vb.net Post login data httpwebrequest and cookies -
im trying login website, works wont redirect when logs in. here code
dim request webrequest = webrequest.create("http://www.jamiehayles.com/test/login.php") request.method = "post" dim postdata string = "username=testing&password=lmao1234" dim bytearray byte() = encoding.utf8.getbytes(postdata) request.contenttype = "application/x-www-form-urlencoded" request.contentlength = bytearray.length dim datastream stream = request.getrequeststream() datastream.write(bytearray, 0, bytearray.length) datastream.close() dim response webresponse = request.getresponse() console.writeline(ctype(response, httpwebresponse).statusdescription) datastream = response.getresponsestream() dim reader new streamreader(datastream) dim responsefromserver string = reader.readtoend() richtextbox1.text = responsefromserver reader.close() datastream.close() response.close()
my question is, need cookies redirect? , if guys?
take @ you're doing response server:
dim responsefromserver string = reader.readtoend() richtextbox1.text = responsefromserver
all you're doing reading , placing content of response text box. you're not responding redirect in way.
a "redirect" means server responds specific status code , header in response. status code tells browser response instruction redirect resource. header contains new resource browser should request.
a web browser checks response code/header , performs new request. code doesn't. shows response user, regardless of is.
so, illustrate, normal redirect happens this:
- browser: i'm requesting resource.
- server: ok, i've received request, resource want on there.
- browser: ok, i'm requesting resource on there.
- server: ok, here go.
your code, however, more this:
- code: i'm requesting resource.
- server: ok, i've received request, resource want on there.
- code: [no further action taken]
the server instructed code, in http-standard way, request new resource. code ignored instruction.
you need check the status code of response, the headers of response, , make new requests accordingly.
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