Java 使用Junit测试InputStream
我正在尝试测试类“UserInput”中的函数“DataFeatures” 不管我在测试中给出什么样的论点,它总是通过的 字段和构造函数Java 使用Junit测试InputStream,java,junit,inputstream,Java,Junit,Inputstream,我正在尝试测试类“UserInput”中的函数“DataFeatures” 不管我在测试中给出什么样的论点,它总是通过的 字段和构造函数 public @Getter class UserInput { FileType type; FileOperation operation; SynchronizationMethod method; String path; private static InputReader in = new InputRead
public @Getter class UserInput {
FileType type;
FileOperation operation;
SynchronizationMethod method;
String path;
private static InputReader in = new InputReader();
public UserInput() {
// Dont need to do anything
}
要测试的功能
void getDataFeatures() {
System.out.println("For encryption press 1");
System.out.println("For decryption press 0");
operation = FileOperation.fromInt(in.nextInt());
System.out.println("For a file choose 1");
System.out.println("For an entire directory choose 0");
type = FileType.fromInt(in.nextInt());
if (type == FileType.DIR) {
System.out.println("For sync press 1");
System.out.println("For async press 0");
method = SynchronizationMethod.fromInt(in.nextInt());
}
}
我的测试
public class UserInputTest {
UserInput UI;
private final ByteArrayOutputStream outContent = new ByteArrayOutputStream();
private final ByteArrayOutputStream errContent = new ByteArrayOutputStream();
private final PrintStream oldStdOut = System.out;
private final PrintStream oldStdErr = System.err;
private final InputStream oldStdIn = System.in;
@Before
public void initlize(){
System.setOut(new PrintStream(outContent));
System.setErr(new PrintStream(errContent));
UI = new UserInput();
}
@Test
public void getDataFeaturesTest() {
String data = "0" + "\n0" + "\n0";
System.setIn(new ByteArrayInputStream(data.getBytes()));
UI.getDataFeatures();
System.out.println(UI.getOperation());
assertThat(UI.getOperation(), is(equalTo(FileOperation.decryption)));
assertThat(UI.getType(), is(equalTo(FileType.FILE)));
assertThat(UI.getMethod(), is(equalTo(SynchronizationMethod.SYNC)));
}
@After
public void cleanUpStreams() {
System.setOut(oldStdOut);
System.setErr(oldStdErr);
System.setIn(oldStdIn);
}
}
注意1:FileOperation、FileType和SynchronizationMethod都是获取1或0的枚举
同步方法示例:
public enum SynchronizationMethod {
SYNC(1), ASYNC(0);
private int method;
private SynchronizationMethod(int meth) {
this.method = meth;
}
public static SynchronizationMethod fromInt(int meth) {
for (SynchronizationMethod SM : SynchronizationMethod.values()) {
if (SM.method == meth) {
return SM;
}
}
throw new IllegalArgumentException("No constant with method " + meth + " found");
}
public String toString(){
if (method == 1){
return "Sync";
}
else if(method == 0){
return "ASync";
}
else{
throw new IllegalArgumentException("No constant with method " + method + " found in toString");
}
}
}
解决方案
public @Getter class UserInput {
FileType type;
FileOperation operation;
SynchronizationMethod method;
String path;
private static InputReader in = new InputReader();
public UserInput() {
// Dont need to do anything
}
问题出在构造函数中的类InputReader中
public InputReader() {
reader = new BufferedReader(new InputStreamReader(System.in));
tokenizer = null;
}
此函数中的读取器和Junit函数中的输入流断开,正如这里的注释所建议的那样猜测-您的UserInput类说:
private static InputReader in = new InputReader();
而您的测试用例说:
String data = "0" + "\n0" + "\n0";
System.setIn(new ByteArrayInputStream(data.getBytes()));
换言之:可能存在断开连接。根据
InputReader
背后的实现,您可能只是从错误的源读取数据。好的,我更改了itUnrelated:我建议不要将枚举基于整数。枚举已经有了从字符串构建常量的方法。你在这里发明轮子。您可以决定将枚举值表示为小写字符串,从“SYNC”创建SYNC(反之亦然)的代码应该更短!严肃地说:无论如何,当你要求用户输入时,不要使用数字。为什么希望用户记住0是异步的。让我让他说“异步”吧!我给出了一个答案,但问题是:你没有提供一个答案。我们不可能重新处理您的问题,因为您只有部分代码产生了问题。如果我的回答有帮助的话,很好。如果没有:请阅读该链接,并张贴符合这些要求的代码!例如,通过对Java Enum现有方法进行一些研究。更重要的是:用户不犯错误,或者他节省了输入几个字符?