I'm trying to unmarshal my xml file:
public Object convertFromXMLToObject(String xmlfile) throws IOException {
FileInputStream is = null;
File file = new File(String.valueOf(this.getClass().getResource("xmlToParse/companies.xml")));
try {
is = new FileInputStream(file);
return getUnmarshaller().unmarshal(new StreamSource(is));
} finally {
if (is != null) {
is.close();
}
}
}
But I get this errors:
java.io.FileNotFoundException: null (No such file or directory)
Here is my structure:
Why I can't get files from resources folder? Thanks.
Update.
After refactoring,
URL url = this.getClass().getResource("/xmlToParse/companies.xml");
File file = new File(url.getPath());
I can see an error more clearly:
java.io.FileNotFoundException: /content/ROOT.war/WEB-INF/classes/xmlToParse/companies.xml (No such file or directory)
It tries to find WEB-INF/classes/
I have added folder there, but still get this error :(
I had the same problem trying to load some XML files into my test classes. If you use Spring, as one can suggest from your question, the easiest way is to use org.springframework.core.io.Resource - the one Raphael Roth already mentioned.
The code is really straight forward. Just declare a field of the type org.springframework.core.io.Resource and annotate it with org.springframework.beans.factory.annotation.Value - like that:
#Value(value = "classpath:xmlToParse/companies.xml")
private Resource companiesXml;
To obtain the needed InputStream, just call
companiesXml.getInputStream()
and you should be okay :)
But forgive me, I have to ask one thing: Why do you want to implement a XML parser with the help of Spring? There are plenty build in :) E.g. for web services there are very good solutions that marshall your XMLs into Java Objects and back...
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("fileName").getFile());
you are suppose to give an absolute path (so add a loading ´/´, where resource-folder is the root-folder):
public Object convertFromXMLToObject(String xmlfile) throws IOException {
FileInputStream is = null;
File file = new File(String.valueOf(this.getClass().getResource("/xmlToParse/companies.xml")));
try {
is = new FileInputStream(file);
return getUnmarshaller().unmarshal(new StreamSource(is));
} finally {
if (is != null) {
is.close();
}
}
}
Related
I created folder src/test/resources/ in root project directory, and inside this I added a file in folder jsons as jsons/server_request.json.
Now I am trying to read this file by calling a the static function in CommonTestUtilityclass given as:
public class CommonTestUtility {
public static String getFileAsString(String fileName) throws IOException {
ClassLoader classLoader = ClassLoader.getSystemClassLoader();
File file = new File(classLoader.getResource(fileName).getFile());
String content = new String(Files.readAllBytes(file.toPath()));
return content;
}
}
Now while calling this function as
class ServerTest {
#Test
void test_loadResource() {
String content = CommonTestUtility.getFileAsString("jsons/server_request.json");
}
}
, It's giving me the error as:
CommonTestUtility - Cannot invoke "java.net.URL.getFile()" because the return value of "java.lang.ClassLoader.getResource(String)" is null.
I tried to include the src/test/resources/ in the run configuration
of Junit ServerTest.java, but still it's not able to find out the
resource
How to resolve this issue?
https://mkyong.com/java/java-read-a-file-from-resources-folder/
This above link might be helpful.
The getResource() method return an URI you need to change
.getFile() function to. toURI().
Simple code
private File getFileFromResource(String fileName) throws URISyntaxException{
ClassLoader classLoader = getClass().getClassLoader();
URL resource = classLoader.getResource(fileName);
if (resource == null) {
throw new IllegalArgumentException("file not found! " + fileName);
} else {
// failed if files have whitespaces or special characters
//return new File(resource.getFile());
return new File(resource.toURI());
}
}
I recreated the same scenario you describe and your code works for me.
Could you double-check that your project looks like mine below? If so, I suspect it might be something with your environment.
I am trying to access a properties file from the src/main/resources folder but when I try to load the file using a relative path it is not getting updated. But it is working fine for an absolute path.
I need the dynamic web project to work across all platforms.
public static void loadUsers() {
try(
FileInputStream in = new FileInputStream("C:\\Users\\SohamGuha\\Documents\\work-coding\\work-coding\\src\\main\\resources\\users.properties")) {
// write code to load all the users from the property file
// FileInputStream in = new FileInputStream("classpath:users.properties");
users.load(in);
System.out.println(users);
in.close();
}
catch(Exception e){
e.printStackTrace();
}
First of all you are using Spring, at least that is what the tags at the bottom say. Secondly C:\\Users\\SohamGuha\\Documents\\work-coding\\work-coding\\src\\main\\resources\\users.properties is the root of your classpath. Instead of loading a File use the Spring resource abstraction.
As this is part of the classpath you can simply use the ClassPathResource to obtain a proper InputStream. This will work regardless of which environment you are in.
try( InputStream in = new ClassPathResource("users.properties").getInputStream()) {
//write code to load all the users from the property file
//FileInputStream in = new FileInputStream("classpath:users.properties");
users.load(in);
System.out.println(users);
} catch(Exception e){
e.printStackTrace();
}
NOTE: you are already using a try with resources so you don't need to close the InputStream that is already handled for you.
Changing things inside your application simply won't work, as this would mean you could change resources (read classes) in your jar which would be quite a security risk! If you want something to be changable you will have to make it a file outside of the classpath and directly on the file-system.
Try the following code
import java.io.FileInputStream;
import java.io.IOException;
public class LoadUsers {
public static void main(String[] args) throws IOException {
try(FileInputStream fis=new FileInputStream("src/main/resources/users.properties")){
Properties users=new Properties();
users.load(fis);
System.out.println(users);
}catch(IOException ioe) {
ioe.printStackTrace();
}
}
}
I am trying to load a properties file from WEb-INF folder in my web application , which is running on Websphere 8.5 . I am using below code to load the file from the location
public class Init {
private final String WEB_INF_DIR_NAME="WEB-INF";
private String web_inf_path;
private final Properties APP_PROPERTIES =null;
InputStream inputStream = null;
public String getWebInfPath() throws IOException {
if (web_inf_path == null) {
web_inf_path = URLDecoder.decode(Init.class.getProtectionDomain().getCodeSource().getLocation().getPath(), "UTF8");
web_inf_path=web_inf_path.substring(0,web_inf_path.lastIndexOf(WEB_INF_DIR_NAME)+WEB_INF_DIR_NAME.length()).substring(1);
}
inputStream = Init.class.getResourceAsStream("/config/localhost/accountservice.properties");
// inputStream = this.getClass().getClassLoader().getResourceAsStream("/config/localhost/accountservice.properties");
if (inputStream != null) {
APP_PROPERTIES.load(inputStream);
}
System.out.println(APP_PROPERTIES.getProperty(AccountServiceDataAccessConstants.INET_LIBRARY_NAME)); // Here i am getting NULL
return web_inf_path;
}
}
I have also tried using servlet context , but its also giving me NULL. I have tried all possible ways to solve it but unfortunately i am not able to do it. I am also giving my folder structure.
Please excuse me if this is a silly question , but i am not really getting any idea about it.
Usually, everything in WebContent is placed in the root of your WAR file. So instead of
inputStream = Init.class.getResourceAsStream("/config/localhost/accountservice.properties");
It would be
inputStream = Init.class.getResourceAsStream("/WEB-INF/config/localhost/accountservice.properties");
The root of the WAR has WEB-INF in it, and then you can descend into your folder structure as normal.
I need to access a file inside the currently executed .jar using a URL.
URL url = BlockConverter.class.getResource("/test.txt");
System.out.println(url.toString());
InputStream is = url.openStream();
This is what I did.
The output is:
jar:file:/C:/Users/User/Desktop/SERVER/plugins/MyJar.jar!/test.txt
My InputStream always ends up throwing an IOException when being initialized, even though the URL is being output correctly.
So how is that possible?
Why can't I open the stream?
EDIT:
Also, please don't answer with "use getResourceAsStream", since it uses the same kind of code:
public InputStream getResourceAsStream(String name) {
URL url = getResource(name);
try {
return url != null ? url.openStream() : null;
} catch (IOException e) {
return null;
}
}
I would open it as a stream directly e.g.
InputStream is = BlockConverter.class.getResourceAsStream("/test.txt");
The above method is the way I normally access resources within a jar (it will open the resource regardless of it being packaged within a jar, or simply as an unpackaged deployment, note)
i want to open a file and return its content. Although it is in the same directory like the class that wants to open the file, the file can't be found. Would be cool if you could help me solving the problem.
Here is the code:
#GET #Produces("text/html") #Path("/{partNO}/") #Consumes("text/html")
public String getPartNoResponseHTML(#PathParam("partNO") String parID) throws WebApplicationException {
PartNoTemplate partNo = getPartNoResponse(parID);
String result = "";
try {
result = readFile(PART_NO_TEMPLATE_FILE);
} catch (FileNotFoundException e) {
e.printStackTrace(System.out);
return e.getMessage() + e.toString();
// throw new WebApplicationException(Response.Status.NOT_FOUND);
} finally {
result = result.replace("{partNO}", parID);
result = result.replace("{inputFormat}", partNo.getFormat().toString());
}
return result;
}
I guess it can't find the file, because its running on tomcat. I'm also using Jersey and JAX-RS. Thank you for your help,
Maxi
If the file is inside the application WAR (or in a jar) you can try by using
InputStream input = servletContext.getClass().getClassLoader().getResourceAsStream("my_filename.txt");
Your problem is similar (I think) with How can I read file from classes directory in my WAR?
Try to get the path of the file from ServletContext.
ServletContext context = //Get the servlet context
In JAX-RS to get servlet context use this:
#javax.ws.rs.core.Context
ServletContext context;
Then get the file from your web application:
File file = new File(context.getRealPath("/someFolder/myFile.txt"));
You don't post the code that actually tries to read the file, but assuming the file is in the classpath (as you mention it's in the same directory as the class) then you can do:
InputStream in = this.getClass().getResourceAsStream("/SomeTextFile.txt");
See here