database server is being apporached.Apart from this rest-call. So maybe the problem is that I use a rest-url straight to the cache-server in stead of using the webserver? Best regards, Simon
S dataToProcess=%request.Content.Read(32000) s dataToJSON=$TR(dataToProcess,"'"," ") s dataToJSON=$TR(dataToProcess,"""","'") s ^simonupload(2)=dataToJSON S dataToJSON=##class(Util.JSON).Decode(dataToJSON) s ^simonupload(3)=dataToJSON s ^simonupload(4)=dataToJSON.GetAt("Opmerkingen") s ^simonupload(5)=dataToJSON.GetAt("Bijlagen").GetAt(1) quit $$$OK }
go to post
Thanks a lot!
go to post
Removed post. (Made stupid mistake.) Thanks Timothy!
go to post
True: mixed-content.
So I'll just hade to find out how to redirect an url to CSP when not using .cls in the url.
go to post
Yes indeed.
go to post
Hmm, i've visited your links but still I don't know what to do, my knowledge on this area is very limited.
As for your reply about not using cache's own webserver: I can't change that situation now. But csp (and weblink)have been installed on the iis-webserver and througt that server the cache's database server is being apporached.Apart from this rest-call. So maybe the problem is that I use a rest-url straight to the cache-server in stead of using the webserver? Best regards, Simon
go to post
Thanks to you all, I got it working now.
Best regards,
Simon
go to post
Where should I get the output HTML-page?
Thanks,
Simon
go to post
Hi Henrique,
I don't have any code for this api yet but I do have one for (for example) JSON:
<Route Url="/newOrder" Method="post" Call="newOrder"/>
ClassMethod newOrder() As %Status {
S dataToProcess=%request.Content.Read(32000)
s dataToJSON=$TR(dataToProcess,"'"," ")
s dataToJSON=$TR(dataToProcess,"""","'")
s ^simonupload(2)=dataToJSON
S dataToJSON=##class(Util.JSON).Decode(dataToJSON)
s ^simonupload(3)=dataToJSON
s ^simonupload(4)=dataToJSON.GetAt("Opmerkingen")
s ^simonupload(5)=dataToJSON.GetAt("Bijlagen").GetAt(1)
quit $$$OK
}
This works for receiving JSON.
But I don't know how to do it with Form Data.