我有WSDL URL http://mycompany.com:8081/isapi/MyService.dll
的Web服务URL http://mycompany.com:8081/isapi/MyService.dll/wsdl/IMyService
,我在WSDL中定义了以下操作:
<operation name="test">
<input message="tns:test4Request"/>
<output message="tns:test4Response"/>
</operation>
我想从Postman或其他类似工具执行/调用此测试程序(甚至浏览器可能对此GET请求有用) - 但我不知道要调用哪个URL。如何从Web服务本身的URL获取Web服务过程的URL?
我在Delphi中创建了Web服务,我想从C#中使用它,但是收到错误消息:
client found response content type of 'text/html; charset=utf-8', but expected 'text/xml'
我试过了:
http://mycompany.com:8081/isapi/MyService.dll/test?Value=OK
但响应不包含正文,它只包含标题:
Connection →Keep-Alive
Content-Length →0
Content-Type →application/x-msdownload
Date →Mon, 14 Nov 2016 09:24:47 GMT
Keep-Alive →timeout=5, max=99
Server →Apache/2.4.17 (Win32) OpenSSL/1.0.2d PHP/5.6.23