I am new to play framework. I want to know how to use the property file in play framework.
My property file is,
conf/test.properties
name=kumar
framework=playframework
so now i want to use test.properties inside my controller class (Application.java).
Please let me know what are the steps i need to do.
All files placed in the conf/ folder are automatically added to your classpath. You should be able to access the your file like this:
Properties prop;
try {
prop = new Properties();
InputStream is = this.getClass().getResourceAsStream("test.properties");
prop.load(is);
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
String name = prop.getProperty("name");
String framework = prop.getProperty("playframework");
Note: I haven't tested any of the above.
Update:
I've just realized that this question is a close duplicate of Load file from '/conf' directory on Cloudbees but since my solution also includes how to access the properties in the file, I'll leave it as is.
Also since your controller method will most likely be static the above might fail to compile. In the answer I referenced they suggested using the facility provided by Play!:
Play.application().resourceAsStream("test.properties")
you can access to property files in play framework by doing:
val ConfigLoader = play.Play.application.configuration
implicit val connector = ContactPoints(Seq(ConfigLoader.getString("cassandra.host"))).keySpace(ConfigLoader.getString("cassandra.keyspace"))
Related
I am trying to load and validate xml files from a directory in the class path at startup of a Spring Boot application. I am seeing the following error which indicates that I am trying to load files using absolute path and not class path:
java.io.FileNotFoundException: class path resource [converters/mapper.xml] cannot be resolved to absolute file path because it does not reside in the file system: jar:file:/opt/core/home/libexec/boss/core-service-2.0.0.jar!/BOOT-INF/lib/core-api-2.0.0.jar!/converters/mapper.xml
Below is a code snippet that loads the files:
..
#Autowired
public FieldsMapTypeConvertersRegistry(#Value("${core.files-location:converters}")
String mapperFilesLocation) {
this.mapperFilesLocation = mapperFilesLocation;
}
..
try {
// ToDo we need to replace this when we enable multi-tenancy
ClassLoader classLoader = ClassUtils.getDefaultClassLoader();
ResourcePatternResolver resolver = new PathMatchingResourcePatternResolver(classLoader);
Resource[] xmlResources = resolver.getResources(mapperFilesLocation + "/*.xml");
for (Resource xmlResource : xmlResources) {
File file = ResourceUtils.getFile(xmlResource.getURL());
registerTypeConverter(file);
}
} catch (IOException e) {
// do stuff
} catch (JAXBException e) {
//do stuff
}
I think the issue is in this statement in the code above:
File file = ResourceUtils.getFile(xmlResource.getURL());
but I am not sure what other ways I can do that. Any help is really appreciated.
I'm just wondering why you are using ResourceUtils.getFile(xmlResource.getURL()) when xmlResource.getFile() is already available to get the File Handle. Ideally speaking, you should be catching the FileNotFoundException inside the catch block and checking the detailed message wrapped inside the exception.
Edit:
The exception is being thrown because the xml is not found in classpath at runtime. Most probably, the file target/converters/mapper.xml is not available.
Try something like this MyService.class.getClassLoader().getResourceAsStream("/file.xml"); and then create File from stream.
Try using commons-io:commons-io:2.7 (Maven artifact) and use the following code:
InputStream inputStream = obj.getClass()
.getClassLoader()
.getResourceAsStream("converters/mapper.xml");
String data = IOUtils.toString(inputStream, "UTF-8");
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 defined a property in gradle.properties file like below:
user.password=mypassword
Can I use it as a variable value in my java statement?
Yes you can, however this is not a good idea nor a good practice. gradle.properties file is meant to keep gradle's properties itself, e.g. JVM args used at build time.
If you need to store user/pass pair in properties file, it should be placed under src/main/resources or other appropriate folder and separate from gradle.properties.
Side note: Not sure if keeping properties file in mobile application is safe in general.
You would have to read the properties file and extract the property first.
Properties prop = new Properties();
InputStream input = null;
try {
input = new FileInputStream("gradle.properties");
// load a properties file
prop.load(input);
// get the property value and print it out
System.out.println(prop.getProperty("user.password"));
} catch (IOException ex) {
ex.printStackTrace();
}
You can find the detailed tutorial here
I am new to servlet . I use the following code in servlet.then deployed to Jboss 4.1 . backup_database_configuration_location is location of properties file.But it can't be find. how I can specify directories in war file ?
Thanks all in advance
try {
backupDatabaseConfiguration = new Properties();
FileInputStream backupDatabaseConfigurationfile = new FileInputStream(backup_database_configuration_location));
backupDatabaseConfiguration.load(backupDatabaseConfigurationfile);
backupDatabaseConfigurationfile.close();
} catch (Exception e) {
log.error("Exception while loading backup databse configuration ", e);
throw new ServletException(e);
}
If it is placed in the webcontent, then use ServletContext#getResourceAsStream():
InputStream input = getServletContext().getResourceAsStream("/WEB-INF/file.properties"));
The getServletContext() method is inherited from HttpServlet. Just call it as-is inside servlet.
If it is placed in the classpath, then use ClassLoader#getResourceAsStream():
InputStream input = Thread.currentThread().getContextClassLoader().getResourceAsStream("file.properties");
The difference with Class#getResourceAsStream() is that you're not dependent on the classloader which loaded the class (which might be a different one than the thread is using, if the class is actually for example an utility class packaged in a JAR and the particular classloader might not have access to certain classpath paths).
Where is your properties file located? Is it directly somewhere in your hard drive, or packaged in a JAR file?
You can try to retrieve the file using the getResourceAsStream() method:
configuration = new Properties();
configuration.load(MyClass.class.getResourceAsStream(backup_database_configuration_location));
(or course, replace MyClass by your current class name)
I want read property file in my java class.
for this i have written:
try {
Properties property = new Properties();
property .load(new FileInputStream("abc.properties"));
String string = property.getProperty("userName");
} catch (Exception e) {
e.printStackTrace();
}
i have placed my abc.properties file in same package where my java class resides. But i am getting FileNotFound exception.
Please let me know how to give path of property file.
Thanks and Regards
The FileInputStream will look for the file in your program's "working directory", not in the same package as your class.
You need to use getClass().getResourceAsStream("abc.properties") to load your properties file from the same package.
If the properties files is next to your class file, you should not use FileInputStream, but Class.getResourceAsStream.
property.load(getClass().getResourceAsStream("abc.properties"));
This will even work when you package everything up as a jar file.
You are better off using ClassLoader.getResourceAsStream. See this article.
let say ur class is TestClass then ur code must change to ,
InputStream input = TestClass.class.getResourceAsStream("abc.properties");
property.load(input);