我有三种方法的WCF服务.其中两个方法返回自定义类型(这些按预期工作),第三个方法将自定义类型作为参数并返回一个布尔值.通过 PHP SOAp客户端调用第三个方法时,它返回“未设置为对象实例的对象引用”异常.

示例自定义类型：

 _
公共类MyClass

PRivate _proPErtyA As Double
<Datamember()> _
Public Property PropertyA() As Double
    Get
        Return _propertyA
    End Get
    Set(ByVal value As Double)
        _propertyA = value
    End Set
End Property

Private _propertyB As Double
<DataMember()> _
Public Property PropertyB() As Double
    Get
        Return _propertyB
    End Get
    Set(ByVal value As Double)
        _propertyB = value
    End Set
End Property

Private _propertyC As Date
<DataMember()> _
Public Property PropertyC() As Date
    Get
        Return _propertyC
    End Get
    Set(ByVal value As Date)
        _propertyC = value
    End Set
End Property

结束班

方法：

公共函数Add(ByVal param As MyClass)As Boolean Implements IService1.Add
    ‘…
结束功能

PHP客户端调用：

$客户端 – >添加(阵列( ‘PARAM’=>阵列(
‘PropertyA’=> 1,
‘PropertyB’=> 2,
‘PropertyC’=> “2009-01-01”
)));

WCF服务适用于.Net客户端,但我是PHP的新手,无法使其工作.

是否可以在PHP中创建“MyClass”的实例.

任何帮助,将不胜感激.

注意：我使用的是PHP 5(适用于Windows的XAMPP 1.7.0).

谢谢

马特

$wsdl = "https://....../ServiceName.svc?wsdl";
$endpoint = "https://...../ServiceName.svc/endpointName";
$client = new SoapClient($wsdl,array('location'=>$endpoint));

$container = new stdClass();

$container->request->PropertyA = 'test 1';
$container->request->PropertyB = 'Test 2';
$container->request->PropertyC = '05/10/2010';

$response = $client->ServiceMethodA($container);

$container->request->OtherCustomType->Property1 = 'Test';