Oh, and also, you'll need to take out the second last line in the
python script.
That's for a Firefox/Python bridge.
On Aug 4, 1:51 pm, The Quill <[EMAIL PROTECTED]> wrote:
> I should have edited this before posting it. I'm not sure whether the
> problem is with the Javascript or with the Python script.
>
> Thanks.
>
> On Aug 4, 1:47 pm, The Quill <[EMAIL PROTECTED]> wrote:
>
> > Hi there,
>
> > I'm trying to get AJAX POST to work properly using a local web.py
> > server. It seems that I can create a successful POST using the first
> > AJAX call. The second call fails, and then every subsequent call
> > works! If I stop the server and restart, I get the same failure on the
> > second call. I'm thinking this is a problem with my python code, but
> > if not, could someone out there please correct me?
>
> > Try it for yourself and you'll see what I mean. Also, you'll need to
> > test this in Firefox and you might need to change the port to suit
> > your configuration.
> > Here's my HTML file:
> > ******************************************************************
> > <html>
> > <head>
> > <title>Firefox to Python and back...</title></head>
> > <script>
> > function alertContents() {
> > if (http_request.readyState == 4) {
> > if (http_request.status == 200) {
> > var string = http_request.responseText;
> > alert(string);
> > } else {
> > alert('There was a problem with the request.');
> > }
> > }
> > POSTDATA = "";
> > }
>
> > var testvalue1 = '5';
> > var testvalue2 = 'Hello World!';
> > var POSTDATA = 'test1=' + escape(testvalue1) + '&test2=' +
> > escape(testvalue2);
> > function makeRequest(url, parameters) {
>
> > try {
>
> > netscape.security.PrivilegeManager.enablePrivilege("UniversalBrowserRead");
> > } catch (e) {
> > alert("Permission UniversalBrowserRead denied.");
> > }
>
> > http_request = false;
> > http_request = new XMLHttpRequest();
> > if (!http_request) {
> > alert('Cannot create XMLHTTP instance');
> > return false;
> > }
> > http_request.open('POST', url, true);
> > http_request.setRequestHeader('content-type', 'text/plain');
> > http_request.onreadystatechange = alertContents;
> > http_request.send(POSTDATA);
> > }
> > </script>
> > <body>
> > <a href="#" onClick='makeRequest("http://127.0.0.1:8080/
> > view","");'>makeRequest</a>
> > </body>
> > </html>
> > ******************************************************************
>
> > Here's my PYTHON file: app.py
> > ******************************************************************
> > import web, os, sys
>
> > #load urls and matching classes
> > urls = (
> > '/view', 'view',
> > '/test', 'test',
> > '/quit','quit'
> > )
> > class view:
> > def POST(self):
> > print('Hello Test!')
> > web.internalerror = web.debugerror
> > if __name__ == "__main__":
> > os.spawnv(os.P_NOWAIT,'runxulapp.bat',['',''])
> > web.run(urls, globals())
> > ******************************************************************
>
> > TIA.
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