如何将文本文件资源读入Java单元测试？

Question

我有一个单元测试,需要使用位于的XML文件src/test/resources/abc.xml.获取文件内容的最简单方法是String什么？

Answer 1

最后,我发现了一个简洁的解决方案,感谢Apache Commons:

package com.example;
import org.apache.commons.io.IOUtils;
public class FooTest {
  @Test 
  public void shouldWork() throws Exception {
    String xml = IOUtils.toString(
      this.getClass().getResourceAsStream("abc.xml"),
      "UTF-8"
    );
  }
}

完美的工作.文件src/test/resources/com/example/abc.xml已加载(我正在使用Maven).

如果替换"abc.xml"为,例如,"/foo/test.xml"将加载此资源:src/test/resources/foo/test.xml

你也可以使用Cactoos:

package com.example;
import org.cactoos.io.ResourceOf;
import org.cactoos.io.TextOf;
public class FooTest {
  @Test 
  public void shouldWork() throws Exception {
    String xml = new TextOf(
      new ResourceOf("/com/example/abc.xml") // absolute path always!
    ).asString();
  }
}

Answer 2

正确的观点:

ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("file/test.xml").getFile());

Answer 3

假设UTF8编码在文件中 - 如果没有,只省略"UTF8"参数,并将在每种情况下使用底层操作系统的默认字符集.

JSE 6中的快捷方式 - 简单且无第三方库!

import java.io.File;
public class FooTest {
  @Test public void readXMLToString() throws Exception {
        java.net.URL url = MyClass.class.getResource("test/resources/abc.xml");
        //Z means: "The end of the input but for the final terminator, if any"
        String xml = new java.util.Scanner(new File(url.toURI()),"UTF8").useDelimiter("\\Z").next();
  }
}

JSE 7(未来)的快捷方式

public class FooTest {
  @Test public void readXMLToString() throws Exception {
        java.net.URL url = MyClass.class.getResource("test/resources/abc.xml");
        java.nio.file.Path resPath = java.nio.file.Paths.get(url.toURI());
        String xml = new String(java.nio.file.Files.readAllBytes(resPath), "UTF8"); 
  }

虽然没有打算用于大量文件.

Answer 4

首先确保将abc.xml其复制到输出目录.然后你应该使用getResourceAsStream():

InputStream inputStream = 
    Thread.currentThread().getContextClassLoader().getResourceAsStream("test/resources/abc.xml");

拥有InputStream后,您只需将其转换为字符串即可.该资源说明了这一点:http://www.kodejava.org/examples/266.html.但是,我会摘录相关代码:

public String convertStreamToString(InputStream is) throws IOException {
    if (is != null) {
        Writer writer = new StringWriter();

        char[] buffer = new char[1024];
        try {
            Reader reader = new BufferedReader(
                    new InputStreamReader(is, "UTF-8"));
            int n;
            while ((n = reader.read(buffer)) != -1) {
                writer.write(buffer, 0, n);
            }
        } finally {
            is.close();
        }
        return writer.toString();
    } else {        
        return "";
    }
}

Answer 5

使用Google Guava:

import com.google.common.base.Charsets;
import com.google.common.io.Resources;

public String readResource(final String fileName, Charset charset) throws Exception {
        try {
            return Resources.toString(Resources.getResource(fileName), charset);
        } catch (IOException e) {
            throw new IllegalArgumentException(e);
        }
}

例:

String fixture = this.readResource("filename.txt", Charsets.UTF_8)

Answer 6

你可以试试:

String myResource = IOUtils.toString(this.getClass().getResourceAsStream("yourfile.xml")).replace("\n","");