I tried to create a quick framework. in that I created below-mentioned classes:
Config file(All browsers path)
configDataProvider java class(reads the above file)
BrowserFactory class(has firefox browser object)
configDataProviderTest class(access data from dconfigDataProvider class)
now its not reading the paths mentioned in config.properties file.
I have provided all correct path and attached screenshots:
Looks like a problem is at your ConfigDataProvider class.
Firstly, you using Maven for building your project. Maven has defined project structure for code sources and for resources:
/src/main/java
/src/main/resorces
Thus, much better to put your .properties file there.
Second, you don't need to set the full path to your config file.
Relative path will be just enough. Something like below:
public class PropertiesFileHandler {
private static Logger log = Logger.getLogger(PropertiesFileHandler.class);
public static final String CONFIG_PROPERTIES = "src/main/resources/config.properties";
public static final String KEY = "browser.type";
public static BrowserType readBrowserType() {
BrowserType browserType = null;
Properties properties = new Properties();
try (InputStream inputStream = new BufferedInputStream(new FileInputStream(CONFIG_PROPERTIES))) {
properties.load(inputStream);
browserType = Enum.valueOf(BrowserType.class, properties.getProperty(KEY));
} catch (FileNotFoundException e) {
log.error("Properties file wasn't found - " + e);
} catch (IOException e) {
log.error("Problem with reading properties file - " + e);
}
return browserType;
}
}
Lastly, if you are building framework you don't need to put everything under src/main/test. This path specifies tests with future possibilities to be executed with maven default lifecycle - mvn test.
The core of your framework can look like:
Two things which I noticed:
Don't give path in your properties path within ""
all the path seperators should be replaced with double backward slash \\ or single forward slash /
Related
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();
}
}
}
Is it possible to access Assets inside the Java code in Play Framework? How?
We access assets from the scala HTML templates this way:
<img src="#routes.Assets.versioned("images/myimage.png")" width="800" />
But I could not find any documentation nor code example to do it from inside the Java code. I just found a controllers.Assets class but it is unclear how to use it. If this is the class that has to be used, should it maybe be injected?
I finally found a way to access the public folder even from a production mode application.
In order to be accessible/copied in the distributed version, public folder need to be mapped that way in build.sbt:
import NativePackagerHelper._
mappings in Universal ++= directory("public")
The files are then accessible in the public folder in the distributed app in production form the Java code:
private static final String PUBLIC_IMAGE_DIRECTORY_RELATIVE_PATH = "public/images/";
static File getImageAsset(String relativePath) throws ResourceNotFoundException {
final String path = PUBLIC_IMAGE_DIRECTORY_RELATIVE_PATH + relativePath;
final File file = new File(path);
if (!file.exists()) {
throw new ResourceNotFoundException(String.format("Asset %s not found", path));
}
return file;
}
This post put me on the right way to find the solution: https://groups.google.com/forum/#!topic/play-framework/sVDoEtAzP-U
The assets normally are in the "public" folder, and I don't know how you want to use your image so I have used ImageIO .
File file = new File("./public/images/nice.png");
boolean exists = file.exists();
String absolutePath = file.getAbsolutePath();
try {
ImageInputStream input = ImageIO.read(file); //Use it
} catch (IOException e) {
e.printStackTrace();
}
System.out.println("EX = "+exists+" - "+absolutePath);
I use gradle which structures projects in maven style so I have the following
src/main/java/Hello.java and src/main/resources/test.properties
My Hello.java look like this
public class Hello {
public static void main(String[] args) {
Properties configProperties = new Properties();
ClassLoader classLoader = Hello.class.getClassLoader();
try {
configProperties.load(classLoader.getResourceAsStream("test.properties"));
System.out.println(configProperties.getProperty("first") + " " + configProperties.getProperty("last"));
} catch (IOException e) {
e.printStackTrace();
}
}
}
This works fine. however I want to be able to point to .properties file outside of my project and I want to it to be flexible enough that I can point to any location without rebuilding the jar every time. Is there a way to this without using a File API and passing file path as an argument to the main method?
You can try this one, which will first try to load properties file from project home directory so that you don't have to rebuild jar, if not found then will load from classpath
public class Hello {
public static void main(String[] args) {
String configPath = "test.properties";
if (args.length > 0) {
configPath = args[0];
} else if (System.getenv("CONFIG_TEST") != null) {
configPath = System.getenv("CONFIG_TEST");
}
File file = new File(configPath);
try (InputStream input = file.exists() ? new FileInputStream(file) : Hello.class.getClassLoader().getResourceAsStream(configPath)) {
Properties configProperties = new Properties();
configProperties.load(input);
System.out.println(configProperties.getProperty("first") + " " + configProperties.getProperty("last"));
} catch (IOException ex) {
ex.printStackTrace();
}
}
}
You can send the properties file path as argument or set the path to an environment variable name CONFIG_TEST
Archaius may be complete overkill for such a simple problem, but it is a great way to manage external properties. It is a library for handling configuration: hierarchies of configuration, configuration from property files, configuration from databases, configuration from user defined sources. It may seem complicated, but you will never have to worry about hand-rolling a half-broken solution to configuration again. The Getting Started page has a section on using a local file as the configuration source.
I'm using Eclipse for EE Developer.
I need to access to a properties file (db.properties) from a class's method (DBQuery.java).
The class is located inside a package inside the src folder.
For the properties file i tried almost everything that i could find over the net to make it work, but looks like i can't.
The properties file is located inside the WebContent folder, and i'll add the code with which i'm trying to load this file:
public class DBQuery {
public static String create_DB_string(){
//the db connection string
String connString = "";
try{
Properties props = new Properties();
FileInputStream fis = new FileInputStream("db.properties");
props.load(fis);
fis.close();
/* creating connString using props.getProperty("String"); */
}
catch (Exception e) {
System.out.println(e.getClass());
}
return connString;
}
}
So my question is, where to put the properties file, and which is the correct way to load it?
You can put this propertie file within your java package for example com/test and use following:
getClass().getResourceAsStream( "com/test/myfile.propertie");
Hope it helps.
I have a properties file which is located under conf folder. conf folder is under the project root directory. I am using the following code.
public class PropertiesTest {
public static void main(String[] args) {
InputStream inputStream = PropertiesTest.class
.getResourceAsStream("/conf/sampleprop.conf");
Properties prop = new Properties();
try {
prop.load(inputStream);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println(prop.getProperty("TEST"));
}
}
But I get nullpointer exception.
I have tried using
InputStream inputStream = PropertiesTest.class
.getResourceAsStream("./conf/sampleprop.conf");
and
InputStream inputStream = PropertiesTest.class
.getResourceAsStream("conf/sampleprop.conf");
But all result in nullpointer exception.
Can anyone please help.
Thanks in advance
Try to recover your working directory first:
String workingDir = System.getProperty("user.dir");
System.out.println("Current working dir: " + workingDir);
and then is simple:
Properties propertiesFile = new Properties();
propertiesFile.load(new FileInputStream(workingDir+ "/yourFilePath"));
String first= propertiesFile.getProperty("myprop.first");
Regards, fabio
The getResourceAsStream() method tries to locate and load the resource using the ClassLoader of the class it is called on. Ideally it can locate the files only the class folders .. Rather you could use FileInputStream with relative path.
EDIT
if the conf folder is under src, then you still be able to access with getResourceAsStream()
InputStream inputStream = Test.class
.getResourceAsStream("../conf/sampleprop.conf");
the path would be relative to the class from you invoke getRes.. method.
If not
try {
FileInputStream fis = new FileInputStream("conf/sampleprop.conf");
Properties prop = new Properties();
prop.load(fis);
System.out.println(prop.getProperty("TEST"));
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
NOTE: this will only work if it is Stand alone application/in eclipse. This will not work if its web based (as the root will be Tomcat/bin, for eg)
I would suggest to copy the configuration file at designated place, then you can acess at ease. At certain extent 'System.getProperty("user.dir")' can be used if you are always copying the file 'tomcat` root or application root. But if the files to be used by external party, ideal to copy in a configurable folder (C:\appconf)
Your code works like a charm! But you might have to add the project root dir to your classpath.
If you work with Maven, place your configuration in src/main/resources/conf/sampleprop.conf
When invoking java directly add the project root dir with the java -classpath parameter. Something like:
java -classpath /my/classes/dir:/my/project/root/dir my.Main