Java 将测试文件放入 JUnit 的简单方法
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Easy way to get a test file into JUnit
提问by benstpierre
Can somebody suggest an easy way to get a reference to a file as a String/InputStream/File/etc type object in a junit test class? Obviously I could paste the file (xml in this case) in as a giant String or read it in as a file but is there a shortcut specific to Junit like this?
有人可以建议一种简单的方法来获取对文件的引用作为 junit 测试类中的 String/InputStream/File/etc 类型对象吗？显然，我可以将文件（在这种情况下为 xml）粘贴为一个巨大的字符串或将其作为文件读取，但是是否有这样的特定于 Junit 的快捷方式？
public class MyTestClass{
@Resource(path="something.xml")
File myTestFile;
@Test
public void toSomeTest(){
...
}
}
回答by Ha.
You can try @Ruleannotation. Here is the example from the docs:
你可以试试@Rule注解。这是文档中的示例：
public static class UsesExternalResource {
Server myServer = new Server();
@Rule public ExternalResource resource = new ExternalResource() {
@Override
protected void before() throws Throwable {
myServer.connect();
};
@Override
protected void after() {
myServer.disconnect();
};
};
@Test public void testFoo() {
new Client().run(myServer);
}
}
You just need to create FileResourceclass extending ExternalResource.
您只需要创建FileResource类扩展ExternalResource.
Full Example
完整示例
import static org.junit.Assert.*;
import org.junit.Rule;
import org.junit.Test;
import org.junit.rules.ExternalResource;
public class TestSomething
{
@Rule
public ResourceFile res = new ResourceFile("/res.txt");
@Test
public void test() throws Exception
{
assertTrue(res.getContent().length() > 0);
assertTrue(res.getFile().exists());
}
}
import java.io.BufferedReader;
import java.io.File;
import java.io.FileOutputStream;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.nio.charset.Charset;
import org.junit.rules.ExternalResource;
public class ResourceFile extends ExternalResource
{
String res;
File file = null;
InputStream stream;
public ResourceFile(String res)
{
this.res = res;
}
public File getFile() throws IOException
{
if (file == null)
{
createFile();
}
return file;
}
public InputStream getInputStream()
{
return stream;
}
public InputStream createInputStream()
{
return getClass().getResourceAsStream(res);
}
public String getContent() throws IOException
{
return getContent("utf-8");
}
public String getContent(String charSet) throws IOException
{
InputStreamReader reader = new InputStreamReader(createInputStream(),
Charset.forName(charSet));
char[] tmp = new char[4096];
StringBuilder b = new StringBuilder();
try
{
while (true)
{
int len = reader.read(tmp);
if (len < 0)
{
break;
}
b.append(tmp, 0, len);
}
reader.close();
}
finally
{
reader.close();
}
return b.toString();
}
@Override
protected void before() throws Throwable
{
super.before();
stream = getClass().getResourceAsStream(res);
}
@Override
protected void after()
{
try
{
stream.close();
}
catch (IOException e)
{
// ignore
}
if (file != null)
{
file.delete();
}
super.after();
}
private void createFile() throws IOException
{
file = new File(".",res);
InputStream stream = getClass().getResourceAsStream(res);
try
{
file.createNewFile();
FileOutputStream ostream = null;
try
{
ostream = new FileOutputStream(file);
byte[] buffer = new byte[4096];
while (true)
{
int len = stream.read(buffer);
if (len < 0)
{
break;
}
ostream.write(buffer, 0, len);
}
}
finally
{
if (ostream != null)
{
ostream.close();
}
}
}
finally
{
stream.close();
}
}
}
回答by Joey Gibson
I know you said you didn't want to read the file in by hand, but this is pretty easy
我知道你说过你不想手动读取文件，但这很容易
public class FooTest
{
private BufferedReader in = null;
@Before
public void setup()
throws IOException
{
in = new BufferedReader(
new InputStreamReader(getClass().getResourceAsStream("/data.txt")));
}
@After
public void teardown()
throws IOException
{
if (in != null)
{
in.close();
}
in = null;
}
@Test
public void testFoo()
throws IOException
{
String line = in.readLine();
assertThat(line, notNullValue());
}
}
All you have to do is ensure the file in question is in the classpath. If you're using Maven, just put the file in src/test/resources and Maven will include it in the classpath when running your tests. If you need to do this sort of thing a lot, you could put the code that opens the file in a superclass and have your tests inherit from that.
您所要做的就是确保有问题的文件在类路径中。如果您使用 Maven，只需将文件放在 src/test/resources 中，Maven 将在运行测试时将其包含在类路径中。如果您需要经常做这类事情，您可以将打开文件的代码放在超类中，并让您的测试从中继承。
回答by slashnick
If you need to actually get a Fileobject, you could do the following:
如果您需要实际获取一个File对象，您可以执行以下操作：
URL url = this.getClass().getResource("/test.wsdl");
File testWsdl = new File(url.getFile());
Which has the benefit of working cross platform, as described in this blog post.
回答by Guido Celada
You can try doing:
你可以尝试这样做：
String myResource = IOUtils.toString(this.getClass().getResourceAsStream("yourfile.xml")).replace("\n","");
回答by gMale
If you want to load a test resource file as a string with just few lines of code and without any extra dependencies, this does the trick:
如果您想将测试资源文件作为字符串加载，只需几行代码并且没有任何额外的依赖项，就可以这样做：
public String loadResourceAsString(String fileName) throws IOException {
Scanner scanner = new Scanner(getClass().getClassLoader().getResourceAsStream(fileName));
String contents = scanner.useDelimiter("\A").next();
scanner.close();
return contents;
}
"\\A" matches the start of input and there's only ever one. So this parses the entire file contents and returns it as a string. Best of all, it doesn't require any 3rd party libraries (like IOUTils).
"\\A" 匹配输入的开头，并且只有一个。所以这会解析整个文件内容并将其作为字符串返回。最重要的是，它不需要任何 3rd 方库（如 IOUTils）。
