模拟 url.openStream() 应该 return InputStream

Question

我的实际 class 接受 URL 作为输入并调用 url.openStream()，这应该 return InputStream。

public static Map<String, Object> parseA(URL url) throws Exception {
   byte[] readData = new byte[25*1024*1024];
    // Here url.openStream() returning null
            InputStream is = url.openStream();

            while((readLength = is.read(readData, 0, 25*1024*1024)) != -1){
                br = new BufferedReader(new InputStreamReader(new 
    ByteArrayInputStream(readData)));

                // All CW_* strings are collected first
}

我的测试class是

 @Rule
    public TemporaryFolder folder= new TemporaryFolder();
 @Test(enabled = true)
    public void parseATest() {

            File file=null;
           try {
             file =folder.newFile("testingData.txt");
        } catch (IOException e1) {
            // TODO Auto-generated catch block
            e1.printStackTrace();
        }

            final URLConnection mockConnection = EasyMock.createMock(URLConnection.class);
            final URLStreamHandler handler = new URLStreamHandler() {

                @Override
                protected URLConnection openConnection(final URL arg0)
                        throws IOException {
                    return mockConnection;
                }
            };
            URL url=null;
            try {
                url = new URL("http://foo.bar", "foo.bar", 80, "", handler);
            } catch (MalformedURLException e1) {
                // TODO Auto-generated catch block
                e1.printStackTrace();
            }
            InputStream is=null;
            try {
                is = new FileInputStream(file);
            } catch (FileNotFoundException e2) {
                // TODO Auto-generated catch block
                e2.printStackTrace();
            } 
            try {
                EasyMock.expect(url.openStream()).andReturn(is).anyTimes();
            } catch (FileNotFoundException e1) {
                // TODO Auto-generated catch block
                e1.printStackTrace();
            } catch (IOException e1) {
                // TODO Auto-generated catch block
                e1.printStackTrace();
            }
        try {
             // imageHeaderParser is object of actual class
            imageHeaderParser.parseA(url);
        } catch (IfmSwimParserException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

    }

此处 TemporaryFolder 用于创建临时文件 file.And 我不想 URL 去 network.It 可以只是虚拟的 URL当我调用 url.openStream() 时，它应该 return 我提到的临时文件流。

Answer 1

文件可以转换为 URL，所以就这样做吧：

Url testUrl = Paths.get("folder",("testingData.txt").toUri().toURL();
Map<String, Object> map = parseA(testUrl);
// assert map content

此外，如果您想测试文件处理行为，则不需要任何模拟。

模拟 url.openStream() 应该 return InputStream

mocking url.openStream() should return InputStream

java

junit

mocking

java-io