src / main / resources中的文件不在classpath

时间:2016-01-19 09:02:57

标签: java spring

我创建了一个简单的程序来检查spring-context是否在src / main / resources文件夹中定义了一个文件。

我有这个文件结构:

project
--> src/main/resources/spring-config.xml
--> src/main/resources/testfile02

我尝试使用此测试类

访问这些文件
public class ClasspathTest {

    public static void main(String args[]) throws URISyntaxException {
        ClassPathXmlApplicationContext context = new ClassPathXmlApplicationContext("classpath:spring-config.xml");

        ClassPathResource testfile02 = new ClassPathResource("classpath:testfile02");

        if (testfile02 != null) {
            try (InputStream inputStream = testfile02.getInputStream();
                Reader streamReader = new InputStreamReader(inputStream, "UTF-8");
                BufferedReader bufferedReader = new BufferedReader(streamReader)) {
                String line;
                while ((line = bufferedReader.readLine()) != null) {
                    System.out.println(line);
                }
            } catch (IOException exc) {
                exc.printStackTrace();
            }
        }

    }

}

如果ClassPathXmlAppContext工作正常,我无法理解为什么在classpathResource.getInputStream()期间我得到FileNotFound异常。

执行日志:

янв 19, 2016 11:57:48 AM org.springframework.context.support.ClassPathXmlApplicationContext prepareRefresh
INFO: Refreshing org.springframework.context.support.ClassPathXmlApplicationContext@6193b845: startup date [Tue Jan 19 11:57:48 MSK 2016]; root of context hierarchy
янв 19, 2016 11:57:48 AM org.springframework.beans.factory.xml.XmlBeanDefinitionReader loadBeanDefinitions
INFO: Loading XML bean definitions from class path resource [spring-config.xml]
java.io.FileNotFoundException: class path resource [classpath:testfile02] cannot be opened because it does not exist

该项目使用gradle构建:

apply plugin: 'java'
apply plugin: 'eclipse'
sourceCompatibility = 1.8
version = '1.0'
repositories {
    mavenCentral()
}
dependencies {
    compile 'org.springframework:spring-context:4.1.6.RELEASE'
}

的.classpath:

<?xml version="1.0" encoding="UTF-8"?>
<classpath>
    <classpathentry kind="src" path="src/main/java"/>
    <classpathentry kind="src" path="src/main/resources"/>
    <classpathentry kind="src" path="src/test/java"/>
    <classpathentry kind="src" path="src/test/resources"/>
    <classpathentry exported="true" kind="con" path="org.eclipse.jdt.launching.JRE_CONTAINER"/>
    <classpathentry exported="true" kind="con" path="org.springsource.ide.eclipse.gradle.classpathcontainer"/>
    <classpathentry kind="output" path="bin"/>
</classpath>

我试图检测系统性的java.class.path,我有春天的{project} / bin和jars,就是这样。没有资源文件夹。 当我访问src / main / resources / spring-config.xml时,如何从src / main / resources访问资源?

2 个答案:

答案 0 :(得分:3)

您不需要添加&#39;类路径:&#39;使用构造函数时的前缀。这有效:

public class ClasspathTest {

    public static void main(String args[]) throws URISyntaxException {
        ClassPathXmlApplicationContext context = new ClassPathXmlApplicationContext("classpath:spring-config.xml");

        ClassPathResource testfile02 = new ClassPathResource("testfile02");

        if (testfile02 != null) {
            try (InputStream inputStream = testfile02.getInputStream();
                Reader streamReader = new InputStreamReader(inputStream, "UTF-8");
                BufferedReader bufferedReader = new BufferedReader(streamReader)) {
                String line;
                while ((line = bufferedReader.readLine()) != null) {
                    System.out.println(line);
                }
            } catch (IOException exc) {
                exc.printStackTrace();
            }
        }

    }

}

答案 1 :(得分:0)

问题是vanilla Java,你运行它的方式不会将src / main / resources添加到类路径中,这是其他运行者使用的惯例。

以下工作正常。

package com.greg;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.Reader;

import org.junit.Test;
import org.junit.runner.RunWith;
import org.springframework.context.support.ClassPathXmlApplicationContext;
import org.springframework.core.io.ClassPathResource;
import org.springframework.test.context.ContextConfiguration;
import org.springframework.test.context.junit4.SpringJUnit4ClassRunner;

@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration(value = { "classpath:spring-config.xml",
        "classpath:testfile02.xml" })
public class ClasspathTest {

    @Test
    public void test1() {
        ClassPathXmlApplicationContext context = new ClassPathXmlApplicationContext(
                "classpath:spring-config.xml");

        ClassPathResource testfile02 = new ClassPathResource("testfile02.xml");

        if (testfile02 != null) {
            try {
                InputStream inputStream = testfile02.getInputStream();
                Reader streamReader = new InputStreamReader(inputStream,
                        "UTF-8");
                BufferedReader bufferedReader = new BufferedReader(streamReader);
                String line;
                while ((line = bufferedReader.readLine()) != null) {
                    System.out.println(line);
                }
            } catch (IOException exc) {
                exc.printStackTrace();
            }
        }

    }
}