First of all i've tried my code by extracting the war file (using maven) on both eclipse and on a tomcat 8.0.33 stand alone on my mac.
I have also tried my code on a windows server 2008 with the same java version 1.8, and it works when i put some variable (which are username and password) hardcoded, but when i make the code to read them from a reousrce file, it is just working on my mac (both eclipse and tomcat stand alone), but not on the server
this is the code to read the resource
private static String getUsername() {
Properties prop = new Properties();
InputStream input = null;
try {
input = new FileInputStream(MyConfiguration.class.getClassLoader()
.getResource("configuration.properties").getFile());
// load a properties file
prop.load(input);
// get the property value and print it out
return prop.getProperty("username");
} catch (IOException ex) {
ex.printStackTrace();
} finally {
if (input != null) {
try {
input.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
return null;
}
where the location of the configuration.properties is in the src/main/resources and on the server, i can see that file is in the correct directory.
i am using maven, i don't know what other information you need to help me, but if you say, i will give you
You may try
input = Thread.currentThread().getContextClassLoader().getResourceAsStream("configuration.properties");
Open your war and the jar file that contains your application. Ensure that the resource you are trying to open is in the jar (in the war) that is getting deployed. Most of the times this has happened to me, it's because I failed to tell my build process that this file needed to be copied over to the deployment. Class files go automatically, but not always resource files.
Related
My project structure looks like below. I do not want to include the config file as a resource, but instead read it at runtime so that we can simply change settings without having to recompile and deploy. my problem are two things
reading the file just isn't working despite various ways i have tried (see current implementation below i am trying)
When using gradle, do i needto tell it how to build/or deploy the file and where? I think that may be part of the problem..that the config file is not getting deployed when doing a gradle build or trying to debug in Eclipse.
My project structure:
myproj\
\src
\main
\config
\com\my_app1\
config.dev.properties
config.qa.properties
\java
\com\myapp1\
\model\
\service\
\util\
Config.java
\test
Config.java:
public Config(){
try {
String configPath = "/config.dev.properties"; //TODO: pass env in as parameter
System.out.println(configPath);
final File configFile = new File(configPath);
FileInputStream input = new FileInputStream(configFile);
Properties prop = new Properties()
prop.load(input);
String prop1 = prop.getProperty("PROP1");
System.out.println(prop1);
} catch (IOException ex) {
ex.printStackTrace();
}
}
Ans 1.
reading the file just isn't working despite various ways i have tried
(see current implementation below i am trying)
With the location of your config file you have depicted,
Change
String configPath = "/config.dev.properties";
to
String configPath = "src\main\config\com\my_app1\config.dev.properties";
However read the second answer first.
Ans 2:
When using gradle, do i needto tell it how to build/or deploy the file
and where? I think that may be part of the problem..that the config
file is not getting deployed when doing a gradle build or trying to
debug in Eclipse.
You have two choices:
Rename your config directory to resources. Gradle automatically builds the resources under "src/main/resources" directory.
Let Gradle know the additional directory to be considered as resources.
sourceSets {
main {
resources {
srcDirs = ["src\main\config\com\my_app1"]
includes = ["**/*.properties"]
}
}
}
reading the file just isn't working despite various ways i have tried (see current implementation below i am trying)
You need to clarify this statement. Are you trying to load properties from an existing file? Because the code you posted that load the Properties object is correct. So probably the error is in the file path.
Anyway, I'm just guessing what you are trying to do. You need to clarify your question. Is your application an executable jar like the example below? Are trying to load an external file that is outside the jar (In this case gradle can't help you)?
If you build a simple application like this as an executable jar
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.Properties;
public class Main {
public static void main(String[]args) {
File configFile = new File("test.properties");
System.out.println("Reading config from = " + configFile.getAbsolutePath());
FileInputStream fis = null;
Properties properties = new Properties();
try {
fis = new FileInputStream(configFile);
properties.load(fis);
} catch (IOException e) {
e.printStackTrace();
return;
} finally {
if(fis != null) {
try {
fis.close();
} catch (IOException e) {}
}
}
System.out.println("user = " + properties.getProperty("user"));
}
}
When you run the jar, the application will try to load properties from a file called test.properties that is located in the application working directory.
So if you have test.properties that looks like this
user=Flood2d
The output will be
Reading config from = C:\test.properties
user = Flood2d
And that's because the jar file and test.properties file is located in C:\ and I'm running it from there.
Some java applications load configuration from locations like %appdata% on Windows or /Library/Application on MacOS. This solution is used when an application has a configuration that can change (it can be changed by manually editing the file or by the application itself) so there's no need to recompile the application with the new configs.
Let me know if I have misunderstood something, so we can figure out what you are trying to ask us.
Your question is slightly vague but I get the feeling that you want the config files(s) to live "outside" of the jars.
I suggest you take a look at the application plugin. This will create a zip of your application and will also generate a start script to start it. I think you'll need to:
Customise the distZip task to add an extra folder for the config files
Customise the startScripts task to add the extra folder to the classpath of the start script
The solution for me to be able to read an external (non-resource) file was to create my config folder at the root of the application.
myproj/
/configs
Doing this allowed me to read the configs by using 'config/config.dev.properies'
I am not familiar with gradle,so I can only give some advices about your question 1.I think you can give a full path of you property file as a parameter of FileInputStream,then load it using prop.load.
FileInputStream input = new FileInputStream("src/main/.../config.dev.properties");
Properties prop = new Properties()
prop.load(input);
// ....your code
I have a certain requirement where I need to copy files from Unix server to Windows Shared Drive. I am developing the necessary code for this in Java. I am a beginner so please excuse me for this basic question.
I have my source path in my config file. So, I am using the below code to import my config file and set my variable. My Project has config.properties file attached to it.
public static String rootFolder = "";
Properties prop = new Properties();
InputStream input = null;
try {
input = new FileInputStream("config.properties");
} catch (FileNotFoundException e) {
e.printStackTrace();
System.out.println("Config files not able to set properly for Dest Folder");
}
try {
prop.load(input);
rootFolder = prop.getProperty("Dest_Root_Path");
System.out.println("Destination Folder is being initialized to - "+rootFolder);
} catch (IOException e) {
e.printStackTrace();
System.out.println("Destination Path not set properly");
}
When I am doing this I am getting an error saying the file is not found.
java.io.FileNotFoundException: config.properties (No such file or directory)
at java.io.FileInputStream.<init>(FileInputStream.java:158)
at java.io.FileInputStream.<init>(FileInputStream.java:113)
Exception in thread "main" java.lang.NullPointerException
at java.util.Properties.load(Properties.java:357)
I am triggering this jar using a unix ksh shell. Please provide guidance to me.
Put your config file somewhere within the classpath. For example, if it's a webapp, in WEB-INF/classes. If it isn't a webapp, create a folder outside the project, put the file there, and set the classpath so the new folder is in it.
Once you have your file in the classpath, get it as a resource with getResourceAsStream():
InputStream is = MyProject.class.getResourceAsStream("/config.properties");
Don't forget the slash / before the filename.
I'm using eclipse. I put a .pdf file in my src folder, i want to open it with the default OS program. The problem is that, if i execute the program with eclipse, i can open it (clicking on a MenuItem) like this :
File memo=new File("src/chap.pdf");
try {
Desktop.getDesktop().open(memo);
} catch (IOException e) {
e.printStackTrace();
}
However, after exporting the project to a jar file, this isn't working anymore. So is there a problem in my code or is there another way to get the file to open when it's in the jar file ?
The src path will not be available after you have exported your program and you should never reference it in any way.
You need to extract the resource from the Jar file and write it to the local disk...
try (InputStream is = getClass().getResourceAsStrea("/chap.pdf")) {
try (BufferedOutputStream os = new BufferedOutputStream(new FileOutputStream(...)) {
// Write contents like you would any file
}
} catch (IOException exp) {
exp.printStackTrace();
}
Then you can use the extracted file with Desktop
I am working on a dynamic web project.
On submit button click (present on my form)I want to create a new file and put some data inside.
I have written only these two lines and I am getting failure to create file
try{
File file = new File("C:/database.txt");
file.createNewFile();
}catch(Exception e){
return "error in creating file";
}
If I run the enire code in normal java class everything works fine. Why so?
Your web project is working on application server. The web application can manage files, which are on that server. Other files are not accessible. (of course localhost server is on your computer, but that it's path is not "C:/", so you can't write there). You can find the path of your server running this code (it also create a test file):
String pathWhereYouFindYourFile = new File("").getAbsolutePath();
System.out.println(pathWhereYouFindYourFile);
File f = new File("test.txt");
try {
f.createNewFile();
} catch (IOException e) {
e.printStackTrace();
}
I'm developing a program with NetBeans 8.0 and JavaFX Scene Builder 2.0 that need store some variables in a file, where admin users can modify it when needed, (like change server IP address, or a number value from a no editable textfield) and if they close and load again the program, the changes made in variables are kept. Like any settings section of a program.
I just try do it with the Properties file, but i have problems to store it in the same folder as .jar file. When the program execute the line new FileOutputStream("configuration.properties"); the file is created at root of the disk. As the folder of the file can be stored anywhere, i not know how indicate the right path.
Creating the properties file in the package of the main project and using getClass().getResourceAsStream("configuration.properties"); i can read it but then i can not write in for change values of variables.
Is there a better method to create a configuration file? Or properties file is the best option for this case?
My other question is whether it is possible to prevent access to the contents of the file or encrypt the content?
PD: I've been testing this part of the code in Linux operating system currently, but the program will be used in Windows 7 when ready.
If you use Maven, you can store your property files in your resources folder, say resources/properties/. When you need to load them, do this:
private Properties createProps(String name)
{
Properties prop = new Properties();
InputStream in = null;
try
{
in = getClass().getResourceAsStream(name);
prop.load(in);
}
catch (IOException ex)
{
System.err.println("failed to load \"" + name + "\": " + ex);
}
finally
{
try
{
if (in != null)
{
in.close();
}
}
catch (IOException ex)
{
System.err.println("failed to close InputStream for \"" + name + "\":\n" + FXUtils.extractStackTrace(ex));
}
}
return prop;
}
Where name is the full path to your properties file within your resources folder. For example, if you store props.properties in resources/properties/, then you would pass in properties/props.properties.
I am not 100% sure if you can carry over this exact procedure to a non-Maven project. You'd need to instruct whatever compiler tool you are using to also include your property files.
As far as your final question goes, in regards to encrypting your properties, I would consider posting that as a separate question (after having done thorough research to try to discover an existing solution that works for you).
At last i found how obtain the absolute path from folder where is .jar file to create properties file in, and read/write it. Here is the code:
File file = new File(System.getProperty("java.class.path"));
File filePath = file.getAbsoluteFile().getParentFile();
String strPath = filePath.toString();
File testFile = new File(strPath+"/configuration.properties");
Tested in Ubuntu 13.04 And Windows 7 and it works.
For encrypt the properties values i found this thread that answer how do it.