Hi Rob,
I don't know if anyone answered this, but you can use the following to get the
posted XML data and then load that into your XML processor.
$req = file_get_contents("php://input");
Regards,
Bevan
On Mon 11/05/09 11:21 AM , Robert martin r...@chreos.co.nz sent:
> Hi all
> I am wanting to send an XML file I have to a PHP script for
> processing. I am wanting to use TidHttp and have the following code
> snippet...
> var
> sResponse: String;
> fsParams: TFileStream;
> begin
> fsParams := TFileStream.Create('.test.xml', fmOpenRead or
> fmShareDenyWrite);
> try
> IdHTTP.Request.ContentType := 'text/xml';
> sResponse :=
> IdHTTP.Post('https://secure.dev.gateway.gov.uk/submission', fsParams);
> ShowMessage(sResponse);
> except
> on E: Exception do
> ShowMessage('Error encountered during POST: ' + E.Message);
> end;
> This was from a web site. Not mine !
> But I wonder what should happen at the PHP end. When my scripts
> processes usually things usually I go 'vardata = $_REQUEST[FieldName]
> to get the value. Than my XML I should do something like
> $xml = new SimpleXMLElement($vardata);
> echo $xml->movie[0]->plot;
> The question I have (which is Delphi related) is. My code doesn't
> pass the XML file as a param (does it?) how should I do this or am I
> misunderstanding.
> Sorry if I have not been clear enough.
> Thanks
> Rob
>
>
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