如何让 @WebService spring 感知

发布于 2024-10-17 18:53:26 字数 2475 浏览 8 评论 0原文

我有一个 Web 服务,我正在尝试将变量自动装配到其中。以下是该类:

package com.xetius.isales.pr7.service;

import java.util.Arrays;
import java.util.List;

import javax.jws.WebService;

import org.springframework.beans.factory.annotation.Autowired;

import com.xetius.isales.pr7.domain.PR7Product;
import com.xetius.isales.pr7.domain.PR7Upgrade;
import com.xetius.isales.pr7.logic.UpgradeControllerInterface;

@WebService(serviceName="ProductRulesService",
            portName="ProductRulesPort",
            endpointInterface="com.xetius.isales.pr7.service.ProductRulesWebService",
            targetNamespace="http://pr7.isales.xetius.com")
public class ProductRulesWebService implements ProductRulesWebServiceInterface {

    @Autowired
    private UpgradeControllerInterface upgradeController;

    @Override
    public List<PR7Product> getProducts() {
        if (upgradeController == null) {
            return Arrays.asList(new PR7Product("Fail"));
        }
        return upgradeController.getProducts();
    }

    @Override
    public List<PR7Upgrade> getUpgrades() {
        if (upgradeController == null) {
            return Arrays.asList(new PR7Upgrade("Fail"));
        }
        return upgradeController.getUpgrades();
    }

    @Override
    public List<PR7Product> getProductsForUpgradeWithName(String upgradeName) {
        if (upgradeController == null) {
            return Arrays.asList(new PR7Product("Fail"));
        }
        return getProductsForUpgradeWithName(upgradeName);
    }

}

但是,当我尝试访问 Web 服务时,我收到返回的失败版本,这意味着upgradeController 没有被自动连接。这是我的 applicationContext:

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="http://www.springframework.org/schema/beans
                        http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
                        http://www.springframework.org/schema/context
                        http://www.springframework.org/schema/context/spring-context-3.0.xsd">
    <context:component-scan base-package="com.xetius.isales.pr7" />
    <context:annotation-config />

    <bean id="upgradeController" class="com.xetius.isales.pr7.logic.UpgradeController" />

</beans>

如何使 @WebService 类能够感知 Spring 并进行自动装配

I have a Web Service which I am trying to Autowire a variable into. Here is the class:

package com.xetius.isales.pr7.service;

import java.util.Arrays;
import java.util.List;

import javax.jws.WebService;

import org.springframework.beans.factory.annotation.Autowired;

import com.xetius.isales.pr7.domain.PR7Product;
import com.xetius.isales.pr7.domain.PR7Upgrade;
import com.xetius.isales.pr7.logic.UpgradeControllerInterface;

@WebService(serviceName="ProductRulesService",
            portName="ProductRulesPort",
            endpointInterface="com.xetius.isales.pr7.service.ProductRulesWebService",
            targetNamespace="http://pr7.isales.xetius.com")
public class ProductRulesWebService implements ProductRulesWebServiceInterface {

    @Autowired
    private UpgradeControllerInterface upgradeController;

    @Override
    public List<PR7Product> getProducts() {
        if (upgradeController == null) {
            return Arrays.asList(new PR7Product("Fail"));
        }
        return upgradeController.getProducts();
    }

    @Override
    public List<PR7Upgrade> getUpgrades() {
        if (upgradeController == null) {
            return Arrays.asList(new PR7Upgrade("Fail"));
        }
        return upgradeController.getUpgrades();
    }

    @Override
    public List<PR7Product> getProductsForUpgradeWithName(String upgradeName) {
        if (upgradeController == null) {
            return Arrays.asList(new PR7Product("Fail"));
        }
        return getProductsForUpgradeWithName(upgradeName);
    }

}

However, when I try to access the web service I am getting the Fail version returned, meaning that upgradeController is not being autowired. Here is my applicationContext:

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="http://www.springframework.org/schema/beans
                        http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
                        http://www.springframework.org/schema/context
                        http://www.springframework.org/schema/context/spring-context-3.0.xsd">
    <context:component-scan base-package="com.xetius.isales.pr7" />
    <context:annotation-config />

    <bean id="upgradeController" class="com.xetius.isales.pr7.logic.UpgradeController" />

</beans>

How do I make it so that the @WebService class is spring aware and autowiring happens

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。

评论(3

不语却知心 2024-10-24 18:53:26

如果您希望自动装配,ProductRulesWebService 需要扩展 SpringBeanAutowiringSupport

扩展该类将允许 UpgradeController 自动装配

If you want autowiring to happen, ProductRulesWebService needs to extend SpringBeanAutowiringSupport

Extending that class will allow UpgradeController to be autowired

不打扰别人 2024-10-24 18:53:26

使用像 CXF 这样的堆栈,它原生支持 Spring,那么你基本上可以做这样的事情:

<bean id="aService" class="com.xetius.isales.pr7.service.ProductRulesWebService " />

<jaxws:endpoint id="aServiceEndpoint" implementor="#aService" address="/aService" />

Use a stack like CXF, which supports Spring natively, then you can essentially do something like this:

<bean id="aService" class="com.xetius.isales.pr7.service.ProductRulesWebService " />

<jaxws:endpoint id="aServiceEndpoint" implementor="#aService" address="/aService" />
紫轩蝶泪 2024-10-24 18:53:26

根据容器版本甚至 Spring,您将有一个简单的解决方案来公开 WSDL,使用:

@PostConstruct
SpringBeanAutowiringSupport.processInjectionBasedOnCurrentContext(this);

Depending on container version or even the Spring, hereinafter you will have an easy solution to expose your WSDL, use:

@PostConstruct
SpringBeanAutowiringSupport.processInjectionBasedOnCurrentContext(this);
~没有更多了~
我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
原文