如何在spring mvc控制器中使用junit返回类型的方法 [英] How to junit return type of a method in spring mvc controller
问题描述
我在我的Spring MVC控制器上做junit -
I am doing junit on my Spring MVC controller -
@RequestMapping(value = "index", method = RequestMethod.GET)
public HashMap<String, String> handleRequest() {
HashMap<String, String> model = new HashMap<String, String>();
String name = "Hello World";
model.put("greeting", name);
return model;
}
以下是上述方法的junit -
And below is my junit for the above method -
public class ControllerTest {
private MockMvc mockMvc;
@Before
public void setup() throws Exception {
this.mockMvc = standaloneSetup(new Controller()).build();
}
@Test
public void test01_Index() {
try {
mockMvc.perform(get("/index")).andExpect(status().isOk());
} catch (Exception e) {
e.printStackTrace();
}
}
}
以上junit工作正常..
Above junit works fine..
但我的问题是如何将 handleRequest
的返回类型设为junit,返回 HashMap
带键和值对..如何验证它是否返回 Hello World
?是否有任何方法可以做到这一点?
But my question is how do I junit the return type of handleRequest
which is returning a HashMap
with key and value pair.. How do I verify that it is returning Hello World
? Is there any method to do that as well?
推荐答案
看看参考Spring参考手册中的示例使用MockMvc测试服务器端代码。假设您正在返回JSON响应:
Take a look at the examples in the Spring reference manual referring to using MockMvc to test server-side code. Assuming you are returning a JSON response:
mockMvc.perform(get("/index"))
.andExpect(status().isOk())
.andExpect(content().contentType("application/json"))
.andExpect(jsonPath("$.greeting").value("Hello World"));
顺便说一下 - 永远不会在 @Test <中捕获并吞下异常/ code>方法,除非你想忽略该异常并防止它失败测试。如果编译器抱怨您的测试方法调用了抛出异常并且您没有处理它的方法,只需将您的方法签名更改为
throws Exception
。
By the way - never catch and swallow an exception in a @Test
method, unless you want to ignore that exception and prevent it from failing the test. If the compiler is complaining that your test method called a method that throws an exception and you didn't handle it, simply change your method signature to throws Exception
.
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