使用 Spring-WS 时如何在客户端动态设置自定义 HTTP 标头（不是 SOAP 标头）？
问问题
47853 次
7 回答
31
public class AddHttpHeaderInterceptor implements ClientInterceptor {
public boolean handleFault(MessageContext messageContext)
throws WebServiceClientException {
return true;
}
public boolean handleRequest(MessageContext messageContext)
throws WebServiceClientException {
TransportContext context = TransportContextHolder.getTransportContext();
HttpComponentsConnection connection =(HttpComponentsConnection) context.getConnection();
connection.addRequestHeader("name", "suman");
return true;
}
public boolean handleResponse(MessageContext messageContext)
throws WebServiceClientException {
return true;
}
}
配置：
<bean id="webServiceTemplate" class="org.springframework.ws.client.core.WebServiceTemplate">
...
<property name="interceptors">
<list>
<bean class="com.blah.AddHttpHeaderInterceptor" />
</list>
</property>
</bean>
于 2011-07-28T21:26:02.663 回答
28
ClientInterceptor适用于静态标头值。但是当每个请求都应该应用不同的值时，就不可能使用它。在这种情况下WebServiceMessageCallback是有帮助的：
final String dynamicParameter = //...
webServiceOperations.marshalSendAndReceive(request,
new WebServiceMessageCallback() {
void doWithMessage(WebServiceMessage message) {
TransportContext context = TransportContextHolder.getTransportContext();
CommonsHttpConnection connection = (CommonsHttpConnection) context.getConnection();
PostMethod postMethod = connection.getPostMethod();
postMethod.addRequestHeader( "fsreqid", dynamicParameter );
}
}
于 2011-11-18T13:00:22.023 回答
12
在使用 spring integration 3 和 spring integration-ws 时，可以使用以下代码来处理请求：
public boolean handleRequest(MessageContext messageContext)
throws WebServiceClientException {
TransportContext context = TransportContextHolder.getTransportContext();
HttpUrlConnection connection = (HttpUrlConnection) context
.getConnection();
connection.getConnection().addRequestProperty("HEADERNAME",
"HEADERVALUE");
return true;
}
Interceptor 可以通过以下方式连接到出站网关：
<ws:outbound-gateway ...
interceptor="addPasswordHeaderInterceptor" >
</ws:outbound-gateway>
<bean id="addPasswordHeaderInterceptor class="com.yourfirm.YourHttpInterceptor" />
于 2014-05-28T13:32:41.683 回答
3
实际上，它是@Tomasz答案的更新版本，但提供了新的 Spring-WS API、Java 8 快捷方式，并关心WebServiceMessageCallback使用单独的方法创建实例。
我相信它更明显和自给自足。
final class Service extends WebServiceGatewaySupport {
/**
* @param URL the URI to send the message to
* @param payload the object to marshal into the request message payload
* @param headers HTTP headers to add to the request
*/
public Object performRequestWithHeaders(String URL, Object payload, Map<String, String> headers) {
return getWebServiceTemplate()
.marshalSendAndReceive(URL, payload, getRequestCallback(headers));
}
/**
* Returns a {@code WebServiceMessageCallback} instance with custom HTTP headers.
*/
private WebServiceMessageCallback getRequestCallback(Map<String, String> headers) {
return message -> {
TransportContext context = TransportContextHolder.getTransportContext();
HttpUrlConnection connection = (HttpUrlConnection)context.getConnection();
addHeadersToConnection(connection, headers);
};
}
/**
* Adds all headers from the {@code headers} to the {@code connection}.
*/
private void addHeadersToConnection(HttpUrlConnection connection, Map<String, String> headers){
headers.forEach((name, value) -> {
try {
connection.addRequestHeader(name, value);
} catch (IOException e) {
e.printStackTrace(); // or whatever you want
}
});
}
}
于 2017-06-24T13:00:39.530 回答
2
使用 java 1.8 的示例方法：如何添加 HTTP 标头：
public void executeObjectWebservice(String id) {
ExecuteObject request = new ExecuteObject();
getWebServiceTemplate().marshalSendAndReceive("http://url/webservice-test/uc4ws",
new ObjectFactory().createExecuteObject(request), new WebServiceMessageCallback() {
public void doWithMessage(WebServiceMessage message) throws IOException {
TransportContext context = TransportContextHolder.getTransportContext();
HttpUrlConnection connection = (HttpUrlConnection) context.getConnection();
connection.addRequestHeader("ID", id);
}
});
}
说明：使用 getWebServiceTemplate().marshalSendAndReceive，例如此处所述：https ://spring.io/guides/gs/sumption-web-service/
第一个参数是 URI，第二个参数是与请求一起发送的对象。作为第三个参数，您可以添加为函数
new WebServiceMessageCallback()
在哪里覆盖public void doWithMessage. 在发送请求之前调用此方法。您可以通过以下方式访问消息并添加请求标头
TransportContext context = TransportContextHolder.getTransportContext();
HttpUrlConnection connection = (HttpUrlConnection) context.getConnection();
connection.addRequestHeader("ID", id);
于 2018-08-24T15:16:28.407 回答
1
Spring 的 webServiceTemplate.marshalSendAndReceive(request) 方法内部使用 HttpComponentsMessageSender 通过网络发送 SOAP 消息，这进一步使用 WebServiceConnection 与服务器建立 http 连接。您所要做的就是编写自己的自定义 HttpComponentsMessageSender 并在 postMethod 中设置 cookie。
客户发件人代码：
package com.swap.ws.sender;
import java.io.IOException;
import java.net.URI;
import javax.annotation.Resource;
import org.apache.http.client.methods.HttpPost;
import org.apache.log4j.Logger;
import org.springframework.stereotype.Service;
import org.springframework.ws.transport.WebServiceConnect ion;
import org.springframework.ws.transport.http.HttpComponen tsConnection;
/**
*
* @author swapnil Z
*/
@Service("urlMessageSender")
public class CustomHttpComponentsMessageSender extends
org.springframework.ws.transport.http.HttpComponen tsMessageSender {
private static Logger _logger = Logger.getLogger("");
@Override
public WebServiceConnection createConnection(URI uri) throws IOException {
String cookie = null;
HttpComponentsConnection conn = (HttpComponentsConnection) super
.createConnection(uri);
HttpPost postMethod = conn.getHttpPost();
cookie = "<Your Custom Cookie>";
postMethod.addHeader("Cookie", cookie);
return conn;
}
}
弹簧配置：
<bean id="messageFactory" class="org.springframework.ws.soap.saaj.SaajSoapMe ssageFactory" />
<bean id="marshaller" class="org.springframework.oxm.jaxb.Jaxb2Marshalle r">
<property name="contextPath" value="com.swap.provision" />
</bean>
<bean id="webServiceTemplate" class="org.springframework.ws.client.core.WebServi ceTemplate">
<constructor-arg ref="messageFactory" />
<property name="marshaller" ref="marshaller"></property>
<property name="unmarshaller" ref="marshaller"></property>
<property name="messageSender" ref="urlMessageSender"/>
<property name="defaultUri" value=<Server URL> />
</bean>
在此之后，我只需获取 bean webServiceTemplate 并调用 marshalSendAndReceive 方法。因此，在进行 HTTP 调用之前，每个请求都会设置其自定义 cookie。
于 2014-04-03T16:23:42.190 回答
-1
以下片段已使用 Spring 4.0 进行了测试。它将 a 附加WebServiceMessageCallback 到 aorg.springframework.ws.client.core.WebServiceTemplate
final String DYNAMICVALUE = "myDynamo";
WebServiceMessageCallback wsCallback = new WebServiceMessageCallback() {
public void doWithMessage(WebServiceMessage message) {
try {
SoapMessage soapMessage = (SoapMessage)message;
SoapHeader header = soapMessage.getSoapHeader();
header.addAttribute(new QName("myHeaderElement"), DYNAMICVALUE);
} catch (Exception e) {
e.printStackTrace();
}
}
};
JAXBElement<MyWsResponse> response = (JAXBElement<MyWsResponse>)
wsTemplate.marshalSendAndReceive(MyWsOP, wsCallback);
于 2014-01-28T12:01:10.750 回答