When my application starts it reads a configuration properties file using the following code:
Properties properties = new Properties();
// parse the config resource
try (InputStream input = getClass().getClassLoader().getResourceAsStream(filename))
{
if (input == null)
{
// throw exception
}
// read the property list (key-value pairs) from the input byte stream
properties.load(input);
}
I am able to read and set individual properties.
The properties file is located in src/main/resources and after I build the application using maven, a copy of it is placed in target/classes. The jar file that is created also has a copy of it in the root directory when I open it up.
I would also like to be able to overwrite the properties file so that next time the application starts up, then it will read the new updated file. How do I achieve this? Is it even possible?
I found this question but no answers.
I've tried this:
try (OutputStream output = new FileOutputStream(filename))
{
properties.store(output, null);
}
which works if I just want to create a new file altogether. I would then have to modify the application so that it reads from a given folder rather than what originated from the resources folder. Is this what I should be doing?
I'm fairly new to Java so please go easy.
Storing the initial, default properties in the jar file, as resources is fine.
But if you want them to be writable, then you need to really store them as a file somewhere on the disk. Typically, under a .yourapp directory (or in a .yourapp file) inside the user's home directory.
So, try finding the file, and if not present, fallback to the resources. When writing, always write to the file.
This is an example code you can use for this. You create a config folder in the project root directory, an inside it you place your app.properties file
package com.yourparckage;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.util.Properties;
public class Config {
/* Create basic object */
private ClassLoader objClassLoader = null;
private Properties commonProperties = new Properties();
public static final String CONFIG_FILE = "config/app.properties";
/**
* This method loads config data from properties file
*
*/
public Config() {
objClassLoader = getClass().getClassLoader();
}
public String readKey(String propertiesFilename, String key)
{
/* Simple validation */
if (propertiesFilename != null && !propertiesFilename.trim().isEmpty() && key != null
&& !key.trim().isEmpty()) {
/* Create an object of FileInputStream */
InputStream objFileInputStream = null;
/**
* Following try-catch is used to support upto 1.6. Use try-with-resource in JDK
* 1.7 or above
*/
try {
/* Read file from resources folder */
objFileInputStream = new FileInputStream(propertiesFilename);
/* Load file into commonProperties */
commonProperties.load(objFileInputStream);
/* Get the value of key */
return String.valueOf(commonProperties.get(key));
} catch (FileNotFoundException ex) {
ex.printStackTrace();
} catch (IOException ex) {
ex.printStackTrace();
} finally {
/* Close the resource */
if (objFileInputStream != null) {
try {
objFileInputStream.close();
} catch (IOException ex) {
ex.printStackTrace();
}
}
}
}
return null;
}
}
Related
I am trying to read. a config from my resources folder in Java project from my deployed code. I am able to read from my local laptop but after deployment as JAR .manifest file, it says path does not exist.
So my Java maven project str: src/main/java/.. and config path as follows:
Java code to read this config where file.exists() always returns false.
Trial 1: When config path is : src/main/resources/config.yaml.
File configPath = new File(Objects.requireNonNull(getClass().getClassLoader().getResource("config.yaml")).getFile());
if (!configPath.exists()) {
Log("ERROR", "Config file does not exist "); // this is printed
}
Trial 2: When config path is src/main/resources/feed/configs/config.yaml.
File dir = new File(Objects.requireNonNull(getClass().getClassLoader().getResource("feed/configs")).getFile());
if (!dir.exists()) {
Log("ERROR", "Config folder does not exist, "ERROR"); // THIS IS PRINTED
return;
}
File[] configFiles = configPath.listFiles(); // NOT EXECUTED AS ABOVE IS RETURNED
Since you have added the maven tag, I am assuming you are using maven.
As the .yaml is inside the resources folder you should be using getResourceAsStream()
/src/main/resources/config.yaml:
first: value1
second: value2
To read the file and its content:
import java.util.Properties;
import java.io.InputStream;
import java.io.IOException;
public class Example {
InputStream inputStream = null;
final Properties properties = new Properties();
public Example() {
try {
inputStream =
this.getClass().getClassLoader().getResourceAsStream("config.yaml");
properties.load(inputStream);
} catch (IOException exception) {
LOG("ERROR", "Config file does not exist ");
} finally {
if (inputStream != null){
try {
inputStream.close();
} catch (Exception e) {
LOG("ERROR", "Failed to close input stream");
}
}
}
}
public printValues(){
LOG("INFO", "First value is: " + properties.getProperty("first"));
}
}
Currently I am trying to read my config file from root of project directory, in order to make this actual configuration I want to move this to external location and then read from there.
Adding a complete path in following code throws out error :
package CopyEJ;
import java.util.Properties;
public class Config
{
Properties configFile;
public Config()
{
configFile = new java.util.Properties();
try {
// configFile.load(this.getClass().getClassLoader().getResourceAsStream("CopyEJ/config.properties"));
Error Statement ** configFile.load(this.getClass().getClassLoader().getResourceAsStream("C://EJ_Service//config.properties"));
}catch(Exception eta){
eta.printStackTrace();
}
}
public String getProperty(String key)
{
String value = this.configFile.getProperty(key);
return value;
}
}
Here's the error:
java.lang.NullPointerException
at java.util.Properties$LineReader.readLine(Properties.java:365)
at java.util.Properties.load(Properties.java:293)
at CopyEJ.Config.<init>(Config.java:13)
at CopyEJ.CopyEJ.main(CopyEJ.java:22)
Exception in thread "main" java.lang.NullPointerException
at java.io.File.<init>(File.java:194)
at CopyEJ.CopyEJ.main(CopyEJ.java:48)
How can I fix this ?
The purpose of method getResourceAsStream is to open stream on some file, which exists inside your jar. If you know exact location of particular file, just open new FileInputStream.
I.e. your code should look like:
try (FileInputStream fis = new FileInputStream("C://EJ_Service//config.properties")) {
configFile.load(fis);
} catch(Exception eta){
eta.printStackTrace();
}
This line requires your config.properties to be in the java CLASSPATH
this.getClass().getClassLoader().getResourceAsStream("C://EJ_Service//config.properties")
When it is not, config.properties won't be accessible.
You can try some other alternative and use the configFile.load() function to read from.
One example would be:
InputStream inputStream = new FileInputStream(new File("C:/EJ_Service/config.properties"));
configFile.load(inputStream);
I'm working on a Java EE project that provide a servlet that store a lot of images (or common files) into glassfish server for specific users.
I'm wondering if there are standard directories to save the files into standard web programming.
For example, I have three users that want upload their files, where I can save them into server?
There are'nt any standard directories. I suggest you to create directory on a server to each user. For example: The user registers, some data goes to the database and also a directory for this user is created. Than this user can upload any file to his own directory.
P.S you can create directories anywhere on the server, then configure the path of directory in server's JNDI resources for look in your application.
You have to create PropertiesObjectFactory class to handle JNDI properties of java.util.Porperties (if you are using glassfish 2). Or yuo can write your custom ObjectFactory also. Glassfish 3 already have this function. It is set into: org.glassfish.resources.custom.factory.PropertiesFactory.
Create directory somewhere on the server. For e.g: /server/glassfish/users
Open glassfish admin console and navigate to: Resources -> JNDI -> Custom Resources, click "New". Provide a JNDI name, for e.g: jndi/users_directories, choose a resource type "java.util.Properties", specify Factory class: org.glassfish.resources.custom.factory.PropertiesFactory, then click "Add property", specify name for e.g: users.directories and in value column copy your directory path. In this case: /server/glassfish/users. Click OK and thats all.
Restart application server.
Make a look up in your application:
public Properties getProperties(String jndiName) {
Properties properties = null;
try {
InitialContext context = new InitialContext();
properties = (Properties) context.lookup(jndiName);
context.close();
} catch (NamingException e) {
LOGGER.error("Naming error occurred while initializing properties from JNDI.", e);
return null;
}
return properties;
}
When you call this method in your application provide a JNDI name you configured in your application server: jndi/users_directories. If you have mapped resources in deploymet descriptor you have to use: java:comp/env/jndi/users_directories.
If you whant to do the same using spring:
<jee:jndi-lookup id="usersDirectories"
jndi-name="jndi/users_directories"/>
Or if you are using glassfish 2, then create a custom PropertiesObjectFactory class:
public class PropertiesObjectFactory implements Serializable, ObjectFactory {
/**
* File property name.
*/
public static final String FILE_PROPERTY_NAME = "org.glassfish.resources.custom.factory.PropertiesFactory.fileName";
/**
* Implemented method from object factory interface.
*
* #param obj object
* #param name name
* #param nameCtx context name
* #param environment environment
* #return file properties
* #throws Exception if error occurs
*/
public Object getObjectInstance(Object obj, Name name, Context nameCtx, Hashtable<?, ?> environment)
throws Exception {
Reference ref = (Reference) obj;
Enumeration<RefAddr> refAddrs = ref.getAll();
String fileName = null;
Properties fileProperties = new Properties();
Properties properties = new Properties();
while (refAddrs.hasMoreElements()) {
RefAddr addr = refAddrs.nextElement();
String type = addr.getType();
String value = (String) addr.getContent();
if (type.equalsIgnoreCase(FILE_PROPERTY_NAME)) {
fileName = value;
} else {
properties.put(type, value);
}
}
if (fileName != null) {
File file = new File(fileName);
if (!file.isAbsolute()) {
file = new File(System.getProperty("com.sun.aas.installRoot") + File.separator + fileName);
}
try {
if (file.exists()) {
try {
FileInputStream fis = new FileInputStream(file);
if (fileName.toUpperCase().endsWith("XML")) {
fileProperties.loadFromXML(fis);
} else {
fileProperties.load(fis);
}
} catch (IOException ioe) {
throw new IOException("IO Exception during properties load : " + file.getAbsolutePath());
}
} else {
throw new FileNotFoundException("File not found : " + file.getAbsolutePath());
}
} catch (FileNotFoundException fnfe) {
throw new FileNotFoundException("File not found : " + file.getAbsolutePath());
}
}
fileProperties.putAll(properties);
return fileProperties;
}
}
Make a .jar file of this class and put it to server global library directory. Provide this factory class for your JNDI resources, restart the server and you are good to use the same look up, covered above.
I have an exe process that is running with a shortcut.
In the "Start in" property of the shortcut I set it to the folder where all app resources are. The process still looks for files at the location of the exe and not the location written in the shortcut.
I can also see it in Process Explorer - the "current directory" is the location of the exe.
Is there a way to change it?
(If I wasn't clear enough -
I want to put my app in a central network location and not in each user folder - but I want it to run - above each user folder by putting a shortcut in each user folder.)
BTW : Why don't I solve it with code writing? Because of third party jars I have in my exe (I am using exe4j to make an exe)
From exe4-j documentation.., it seems this can be configured in exe4j project.
Working directory
For some applications (especially GUI applications) you might want to change the working directory
to a specific directory relative to the executable, for example to read config files that are in a fixed
location. To do so, please select the Change working directory to: checkbox and enter a
directory relative to the executable in the adjacent text field. To change the current directory to the
same directory where the executable is located, please enter a single dot.
One alternative is to use a System Property. Just create a shortcut like this:
java -Dmyproperty="\\myserver\myfolder" -jar yourjar.jar
And get this property on your program:
System.getProperty("myproperty");
You can also set multiple System Properties.
I would start the java application via a cmd or bat file, then change to the work dir before you call javaw. If you don't do any thing special in your java application code all the paths in it will be relative to the place where you started java.
Jess
You can hack the classpath programatically which would allow you to specify a specific folder or series of folders to access the data.
import java.io.IOException;
import java.io.File;
import java.net.URLClassLoader;
import java.net.URL;
import java.lang.reflect.Method;
public class ClassPathHacker {
private static final Class[] parameters = new Class[]{URL.class};
public static void addFile(String s) throws IOException {
File f = new File(s);
addFile(f);
}//end method
public static void addFile(File f) throws IOException {
addURL(f.toURI().toURL());
}//end method
public static void addURL(URL u) throws IOException {
URLClassLoader sysloader = (URLClassLoader) ClassLoader.getSystemClassLoader();
Class sysclass = URLClassLoader.class;
try {
Method method = sysclass.getDeclaredMethod("addURL", parameters);
method.setAccessible(true);
method.invoke(sysloader, new Object[]{u});
} catch (Throwable t) {
t.printStackTrace();
throw new IOException("Error, could not add URL to system classloader");
}//end try catch
}//end method
}//end class
with the property loader file of
import java.io.InputStream;
import java.util.Enumeration;
import java.util.Locale;
import java.util.Properties;
import java.util.ResourceBundle;
public abstract class PropertyLoader
{
/**
* Looks up a resource named 'name' in the classpath. The resource must map
* to a file with .properties extention. The name is assumed to be absolute
* and can use either "/" or "." for package segment separation with an
* optional leading "/" and optional ".properties" suffix. Thus, the
* following names refer to the same resource:
* <pre>
* some.pkg.Resource
* some.pkg.Resource.properties
* some/pkg/Resource
* some/pkg/Resource.properties
* /some/pkg/Resource
* /some/pkg/Resource.properties
* </pre>
*
* #param name classpath resource name [may not be null]
* #param loader classloader through which to load the resource [null
* is equivalent to the application loader]
*
* #return resource converted to java.util.Properties [may be null if the
* resource was not found and THROW_ON_LOAD_FAILURE is false]
* #throws IllegalArgumentException if the resource was not found and
* THROW_ON_LOAD_FAILURE is true
*/
public static Properties loadProperties (String name, ClassLoader loader)
{
if (name == null)
throw new IllegalArgumentException ("null input: name");
if (name.startsWith ("/"))
name = name.substring (1);
if (name.endsWith (SUFFIX))
name = name.substring (0, name.length () - SUFFIX.length ());
Properties result = null;
InputStream in = null;
try
{
if (loader == null) loader = ClassLoader.getSystemClassLoader ();
if (LOAD_AS_RESOURCE_BUNDLE)
{
name = name.replace ('/', '.');
// Throws MissingResourceException on lookup failures:
final ResourceBundle rb = ResourceBundle.getBundle (name,
Locale.getDefault (), loader);
result = new Properties ();
for (Enumeration keys = rb.getKeys (); keys.hasMoreElements ();)
{
final String key = (String) keys.nextElement ();
final String value = rb.getString (key);
result.put (key, value);
}
}
else
{
name = name.replace ('.', '/');
if (! name.endsWith (SUFFIX))
name = name.concat (SUFFIX);
// Returns null on lookup failures:
in = loader.getResourceAsStream(name);
if (in != null)
{
result = new Properties ();
result.load (in); // Can throw IOException
}
}
}
catch (Exception e)
{
result = null;
}
finally
{
if (in != null) try { in.close (); } catch (Throwable ignore) {}
}
if (THROW_ON_LOAD_FAILURE && (result == null))
{
throw new IllegalArgumentException ("could not load [" + name + "]"+
" as " + (LOAD_AS_RESOURCE_BUNDLE
? "a resource bundle"
: "a classloader resource"));
}
return result;
}
/**
* A convenience overload of {#link #loadProperties(String, ClassLoader)}
* that uses the current thread's context classloader.
*/
public static Properties loadProperties (final String name)
{
return loadProperties (name,
Thread.currentThread ().getContextClassLoader ());
}
private static final boolean THROW_ON_LOAD_FAILURE = true;
private static final boolean LOAD_AS_RESOURCE_BUNDLE = false;
private static final String SUFFIX = ".properties";
} // End of class
then you can add a path as follows
try {
//First Load up the properties and populate the config
ClassPathHacker.addFile("/pathtomyapp");
} catch (IOException ex) {
ex.printStackTrace();
}
properties = PropertyLoader.loadProperties("myapp");
or you can also use getResourceBundle to get your resources, this is just one example of hacking the classpath to allow files to be available, you can always just add the classpath programatically and let the jar files you need to be available to reside there, so if you always ensure that the app network path is Q: you can add Q:\ to the classpath.
I need to read a properties files that's buried in my package structure in com.al.common.email.templates.
I've tried everything and I can't figure it out.
In the end, my code will be running in a servlet container, but I don't want to depend on the container for anything. I write JUnit test cases and it needs to work in both.
When loading the Properties from a Class in the package com.al.common.email.templates you can use
Properties prop = new Properties();
InputStream in = getClass().getResourceAsStream("foo.properties");
prop.load(in);
in.close();
(Add all the necessary exception handling).
If your class is not in that package, you need to aquire the InputStream slightly differently:
InputStream in =
getClass().getResourceAsStream("/com/al/common/email/templates/foo.properties");
Relative paths (those without a leading '/') in getResource()/getResourceAsStream() mean that the resource will be searched relative to the directory which represents the package the class is in.
Using java.lang.String.class.getResource("foo.txt") would search for the (inexistent) file /java/lang/String/foo.txt on the classpath.
Using an absolute path (one that starts with '/') means that the current package is ignored.
To add to Joachim Sauer's answer, if you ever need to do this in a static context, you can do something like the following:
static {
Properties prop = new Properties();
InputStream in = CurrentClassName.class.getResourceAsStream("foo.properties");
prop.load(in);
in.close()
}
(Exception handling elided, as before.)
The following two cases relate to loading a properties file from an example class named TestLoadProperties.
Case 1: Loading the properties file using ClassLoader
InputStream inputStream = TestLoadProperties.class.getClassLoader()
.getResourceAsStream("A.config");
properties.load(inputStream);
In this case the properties file must be in the root/src directory for successful loading.
Case 2: Loading the properties file without using ClassLoader
InputStream inputStream = getClass().getResourceAsStream("A.config");
properties.load(inputStream);
In this case the properties file must be in the same directory as the TestLoadProperties.class file for successful loading.
Note: TestLoadProperties.java and TestLoadProperties.class are two different files. The former, .java file, is usually found in a project's src/ directory, while the latter, .class file, is usually found in its bin/ directory.
public class Test{
static {
loadProperties();
}
static Properties prop;
private static void loadProperties() {
prop = new Properties();
InputStream in = Test.class
.getResourceAsStream("test.properties");
try {
prop.load(in);
in.close();
} catch (IOException e) {
e.printStackTrace();
}
}
public class ReadPropertyDemo {
public static void main(String[] args) {
Properties properties = new Properties();
try {
properties.load(new FileInputStream(
"com/technicalkeeda/demo/application.properties"));
System.out.println("Domain :- " + properties.getProperty("domain"));
System.out.println("Website Age :- "
+ properties.getProperty("website_age"));
System.out.println("Founder :- " + properties.getProperty("founder"));
// Display all the values in the form of key value
for (String key : properties.stringPropertyNames()) {
String value = properties.getProperty(key);
System.out.println("Key:- " + key + "Value:- " + value);
}
} catch (IOException e) {
System.out.println("Exception Occurred" + e.getMessage());
}
}
}
Assuming your using the Properties class, via its load method, and I guess you are using the ClassLoader getResourceAsStream to get the input stream.
How are you passing in the name, it seems it should be in this form: /com/al/common/email/templates/foo.properties
I managed to solve this issue with this call
Properties props = PropertiesUtil.loadProperties("whatever.properties");
Extra, you have to put your whatever.properties file in /src/main/resources
Nobody mentions the similar but even simpler solution than above with no need to deal with the package of the class. Assuming myfile.properties is in the classpath.
Properties properties = new Properties();
InputStream in = ClassLoader.getSystemResourceAsStream("myfile.properties");
properties.load(in);
in.close();
Enjoy
use the below code please :
Properties p = new Properties();
StringBuffer path = new StringBuffer("com/al/common/email/templates/");
path.append("foo.properties");
InputStream fs = getClass().getClassLoader()
.getResourceAsStream(path.toString());
if(fs == null){
System.err.println("Unable to load the properties file");
}
else{
try{
p.load(fs);
}
catch (IOException e) {
e.printStackTrace();
}
}