I have tried DeploymentException & Class Not Found on WebLogic Admin Server 11g among other examples, but I have not been able to get Spring to read external properties file.
I have this code, which read a properties file. Putting the file (PROPERTIES_FILES) in src/main/resources, the app i deployed fine. Moving it to an external folder in the file system, it failes to deploy.
I tried http://www.mkyong.com/java/how-to-print-out-the-current-project-classpath/ to pribt the classpaths and get:
/C:/Oracle/Middleware/patch_wls1034/profiles/default/sys_manifest_classpath/weblogic_patch.jar
/C:/Oracle/Middleware/patch_oepe1040/profiles/default/sys_manifest_classpath/weblogic_patch.jar /C:/Oracle/Middleware/patch_ocp360/profiles/default/sys_manifest_classpath/weblogic_patch.jar
/C:/Oracle/Middleware/jdk160_21/lib/tools.jar
/C:/Oracle/Middleware/wlserver_10.3/server/lib/weblogic_sp.jar
/C:/Oracle/Middleware/wlserver_10.3/server/lib/weblogic.jar
/C:/Oracle/Middleware/modules/features/weblogic.server.modules_10.3.4.0.jar
/C:/Oracle/Middleware/wlserver_10.3/server/lib/webservices.jar
/C:/Oracle/Middleware/modules/org.apache.ant_1.7.1/lib/ant-all.jar
/C:/Oracle/Middleware/modules/net.sf.antcontrib_1.1.0.0_1-0b2/lib/ant-contrib.jar
/C:/Oracle/Middleware/wlserver_10.3/common/derby/lib/derbyclient.jar
/C:/Oracle/Middleware/wlserver_10.3/server/lib/xqrl.jar
Trying to read test.properties
Found false
Could not find properties file: test.properties
My implementation looks like this:
ClassLoader cl = ClassLoader.getSystemClassLoader();
URL[] urls = ((URLClassLoader)cl).getURLs();
for(URL url: urls){
logger.debug(url.getFile());
}
logger.debug("Trying to read {}", PROPERTIES_FILES);
Resource resource = new ClassPathResource(PROPERTIES_FILES);
logger.debug("Found {}", resource.exists());
try {
props = PropertiesLoaderUtils.loadProperties(resource);
} catch (IOException e) {
logger.error("Could not find properties file: " + PROPERTIES_FILES, e);
}
I have the same problem to move another properties file from in servlet-dispatcher:
<context:property-placeholder location="classpath*:test.properties"/>
But I guess it's the same problem.
I am on Windows.
Can anybody help me?
Did a full system restart and it worked.
Related
I have a Spring Boot project and everything works fine locally. Now when i create a runnable jar to run it via jenkins then it is not able to load Property file.
Following is the code where PropertyPlaceholder is configured:
#Bean
public static EncryptablePropertyPlaceholderConfigurer propertyPlaceholderConfigurerEncrypted() {
String env = System.getProperty("spring.profiles.active") != null ? System.getProperty("spring.profiles" +
".active") : "ci";
EncryptablePropertyPlaceholderConfigurer ppc =
new EncryptablePropertyPlaceholderConfigurer(getStandardPBEStringEncryptor());
ppc.setLocations(new ClassPathResource("application.properties"),
new ClassPathResource("application-" + env + ".properties"));
return ppc;
}
In order to debug i added following code within this:
try {
String s = new String(Files.readAllBytes(new ClassPathResource("application-" + env + ".properties").getFile().toPath()));
LOG.info(s);
} catch (IOException e) {
LOG.error("Unable to read file",e);
}
And it gives this error :
java.io.FileNotFoundException: class path resource [application-qa1.properties] cannot be resolved to absolute file path because it does not reside in the file system: jar:file:/var/hudson/workspace/pv/target/T-S-21.3.40.jar!/BOOT-INF/classes!/application-qa1.properties
17:23:44 at org.springframework.util.ResourceUtils.getFile(ResourceUtils.java:217)
I have confirmed that file is located in jar at this location BOOT-INF/classes/application-qa1.properties
So effectively issue is caused due to second exclamation mark showing up in path while loading file from jar /var/hudson/workspace/pv/target/T-S-21.3.40.jar!/BOOT-INF/classes!/application-qa1.properties
Ideally exclamation mark should appear only after jar name.
Can someone please advise on how to address this issue.
You can't read files from a JAR like this. You have to use getResourceAsStream like this:
InputStream is = this.getClass().getClassLoader().getResourceAsStream("application-" + env + ".properties");
I have been searching for 3 days now nonstop looking at every post I can find. My program runs on IntelliJ, but cannot run on an executable. Your help would be appreciated :)
More importantly, where can I find a in depth user-friendly tutorial? Is there a course or book I can pay for? On Udemy, the java classes completely fail to mention I/O such as classpath and URI. TutorialsPoint briefly goes over I/O buts its not indepth. Did I miss something? Is there an easier way to do all this??
Similar posts that have not worked for me:
Java Jar file: use resource errors: URI is not hierarchical
https://stackoverflow.com/a/27149287/155167
I am trying to load an excel file. I am using Maven. Apache POI says it needs a File. So InputStream does not work. http://poi.apache.org/components/spreadsheet/quick-guide.html#FileInputStream
When I java -jar jarFile, it gives me the error:
C:\Users\1010\Documents\Personal\MonsterManager>java -jar monsterManagerVer3.jar
Exception in thread "main" java.lang.IllegalArgumentException: URI is not hierarchical
at java.base/java.io.File.<init>(File.java:421)
at LoadExcel.<init>(LoadExcel.java:62)
at monsterRunner.<init>(monsterRunner.java:13)
at monsterRunner.main(monsterRunner.java:24)
Here is the code
public LoadExcel() throws IOException, URISyntaxException, InvalidFormatException {
mNames = null;
URL ins = this.getClass().getResource("/excel_List.xlsx");
if (ins == null)
throw new FileNotFoundException("The XLSX file didn't exist on the classpath");
ins.toString().replace(" ","%20"); //<-Runs without this part
File file = new File(ins.toURI()); //this is line 62
// File f = new File(getClass().getResource("/MyResource").toExternalForm());
//String path = this.getClass().getClassLoader().getResource("/excel_List.xlsx").toExternalForm();
//File file = new File(path);
OPCPackage pkg = OPCPackage.open(file);
XSSFWorkbook wb = new XSSFWorkbook(pkg);
//FileInputStream fis=new FileInputStream(new File(excelFile));
// XSSFWorkbook wb = new XSSFWorkbook(fis);
XSSFSheet sheet = wb.getSheetAt(0);
loadExcel(sheet);
cacheNames();
// fis.close();
wb.close();
}
If it helps here is the path to the excel file:
src\main\resources\excel_List.xlsx
UPDATE:
so I took the excel file out of the resources folder
\nameOfMyProgram\excel_List.xlsx
and now I get this error.
I tried several versions of using the classLoader, Class and Thread to solve this error from Different ways of loading a file as an InputStream
but I still cannot get it to compile.
Error and my code
If you have to use File object do not put xls-file into resources directory.
Maven puts all files from resources directory into jar.
Java can not create File object based on file in jar-file.
Put your xls-file somewhere in file system and create File object based on its URL.
Since your xls-file is not a resource do not use getResource.
Its URL is its full filename (with path).
This code below works with jar executable
String path = new File("excel_List.xlsx").getAbsoluteFile().toString();
File file = new File(path);
if(!file.exists()){
JOptionPane.showMessageDialog(null,"File not found!");
System.exit(0);
}
OPCPackage pkg = OPCPackage.open(file);
XSSFWorkbook wb = new XSSFWorkbook(pkg);
My grade java project has the following structure.
As you can see the resources folder is in present in the class path.
But when I run the following in a class under java folder
new File("somefile.txt").exists()
I get FileNotFoundException.
Could anyone help me find why I am not able to access this file.
This is in the class path.
You can use.
ClassLoader classLoader = getClass().getClassLoader();
String filePath= classLoader.getResource("filename").getFile();
new File(filePath).exists();
For more info, you can go through this tutorial.
You can resolve your issue like below
Properties prop = new Properties();
InputStream inputStream = getClass().getClassLoader().getResourceAsStream("somefile.txt");
if (inputStream != null) {
prop.load(inputStream);
} else {
throw new FileNotFoundException("Property file '" + fileName + "' not found in the classpath");
}
I found it from the post How to read properties file in Java
my spring boot project structure is like this
src
|-main
|--|-java
|--|-resources
static
|-css
|-images
|-js
now I want to write a file into the static/images folder
I tried to new File like
BufferedOutputStream stream =new BufferedOutputStream(new FileOutputStream(new File("static/images")));it will throw "No such file or directory" exception
but in other html file I can get the js by "js/jsFile.js"
new File("static/images") is right
I used new File("/static/images") so I got an Exception
I was in the situation where using Spring Boot I had to save the image into one directory which is accessed statically.
Below code worked perfect
byte[] imageByteArray ....
String fileName = "image.png";
String fileLocation = new File("static\\images").getAbsolutePath() + "\\" + fileName;
FileOutputStream fos = new FileOutputStream(fileLocation);
fos.write(imageByteArray);
fos.close();
Hope it helped.
#zhuochen shen is correct.
The thing is to happen me is Eclipse doesn't show write file. So I looked at file explore. File is writing correctly.
try {
Path path=Paths.get("static/images/"+productDto.getImage().getOriginalFilename());
Files.write(path,productDto.getImage().getBytes());
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
enter image description here
enter image description here
I'm developing a project with Maven. In a class to send e-mails, in run and dev modes, I get the following error: Caused by: java.io.FileNotFoundException: jQuery/images/logo.png (Ficheiro ou directoria inexistente) ==> translation = File or directory not found.
I've tryed lots of paths, like "./jQuery/images/logo.png", "/jQuery/images/logo.png" and others. The full relative path is: "src/main/webapp/jQuery/images/logo.png".
In "target" folder, the path is "project-1.0-SNAPSHOT/jQuery/images/logo.png".
Inside war file, is "jQuery/images/logo.png".
I don't think it's important, but I'm using NetBeans 7.1.1 as IDE.
I found that the absolute path returned in runtime is "/home/user/apache-tomcat-7.0.22/bin/jQuery/images/logo.png"!... It's not the project path!
How can I get a file in webapp folder and descendents from a Java class, in a Maven project?
The code is:
MimeBodyPart attachmentPart = null;
FileDataSource fileDataSource = null;
for (File a : attachments) {
System.out.println(a.getAbsolutePath());
attachmentPart = new MimeBodyPart();
fileDataSource = new FileDataSource(a) {
#Override
public String getContentType() {
return "application/octet-stream";
}
};
attachmentPart.setDataHandler(new DataHandler(fileDataSource));
attachmentPart.setFileName(fileDataSource.getName());
multipart.addBodyPart(attachmentPart);
}
msg.setContent(multipart);
msg.saveChanges();
Transport transport = session.getTransport("smtp");
transport.connect(host, from, "password");
transport.sendMessage(msg, msg.getAllRecipients());
The path .../apache-tomcat-7.0.22/bin/jQuery/... is really odd. bin contains scripts for Tomcat, it should not contain any code nor any resources related to your application. Even if you deploy your app in the context /bin, the resources would end up under `.../apache-tomcat-7.0.22/webapps/bin/...
This looks like a mistake made in an earlier deployment.
Now back to your question. To get the path of resource of your web app, use this code:
String absolutePath = getServletContext().getRealPath("/jQuery/images/logo.png");
(docs)