ben*_*rre 75 java junit unit-testing junit4
有人可以建议一种简单的方法来获取对junit测试类中的String/InputStream/File/etc类型对象的文件引用吗?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,或者将其作为文件读取,但是有一个特定于Junit的快捷方式吗?
public class MyTestClass{
@Resource(path="something.xml")
File myTestFile;
@Test
public void toSomeTest(){
...
}
}
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Ha.*_*Ha. 81
您可以尝试@Rule注释.以下是文档中的示例:
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);
}
}
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您只需要创建FileResource类扩展ExternalResource.
完整的例子
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());
}
}
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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();
}
}
}
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sla*_*ick 76
如果您需要实际获取File对象,则可以执行以下操作:
URL url = this.getClass().getResource("/test.wsdl");
File testWsdl = new File(url.getFile());
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Joe*_*son 14
我知道你说你不想手工阅读文件,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());
}
}
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您所要做的就是确保有问题的文件位于类路径中.如果您正在使用Maven,只需将文件放在src/test/resources中,Maven会在运行测试时将其包含在类路径中.如果你需要做很多这样的事情,你可以把打开文件的代码放在一个超类中并让你的测试继承.
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