I'm using Spring Boot and json-schema-validator
. I'm trying to read a file called jsonschema.json
from the resources
folder. I've tried a few different ways but I can't get it to work. This is my code.
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("jsonschema.json").getFile());
JsonNode mySchema = JsonLoader.fromFile(file);
This is the location of the file.
And here I can see the file in the classes
folder.
But when I run the code I get the following error.
jsonSchemaValidator error: java.io.FileNotFoundException: /home/user/Dev/Java/Java%20Programs/SystemRoutines/target/classes/jsonschema.json (No such file or directory)
What is it I'm doing wrong in my code?
After spending a lot of time trying to resolve this issue, finally found a solution that works. The solution makes use of Spring's ResourceUtils. Should work for json files as well.
Thanks for the well written page by Lokesh Gupta : Blog
package utils;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.util.ResourceUtils;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.Properties;
import java.io.File;
public class Utils {
private static final Logger LOGGER = LoggerFactory.getLogger(Utils.class.getName());
public static Properties fetchProperties(){
Properties properties = new Properties();
try {
File file = ResourceUtils.getFile("classpath:application.properties");
InputStream in = new FileInputStream(file);
properties.load(in);
} catch (IOException e) {
LOGGER.error(e.getMessage());
}
return properties;
}
}
To answer a few concerns on the comments :
Pretty sure I had this running on Amazon EC2 using java -jar target/image-service-slave-1.0-SNAPSHOT.jar
Look at my github repo : https://github.com/johnsanthosh/image-service to figure out the right way to run this from a JAR.
Very short answer: you are looking for the resource in the scope of a classloader's class instead of your target class. This should work:
File file = new File(getClass().getResource("jsonschema.json").getFile());
JsonNode mySchema = JsonLoader.fromFile(file);
Also, that might be helpful reading:
PS there is a case when a project compiled on one machine and after that launched on another or inside Docker. In such a scenario path to your resource folder would be invalid and you would need to get it in runtime:
ClassPathResource res = new ClassPathResource("jsonschema.json");
File file = new File(res.getPath());
JsonNode mySchema = JsonLoader.fromFile(file);
Update from 2020
On top of that if you want to read resource file as a String, for example in your tests, you can use these static utils methods:
public static String getResourceFileAsString(String fileName) {
InputStream is = getResourceFileAsInputStream(fileName);
if (is != null) {
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
return (String)reader.lines().collect(Collectors.joining(System.lineSeparator()));
} else {
throw new RuntimeException("resource not found");
}
}
public static InputStream getResourceFileAsInputStream(String fileName) {
ClassLoader classLoader = {CurrentClass}.class.getClassLoader();
return classLoader.getResourceAsStream(fileName);
}
Example of usage:
String soapXML = getResourceFileAsString("some_folder_in_resources/SOPA_request.xml");
if you have for example config folder under Resources folder I tried this Class working perfectly hope be useful
File file = ResourceUtils.getFile("classpath:config/sample.txt")
//Read File Content
String content = new String(Files.readAllBytes(file.toPath()));
System.out.println(content);
花了太多时间回到这个页面,所以就把这个留在这里:
File file = new ClassPathResource("data/data.json").getFile();
Simplest way to read file is:
Resource resource = new ClassPathResource("jsonSchema.json");
FileInputStream file = new FileInputStream(resource.getFile());
See my answer here: https://stackoverflow.com/a/56854431/4453282
import org.springframework.core.io.Resource;
import org.springframework.core.io.ResourceLoader;
Use these 2 imports.
Declare
@Autowired
ResourceLoader resourceLoader;
Use this in some function
Resource resource=resourceLoader.getResource("classpath:preferences.json");
In your case, as you need the file you may use following
File file = resource.getFile()
Reference: http://frugalisminds.com/spring/load-file-classpath-spring-boot/ As already mentioned in previous answers don't use ResourceUtils it doesn't work after deployment of JAR, this will work in IDE as well as after deployment
To reliably get a file from the resources in Spring Boot application:
InputStream
, URL
instead of File
resources
public class SpringBootResourcesApplication {
public static void main(String[] args) throws Exception {
ClassPathResource resource = new ClassPathResource("/hello", SpringBootResourcesApplication.class);
try (InputStream inputStream = resource.getInputStream()) {
String string = new String(inputStream.readAllBytes(), StandardCharsets.UTF_8);
System.out.println(string);
}
}
}
ClassPathResource
is Spring's implementation of Resource
- the abstract way to load resource . It is instantiated using the ClassPathResource(String, Class<?>)
constructor:
/hello
is a path to the file
ClassLoader
instead of Class
, the slash can be omittedClass
to load the resource by
Class
instead of ClassLoader
, because ClassLoader.getResource
differs from Class.getResource
in JPMSProject structure:
├── mvnw ├── mvnw.cmd ├── pom.xml └── src └── main ├── java │ └── com │ └── caco3 │ └── springbootresources │ └── SpringBootResourcesApplication.java └── resources ├── application.properties └── hello
The example above works from both IDE and jar
File
InputStream
and URL
File
because it is not always possible to get it from a classpath resource
public class SpringBootResourcesApplication { public static void main(String[] args) throws Exception { ClassLoader classLoader = SpringBootResourcesApplication.class.getClassLoader(); File file = new File(classLoader.getResource("hello").getFile()); Files.readAllLines(file.toPath(), StandardCharsets.UTF_8).forEach(System.out::println); } }
but fails with: java.nio.file.NoSuchFileException: file:/home/caco3/IdeaProjects/spring-boot-resources/target/spring-boot-resources-0.0.1-SNAPSHOT.jar./BOOT-INF/classes./hello at java.base/sun.nio.fs.UnixException:translateToIOException(UnixException.java.92) at java.base/sun.nio.fs.UnixException:rethrowAsIOException(UnixException.java.111) at java.base/sun.nio.fs.UnixException:rethrowAsIOException(UnixException.java:116)
when Spring Boot jar runInputStream
or URL
JsonLoader.fromFile
from the question could be replaced with JsonLoader.fromURL
method: it accepts URL
Spring Framework enables access to classpath resources through ClassPathResource
You can use it:
resources
@Value
: @SpringBootApplication public class SpringBootResourcesApplication implements ApplicationRunner { @Value("classpath:/hello") // Do not use field injection private Resource resource; public static void main(String[] args) throws Exception { SpringApplication.run(SpringBootResourcesApplication.class, args); } @Override public void run(ApplicationArguments args) throws Exception { try (InputStream inputStream = resource.getInputStream()) { String string = new String(inputStream.readAllBytes(), StandardCharsets.UTF_8); System.out.println(string); } } }
ResourceLoader
: @SpringBootApplication public class SpringBootResourcesApplication implements ApplicationRunner { @Autowired // do not use field injection private ResourceLoader resourceLoader; public static void main(String[] args) throws Exception { SpringApplication.run(SpringBootResourcesApplication.class, args); } @Override public void run(ApplicationArguments args) throws Exception { Resource resource = resourceLoader.getResource("/hello"); try (InputStream inputStream = resource.getInputStream()) { String string = new String(inputStream.readAllBytes(), StandardCharsets.UTF_8); System.out.println(string); } } }
stuck in the same issue, this helps me
URL resource = getClass().getClassLoader().getResource("jsonschema.json");
JsonNode jsonNode = JsonLoader.fromURL(resource);
Below is my working code.
List<sampleObject> list = new ArrayList<>();
File file = new ClassPathResource("json/test.json").getFile();
ObjectMapper objectMapper = new ObjectMapper();
sampleObject = Arrays.asList(objectMapper.readValue(file, sampleObject[].class));
Hope it helps one!
The simplest method to bring a resource from the classpath in the resources directory parsed into a String is the following one liner.
As a String(Using Spring Libraries):
String resource = StreamUtils.copyToString(
new ClassPathResource("resource.json").getInputStream(), defaultCharset());
This method uses the StreamUtils utility and streams the file as an input stream into a String in a concise compact way.
If you want the file as a byte array you can use basic Java File I/O libraries:
As a byte array(Using Java Libraries):
byte[] resource = Files.readAllBytes(Paths.get("/src/test/resources/resource.json"));
Here is my solution. May help someone;
It returns InputStream, but i assume you can read from it too.
InputStream is = Thread.currentThread().getContextClassLoader().getResourceAsStream("jsonschema.json");
For me, the bug had two fixes.
I was using Spring boot as jar and deployed to aws ec2 Java variant of the solution is as below :
package com.test;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.stream.Collectors;
import java.util.stream.Stream;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.context.support.ClassPathXmlApplicationContext;
import org.springframework.core.io.Resource;
public class XmlReader {
private static Logger LOGGER = LoggerFactory.getLogger(XmlReader.class);
public static void main(String[] args) {
String fileLocation = "classpath:cbs_response.xml";
String reponseXML = null;
try (ClassPathXmlApplicationContext appContext = new ClassPathXmlApplicationContext()){
Resource resource = appContext.getResource(fileLocation);
if (resource.isReadable()) {
BufferedReader reader =
new BufferedReader(new InputStreamReader(resource.getInputStream()));
Stream<String> lines = reader.lines();
reponseXML = lines.collect(Collectors.joining("\n"));
}
} catch (IOException e) {
LOGGER.error(e.getMessage(), e);
}
}
}
Spring provides ResourceLoader
which can be used to load files.
@Autowired
ResourceLoader resourceLoader;
// path could be anything under resources directory
File loadDirectory(String path){
Resource resource = resourceLoader.getResource("classpath:"+path);
try {
return resource.getFile();
} catch (IOException e) {
log.warn("Issue with loading path {} as file", path);
}
return null;
}
Referred to this link .
If you are using maven resource filter in your proyect, you need to configure what kind of file is going to be loaded in pom.xml. If you don't, no matter what class you choose to load the resource, it won't be found.
pom.xml
<resources>
<resource>
<directory>${project.basedir}/src/main/resources</directory>
<filtering>true</filtering>
<includes>
<include>**/*.properties</include>
<include>**/*.yml</include>
<include>**/*.yaml</include>
<include>**/*.json</include>
</includes>
</resource>
</resources>
Below works in both IDE and running it as a jar in the terminal,
import org.springframework.core.io.Resource;
@Value("classpath:jsonschema.json")
Resource schemaFile;
JsonSchemaFactory factory = JsonSchemaFactory.getInstance(SpecVersion.VersionFlag.V4);
JsonSchema jsonSchema = factory.getSchema(schemaFile.getInputStream());
just to add my solution as another 2 cents together with all other answers. I am using the Spring DefaultResourceLoader to get a ResourceLoader . Then the Spring FileCopyUtils to get the content of the resource file to a string.
import static java.nio.charset.StandardCharsets.UTF_8;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.Reader;
import java.io.UncheckedIOException;
import org.springframework.core.io.DefaultResourceLoader;
import org.springframework.core.io.Resource;
import org.springframework.core.io.ResourceLoader;
import org.springframework.util.FileCopyUtils;
public class ResourceReader {
public static String readResourceFile(String path) {
ResourceLoader resourceLoader = new DefaultResourceLoader();
Resource resource = resourceLoader.getResource(path);
return asString(resource);
}
private static String asString(Resource resource) {
try (Reader reader = new InputStreamReader(resource.getInputStream(), UTF_8)) {
return FileCopyUtils.copyToString(reader);
} catch (IOException e) {
throw new UncheckedIOException(e);
}
}
}
Here is a solution with ResourceUtils
and Java 11 Files.readString
which takes care of UTF-8 encoding and resource closing
import org.json.JSONObject;
import org.springframework.util.ResourceUtils;
public JSONObject getJsonData() throws IOException {
//file path : src/main/resources/assets/data.json
File file = ResourceUtils.getFile("classpath:assets/data.json");
String data = Files.readString(file.toPath());
return new JSONObject(data);
}
But after deploying the application on OpenShift, the resource is not reachable. So the correct solution is
import static java.nio.charset.StandardCharsets.UTF_8;
import static org.springframework.util.FileCopyUtils.copyToByteArray;
import org.springframework.core.io.ClassPathResource;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;
public JsonNode getJsonData() throws IOException {
ClassPathResource classPathResource = new
ClassPathResource("assets/data.json");
byte[] byteArray =
copyToByteArray(classPathResource.getInputStream());
return new ObjectMapper() //
.readTree(new String(byteArray, UTF_8));
}
i think the problem lies within the space in the folder-name where your project is placed. /home/user/Dev/Java/Java%20Programs/SystemRoutines/target/classes/jsonschema.json
there is space between Java Programs.Renaming the folder name should make it work
如果您使用的是spring
和jackson
(大多数较大的应用程序会),那么使用一个简单的 oneliner:
JsonNode json = new ObjectMapper().readTree(new ClassPathResource("filename").getFile());
I had same issue and because I just had to get file path to send to file input stream, I did this way.
String pfxCertificate ="src/main/resources/cert/filename.pfx";
String pfxPassword = "1234";
FileInputStream fileInputStream = new FileInputStream(pfxCertificate));
Using Spring ResourceUtils.getFile() you don't have to take care absolute path :)
private String readDictionaryAsJson(String filename) throws IOException {
String fileContent;
try {
File file = ResourceUtils.getFile("classpath:" + filename);
Path path = file.toPath();
Stream<String> lines = Files.lines(path);
fileContent = lines.collect(Collectors.joining("\n"));
} catch (IOException ex) {
throw ex;
}
return new fileContent;
}
Try this:
In application.properties
app.jsonSchema=classpath:jsonschema.json
On your Properties pojo:
NOTE : You can use any prefered way of reading configs from application.properties.
@Configuration
@ConfigurationProperties(prefix = "app")
public class ConfigProperties {
private Resource jsonSchema;
// standard getters and setters
}
In your class, read the resource from the Properties Pojo:
//Read the Resource and get the Input Stream
try (InputStream inStream = configProperties.getJsonSchema().getInputStream()) {
//From here you can manipulate the Input Stream as desired....
//Map the Input Stream to a Map
ObjectMapper mapper = new ObjectMapper();
Map <String, Object> jsonMap = mapper.readValue(inStream, Map.class);
//Convert the Map to a JSON obj
JSONObject json = new JSONObject(jsonMap);
} catch (Exception e) {
e.printStackTrace();
}
You need to sanitize the path and replace %20 with a space, or rename your directory. Then it should work.
FileNotFoundException: /home/user/Dev/Java/Java%20Programs/SystemRoutines/target/classes/jsonschema.json
Nowadays, in 2023, Java users should be able to read a classpath file more easily. With a simple instruction such as new File("classpath:path-to-file")
.
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