我正在研究/比较CXF和Spring-WS的Web服务?我需要既作为WS的提供者又作为消费者.简而言之,我被告知Spring-WS更易于配置,但CXF更容易启动和运行.这个问题是主观的,但有助于我的研究.
我正在使用Spring WebServiceTemplate进行webservice调用,该调用使用JAXB生成请求XML.我的要求需要所有元素(包括root)在SOAP请求中具有名称空间前缀(只有一个名称空间).
例如:
<ns1:Login xmlns:ns1="www.example.com/a">
<ns1:username>abc</ns1:username>
<ns1:password>abc</ns1:password>
</ns1:Login>
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但我得到了
<Login xmlns="www.example.com/a">
<username>abc<username>
<password>abc<password>
</Login>
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xsd:
<?xml version="1.0" encoding="UTF-8"?>
<xs:schema targetNamespace="www.example.com/a" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:ilreq="www.example.com/a" elementFormDefault="qualified" attributeFormDefault="unqualified">
<xs:complexType name="Login">
<xs:sequence>
<xs:element name="username" type="xs:string"/>
<xs:element name="password" type="xs:string"/>
</xs:sequence>
</xs:complexType>
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从XSD生成的Java类
@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "Login", propOrder = {
"username",
"password"
})
@XmlRootElement
public class Login {
@XmlElement(required = true)
protected String username;
@XmlElement(required = true)
protected String password;
......
}
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package-info.java
@javax.xml.bind.annotation.XmlSchema(
namespace = "www.example.com/a",
elementFormDefault = javax.xml.bind.annotation.XmlNsForm.QUALIFIED)
package authenticator.beans.login;
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想知道如何使用Namespace前缀生成包含root的所有元素的请求XML.
我正在尝试向现有的Web服务发送请求.此网络服务不受我的约束.此Web服务的安全策略要求我在SOAP请求中发送完整的证书链.我的证书链包含3个证书.证书链的设置没有问题,因为我能够测试它的有效性(并且已经这样做了).
此设置的安全配置(=在请求中发送完整的证书链)是:
<xwss:Sign id="signature">
<xwss:X509Token
certificateAlias="alias"
keyReferenceType="Direct"
valueType="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-x509-token-profile-1.0#X509PKIPathv1" />
</xwss:Sign>
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我试图通过Spring-WS实现这一目标.Spring-WS使用spring-ws-security来保证安全性.Spring-ws-security代表xws-security.
<dependency>
<groupId>org.springframework.ws</groupId>
<artifactId>spring-ws-security</artifactId>
<version>2.1.0.RELEASE</version>
<exclusions>
<exclusion>
<groupId>org.apache.ws.security</groupId>
<artifactId>wss4j</artifactId>
</exclusion>
<exclusion>
<groupId>com.sun.xml.wsit</groupId>
<artifactId>xws-security</artifactId>
</exclusion>
</exclusions>
</dependency>
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Xws-security有两种版本:
<dependency>
<groupId>com.sun.xml.wsit</groupId>
<artifactId>xws-security</artifactId>
<version>1.3.1</version>
</dependency>
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和
<dependency>
<groupId>com.sun.xml.wss</groupId>
<artifactId>xws-security</artifactId>
<version>3.0</version>
</dependency>
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第一个由Spring WS Security使用.第二个是遗产.
在xws-security中应用我的XWSS配置是在名为BinarySecurityToken的类中完成的.BinarySecurityToken有一个名为的字段
valueType
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valueType的JavaDoc表示它支持X509PKIPathv1(以及其他).但是,正如这个二传手所说:
protected void setValueType(String valueType) {
if (!(MessageConstants.X509v3_NS.equals(valueType)||MessageConstants.X509v1_NS.equals(valueType))) {
log.log(Level.SEVERE,"WSS0342.valtype.invalid");
throw new RuntimeException("Unsupported value type: " + valueType);
}
this.valueType = valueType;
}
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MessageConstants类(甚至)没有X509PKIPathv1的静态.当我运行我的代码时,我得到了预期的结果:
Unsupported value type: http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-x509-token-profile-1.0#X509PKIPathv1
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我能够查看遗产的源代码com.sun.xml.wss.xws-security:3.0
.尽管我付出了努力,但我还没有找到它的源代码com.sun.xml.wsit.xws-security-1.3.1
.不过我相信代码是一样的.我尝试了两个库,两者都给了我相同的例外.我尝试了它,使用默认的spring-ws-security并对两个库使用显式依赖声明(一次一个).
我的问题:
我正在org.springframework.ws.client.core.WebServiceTemplate
用于进行Web服务调用.如何配置呼叫超时.
在使用Spring-WS时,如何在客户端动态设置自定义HTTP头(而不是SOAP头)?
我有一个基于Maven的Spring-WS客户端项目,我想打包成一个jar.在eclipse中,一切都运行正常.当我尝试将其打包为可执行jar时,我得到ClassNotFound异常,因为Spring jar没有包含在我的应用程序jar中.
所以我添加了maven-shade-plugin来包含我的应用程序jar中的所有依赖项.当我查看我的app jar时,我看到包含所有依赖项的所有类文件(所有库jar都被爆炸).
<build>
<plugins>
<plugin>
<artifactId>maven-compiler-plugin</artifactId>
<configuration>
<source>1.6</source>
<target>1.6</target>
</configuration>
</plugin>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-jar-plugin</artifactId>
<version>2.4</version>
<configuration>
<archive>
<manifest>
<addClasspath>true</addClasspath>
<mainClass>com.cws.cs.Client</mainClass>
</manifest>
</archive>
</configuration>
</plugin>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-shade-plugin</artifactId>
<version>1.7</version>
<executions>
<execution>
<phase>package</phase>
<goals>
<goal>shade</goal>
</goals>
</execution>
</executions>
</plugin>
</plugins>
</build>
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我的问题是在打包过程中,我的多个spring依赖项有不同的META-INF/spring.schemas文件相互覆盖.因此,我的最后一个jar有一个不完整的spring.schemas文件.
因此,当我尝试运行我的可执行jar时,我收到Spring错误消息,因为spring.schemas文件不完整(Spring-WS的jar覆盖了Spring-core的spring.schemas文件),因此无法找到文件.
我的可执行jar的META-INF/spring.schemas:
http\://www.springframework.org/schema/web-services/web-services-1.5.xsd=/org/springframework/ws/config/web-services-1.5.xsd
http\://www.springframework.org/schema/web-services/web-services-2.0.xsd=/org/springframework/ws/config/web-services-2.0.xsd
http\://www.springframework.org/schema/web-services/web-services.xsd=/org/springframework/ws/config/web-services-2.0.xsd
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而不是Spring-beans.jar META-INF/spring.schemas:
http\://www.springframework.org/schema/beans/spring-beans-2.0.xsd=org/springframework/beans/factory/xml/spring-beans-2.0.xsd
http\://www.springframework.org/schema/beans/spring-beans-2.5.xsd=org/springframework/beans/factory/xml/spring-beans-2.5.xsd
http\://www.springframework.org/schema/beans/spring-beans-3.0.xsd=org/springframework/beans/factory/xml/spring-beans-3.0.xsd
http\://www.springframework.org/schema/beans/spring-beans-3.1.xsd=org/springframework/beans/factory/xml/spring-beans-3.1.xsd
http\://www.springframework.org/schema/beans/spring-beans.xsd=org/springframework/beans/factory/xml/spring-beans-3.1.xsd
http\://www.springframework.org/schema/tool/spring-tool-2.0.xsd=org/springframework/beans/factory/xml/spring-tool-2.0.xsd
http\://www.springframework.org/schema/tool/spring-tool-2.5.xsd=org/springframework/beans/factory/xml/spring-tool-2.5.xsd
http\://www.springframework.org/schema/tool/spring-tool-3.0.xsd=org/springframework/beans/factory/xml/spring-tool-3.0.xsd
http\://www.springframework.org/schema/tool/spring-tool-3.1.xsd=org/springframework/beans/factory/xml/spring-tool-3.1.xsd
http\://www.springframework.org/schema/tool/spring-tool.xsd=org/springframework/beans/factory/xml/spring-tool-3.1.xsd
http\://www.springframework.org/schema/util/spring-util-2.0.xsd=org/springframework/beans/factory/xml/spring-util-2.0.xsd
http\://www.springframework.org/schema/util/spring-util-2.5.xsd=org/springframework/beans/factory/xml/spring-util-2.5.xsd
http\://www.springframework.org/schema/util/spring-util-3.0.xsd=org/springframework/beans/factory/xml/spring-util-3.0.xsd
http\://www.springframework.org/schema/util/spring-util-3.1.xsd=org/springframework/beans/factory/xml/spring-util-3.1.xsd
http\://www.springframework.org/schema/util/spring-util.xsd=org/springframework/beans/factory/xml/spring-util-3.1.xsd
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我很难过.我不确定是否/如何将所有内容打包为单个可执行jar.我不知道这是一个阴影插件配置问题,还是我试图做一些不可能的事情.我不得不手动创建自己的spring.schemas文件(其他的串联)似乎是不正确的.
我可能有点跳了一下枪.在挖掘插件的更多信息时,我注意到我之前错过的AppendingTransformer.但是,我关心的是如何知道哪些其他文件有同样的问题?我发现/抓住了这个特殊的Spring问题.我不知道任何其他可能正在做类似事情的库......
任何建议,将不胜感激.
使用spring ws获取StreamResult,如下所示
StreamSource source = new StreamSource(new StringReader(MESSAGE));
StreamResult result = new StreamResult(System.out);
webServiceTemplate.sendSourceAndReceiveToResult("http://someUri",
source, new SoapActionCallback("someCallBack"), result);
return result;
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我得到了结果,但我想把它提取到某种xml甚至是字符串(只是想看到内容以便生成响应).
我怎样才能做到这一点?
Soap UI具有以下选项"使用单个证书进行签名"
什么是相应的spring-ws配置?
我正在使用jdbcTemplate.queryForList以下列方式执行命名查询:
List<Conversation> conversations = jdbcTemplate.queryForList(
SELECT_ALL_CONVERSATIONS_SQL_FULL,
new Object[] {userId, dateFrom, dateTo});
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SQL查询是:
private final String SELECT_ALL_CONVERSATIONS_SQL_FULL =
"select conversation.conversationID, conversation.room, " +
"conversation.isExternal, conversation.startDate, " +
"conversation.lastActivity, conversation.messageCount " +
"from openfire.ofconversation conversation " +
"WHERE conversation.conversationid IN " +
"(SELECT conversation.conversationID " +
"FROM openfire.ofconversation conversation, " +
"openfire.ofconparticipant participant " +
"WHERE conversation.conversationID = participant.conversationID " +
"AND participant.bareJID LIKE ? " +
"AND conversation.startDate between ? AND ?)";
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但是,当以下列方式提取列表的内容时:
for (Conversation conversation : conversations) {
builder.append(conversation.getId());
builder.append(","); …
Run Code Online (Sandbox Code Playgroud) 我计划使用spring 4和java 7为复杂的高性能导向应用程序构建一个ResT完整Web服务.经过研究,我发现了以下选项.
我的选择是Spring WS,但Spring MVC REST不符合JAX-RS(如果我没有错).源 - Spring MVC REST不符合JAX-RS.有关系吗?
题:
spring-ws ×10
java ×9
spring ×5
cxf ×1
header ×1
http ×1
jax-rs ×1
jaxb ×1
jdbctemplate ×1
maven ×1
rest ×1
soap ×1
soapui ×1
spring-mvc ×1
web-services ×1
ws-security ×1
x509 ×1
xml ×1