使用 spring 测试类时模拟配置值

发布于 2024-11-28 19:41:49 字数 1183 浏览 5 评论 0原文

我正在尝试测试一个像这样的类，

@Controller
public class FailureController {

    @Value("#{configValues.defaultRedirectUrl}")
    private String defaultRedirectUrl;
public FailureController (String defaultRedirectUrl) {
this.defaultRedirectUrl = defaultRedirectUrl;
}
...

问题是如果不为测试类创建一个特殊的构造函数（如上所示），我就无法测试这个类，这最终会初始化 中的 defaultRedirectUrl失败控制器。

在测试期间如何在不创建构造函数（来自 spring 上下文）的情况下测试它。我的主要目标是在运行测试时在没有构造函数的情况下初始化 FailureController 中的值。

是否有可能以某种方式在测试期间加载 spring 上下文并初始化 FailureController 中的字段这就是我一直在做的事情，但它不起作用，defaultRedirectUrl在FailureController中仍然是null。

@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration(locations = { "/applicationContext-test.xml" })
public class FailureControllerTest {
    @Autowired
    FailureController failureController;

applicationContext-test.xml

...
        <bean id="failureController" class="se.synergica.watchtower.controllers.FailureController">
        </bean>

        <import resource="spring-config-test.xml" />
    </beans>

谢谢。阿尔

原文

I am trying to test a class like

@Controller
public class FailureController {

    @Value("#{configValues.defaultRedirectUrl}")
    private String defaultRedirectUrl;
public FailureController (String defaultRedirectUrl) {
this.defaultRedirectUrl = defaultRedirectUrl;
}
...

The problem is that I can't test this class without creating a special constructor for test class (like above), which is eventually to initializes the defaultRedirectUrl in FailureController.

How can I test it without creating a constructor (that are to be come from spring context) during the test. My main objective is to initialize the values in FailureController without a constructor when running a test.

Is it possible in some way that spring context gets loaded during the test and initializes the fields in FailureController
this is what i have been doing but its not working, defaultRedirectUrl remains null in FailureController.

@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration(locations = { "/applicationContext-test.xml" })
public class FailureControllerTest {
    @Autowired
    FailureController failureController;

applicationContext-test.xml

...
        <bean id="failureController" class="se.synergica.watchtower.controllers.FailureController">
        </bean>

        <import resource="spring-config-test.xml" />
    </beans>

thank you.
al

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美羊羊 2024-12-05 19:41:49

我将在您的测试类路径（标准项目中的 src/test/resources）中创建一个单独的 applicationContext.xml 文件。在那里我将设置一个带有虚拟值的 configValues 对象。
您的被测类也应该在该上下文中实例化。

然后在 junit 测试中加载这样的测试上下文：

@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration("/applicationContext-test.xml")
public class MyTest {

    @Autowired
    private AClassToBeTest subject;

    //...perform test on subject
}

有关 spring 测试的更多信息： http://static.springsource.org/spring/docs/3.0.x/reference/testing.html

I would create a separate applicationContext.xml file in your test-classpath (src/test/resources in standard projects). There I would set up a configValues object with dummy values.
Your class under test should also be instanciated in that context.

Then load the test context like this in a junit test:

@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration("/applicationContext-test.xml")
public class MyTest {

    @Autowired
    private AClassToBeTest subject;

    //...perform test on subject
}

more on spring testing: http://static.springsource.org/spring/docs/3.0.x/reference/testing.html

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