【发布时间】:2018-09-05 15:45:50
【问题描述】:
我正在尝试使用 java 8、windows 10、intelliJ、junit5、mockito 创建未来应用程序的骨架。 我在使用 mockito 运行测试时遇到问题,这是我使用间谍编写的第一个测试,但它不起作用。
似乎when(client_spy.try_to_connect(anyString(),anyInt())).thenReturn(null); 语句导致了方法的执行,这是意料之外的:我只想告诉mockito 返回null 是调用了try_to_connect 方法。
因此,该方法在测试中被调用,并且没有参数,所以我在这个方法中得到一个错误。
这里是测试类(部分):
@ExtendWith(MockitoExtension.class)
public class unitTestsTry2 {
private static final String LOCAL_IP = "localhost";
private static final String REMOTE_IP = "localhost";
private static final String LOCAL_PORT = "1001";
private static final String REMOTE_PORT = "1002";
private static ApplicationRunner apr=new ApplicationRunner();
@Spy
ClientExtremity client_spy=new ClientExtremity();
(...)
@Test
void startupAndTryToConnect() {
apr.client_endpoint = client_spy;
when(client_spy.try_to_connect(anyString(),anyInt()))
.thenReturn(null); <--- THE FAULTLY LINE
apr.main(new String[]{LOCAL_IP,LOCAL_PORT,REMOTE_IP,REMOTE_PORT});
verify(((ClientOrders)apr.client_endpoint)).try_to_connect(REMOTE_IP,Integer.parseInt(REMOTE_PORT));
}
}
这是我的 gradle 文件:
buildscript {
dependencies {
classpath group: 'de.dynamicfiles.projects.gradle.plugins', name: 'javafx-gradle-plugin', version: '8.8.2'
}
repositories {
mavenLocal()
mavenCentral()
}
}
plugins {
id 'java'
}
group 'lorry'
version '1'
sourceCompatibility = 1.8
repositories {
mavenLocal()
mavenCentral()
maven { url "https://dl.bintray.com/mockito/maven" }
}
apply plugin: 'javafx-gradle-plugin'
test {
useJUnitPlatform()
}
compileJava.dependsOn clean
jfx {
// minimal requirement for jfxJar-task
mainClass = 'lorry.ApplicationRunner'
// minimal requirement for jfxNative-task
vendor = 'YourName'
launcherArguments = ["localhost", "1001", "localhost", "1002"]
// gradle jfxRun
runJavaParameter = null // String
//runAppParameter = "localhost" "1001" "localhost" "1002" // String
jfxMainAppJarName = "chat.jar"
}
dependencies {
//testCompile group: 'junit', name: 'junit', version: '4.12'
compile group: 'javax.websocket', name: 'javax.websocket-api', version: '1.1'
compile group: 'com.google.code.gson', name: 'gson', version: '2.8.5'
//testImplementation('org.junit.jupiter:junit-jupiter-api:5.2.0')
//testRuntimeOnly('org.junit.jupiter:junit-jupiter-engine:5.2.0')
def final junitVersion = "5.2.0"
compile group: 'com.google.inject', name: 'guice', version: '4.1.0'
//compile group: 'com.google.code.gson', name: 'gson', version: '2.8.2'
compile group: 'org.junit.jupiter', name: 'junit-jupiter-api', version: junitVersion
compile group: 'org.assertj', name: 'assertj-core', version: '3.9.0'
compile group: 'org.apache.commons', name: 'commons-lang3', version: '3.7'
testCompile group: 'org.junit.jupiter', name: 'junit-jupiter-params', version: junitVersion
testCompile group: 'org.mockito', name: 'mockito-core', version: '2.21.0'
testCompile 'org.mockito:mockito-junit-jupiter:2.21.0'
testCompile group:'org.junit.jupiter',name:'junit-jupiter-api',version: junitVersion
testRuntime group: 'org.junit.jupiter', name: 'junit-jupiter-engine', version: junitVersion
compile 'org.hamcrest:hamcrest-all:1.3'
testCompile "org.testfx:testfx-core:4.0.13-alpha"
testCompile 'org.testfx:testfx-junit5:4.0.13-alpha'
compile group: 'org.glassfish.tyrus', name: 'tyrus-server', version: '1.13.1'
// https://mvnrepository.com/artifact/org.glassfish.tyrus/tyrus-client
compile group: 'org.glassfish.tyrus', name: 'tyrus-client', version: '1.13.1'
}
jar {
baseName = 'Chat'
version = ''
manifest {
attributes(
'Class-Path': configurations.compile.collect { it.getName() }.join(' '),
'Main-Class': 'lorry.Chat'
)
}
}
编辑 谢谢您的回答。我改变了我的代码:
@Spy
public ClientExtremity client_spy=new ClientExtremity();
@Test
void startupAndTryToConnect() {
apr.client_endpoint = client_spy;
when(client_spy).try_to_connect(anyString(),anyInt()).thenReturn(null);
apr.main(new String[]{LOCAL_IP,LOCAL_PORT,REMOTE_IP,REMOTE_PORT});
verify(((ClientOrders)apr.client_endpoint)).try_to_connect(REMOTE_IP,Integer.parseInt(REMOTE_PORT));
}
但 intelliJ 告诉我:错误:(73, 25) java: 找不到符号 符号:方法 try_to_connect(java.lang.String,int) 位置:接口org.mockito.stubbing.OngoingStubbing
拥有以下客户端:
@ClientEndpoint(encoders = MessageEncoder.class, decoders = MessageDecoder.class)
public class ClientExtremity implements ClientOrders {
@OnMessage
public void onMessage(Message message) {
}
@Override
public Session try_to_connect(String remote_ip, Integer remote_port) {
ClientManager client = ClientManager.createClient();
Session session;
try {
session = client.connectToServer(ClientExtremity.class, new URI(
format("wss://%1$2s:%2$2d/chat",remote_ip,remote_port)
));
} catch (Exception e) {
e.printStackTrace();
session=null;
}
return session;
}
}
我不明白为什么在测试方法中声明try_to_connect失败。
编辑 2 这是applicationRunner:
public class ApplicationRunner {
public static String localIP, remoteIP;
public static int localPort, remotePort;
public static Fenetre fenetre = new Fenetre();
public static Session session=null;
public static ClientExtremity client_endpoint=new ClientExtremity();
public static void init(String[] args) {
localIP = args[0];
localPort = Integer.parseInt(args[1]);
remoteIP = args[2];
remotePort = Integer.parseInt(args[3]);
try {
new Thread(fenetre).start();
} catch (Exception e) {
e.printStackTrace();
}
session = ((ClientOrders) client_endpoint).try_to_connect(localIP,localPort);
}
public static void main (String[]args){
init(args);
}
}
【问题讨论】:
-
我强烈建议您阅读
Mockito.spy()的javadoc,它详细解释了将when() 与间谍一起使用是危险的并且可能导致这种行为。关于改进您的帖子,而不是显示您的 build.gradle,您应该添加您的ApplicationRunner。