0
웹 서비스 요청 입력 매개 변수에 기본값을 지정할 수 있는지 궁금합니다. 나는이 같은 pensionType의 값을 확인 코드 측에서웹 서비스 요청 입력 매개 변수에 기본값을 할당 할 수 있습니까?
<element name="pensionType" default="0" type="int">
:
if (pensionType!=0)
{ TODO code here}
else
{ return warning that no data found}
내가 모질라 포스터이 테스트 다음은 WSDL 내 입력 매개 변수입니다. 그래서 이것으로 저는 pensionType에 어떤 가치도 제공하지 않으면 경고를받을 것이라고 예상했습니다.
System.Web.Services.Protocols.SoapException: Server was unable to read request. ---> System.InvalidOperationException: There is an error in XML document (13, 39). ---> System.FormatException: Input string was not in a correct format. at System.Number.StringToNumber(String str, NumberStyles options, NumberBuffer& number, NumberFormatInfo info, Boolean parseDecimal) at System.Number.ParseInt32(String s, NumberStyles style, NumberFormatInfo info) at System.Xml.XmlConvert.ToInt32(String s) at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationReader1.Read44_getFileTypesRequest(Boolean isNullable, Boolean checkType) at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationReader1.Read45_getFileTypes(Boolean isNullable, Boolean checkType) at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationReader1.Read62_getFileTypes() at Microsoft.Xml.Serialization.GeneratedAssembly.ArrayOfObjectSerializer24.Deserialize(XmlSerializationReader reader) at System.Xml.Serialization.XmlSerializer.Deserialize(XmlReader xmlReader, String encodingStyle, XmlDeserializationEvents events) --- End of inner exception stack trace --- at System.Xml.Serialization.XmlSerializer.Deserialize(XmlReader xmlReader, String encodingStyle, XmlDeserializationEvents events) at System.Xml.Serialization.XmlSerializer.Deserialize(XmlReader xmlReader, String encodingStyle) at System.Web.Services.Protocols.SoapServerProtocol.ReadParameters() --- End of inner exception stack trace --- at System.Web.Services.Protocols.SoapServerProtocol.ReadParameters() at System.Web.Services.Protocols.WebServiceHandler.CoreProcessRequest(