Python Web Code de service:service web Appel Python en utilisant PHP
import web
from soaplib.wsgi_soap import SimpleWSGISoapApp
from soaplib.service import soapmethod
from soaplib.serializers import primitive as soap_types
urls = ("/hello", "HelloService",
"/hello.wsdl", "HelloService",
)
render = web.template.Template("$def with (var)\n$:var")
class SoapService(SimpleWSGISoapApp):
"""Class for webservice """
@soapmethod(soap_types.String,_returns=soap_types.String)
def hello(self,message):
""" Method for webservice"""
return "Hello world "+message
class HelloService(SoapService):
"""Class for web.py """
def start_response(self,status, headers):
web.ctx.status = status
for header, value in headers:
web.header(header, value)
def GET(self):
response = super(SimpleWSGISoapApp, self).__call__(web.ctx.environ, self.start_response)
return render("\n".join(response))
def POST(self):
response = super(SimpleWSGISoapApp, self).__call__(web.ctx.environ, self.start_response)
return render("\n".join(response))
app=web.application(urls, globals())
if __name__ == "__main__":
app.run()
Code PHP:
<?php
@ini_set("soap.wsdl_cache_enabled", "0");
$client = new SoapClient('http://localhost:8080/hello.wsdl');
echo("<pre>");
var_dump($client->__getFunctions());
echo("</pre>");
$params = array('World');
try {
print_r($client->__soapCall('hello', $params));
} catch (SoapFault $exception) {
echo $exception;
}
?>
Quand je lance mon code php, il rapporte des informations ci-dessous
SoapFault exception: [helloFault] hello() prend exactement 2 arguments (1 donné) dans C: \ website \ cosmétique \ src \ test02.php: 15 Trace de pile: # 0 C: \ website \ cosmétique \ src \ test02.php (15): SoapClient -> __ soapCall ('bonjour', Tableau) # 1 {main}
Comment résoudre ce problème? Merci.
Pour regarder WSDL (hello.wsdl) dans ce cas est utile. Postez-le. – dmitry