Develop Reference

Deployed WAR can't access a file

I have a spring application, and i'm trying to access a json file with the following code :
try (FileReader reader = new FileReader("parameters.json")) {
Object obj = jsonParser.parse(reader);
parameterList = (JSONArray) obj;
}
I have put the parameter.json file in the project folder and I'm accesing the file data from an angular app through a rest api, and that works fine when I run the application on local machine, but when I deploy the war file on tomcat, my application can't load the file, should I put parameter.json file somewhere else on tomcat or what is the best solution for it.

Your question states you are attempting to access a file called parameter.json, while your code excerpt shows parameters.json. Perhaps that discrepancy indicates a typo in your source code?
If not, there are various ways to access a file from the classpath in Spring, with the first step for each being to ensure the file is in the project's src/main/resources directory.
You can then use one of the Spring utility classes ClassPathResource, ResourceLoader or ResourceUtils to get to the file. The easiest approach, though, may be to put your properties in a .properties file (default file name application.properties) and access the values using Spring's #Value annotation:
#Value("${some.value.in.the.file}")
private String myValue;
You can use other file names as well by utilizing #PropertySource:
#Configuration
#PropertySource(value = {"classpath:application.properties",
"classpath:other.properties"})
public class MyClass {
#Value("${some.value.in.the.file}")
private String myValue;
...
}

Make sure your parameters.josn filename is exactly same in the code.
Move you parameters.json file in the resources folder and then use the classpath with the filename.
try (FileReader reader = new FileReader("classpath:parameters.json")) {
Object obj = jsonParser.parse(reader);
parameterList = (JSONArray) obj;
}

Try to put the file under resources folder in your spring project. You should be able to access the file from that location.

FileReader is looking for a full-fledged file system like the one on your computer, but when your WAR is deployed, there just isn't one, so you have to use a different approach. You can grab your file directly from your src/main/resources folder like this
InputStream inputStream = getClass().getResourceAsStream("/parameters.json");

Java/Gradle reading external config files

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

How to read the resource file? (google cloud dafaflow)

My Dataflow pipeline needs to read a resource file GeoLite2-City.mmdb. I added it to my project and ran the pipeline. I confirmed that the project package zip file exists in the staging bucket on GCS.
However, when I try to read the resource file GeoLite-City.mmdb, I get a FileNotFoundException. How can I fix this? This is my code:
String path = myClass.class.getResource("/GeoLite2-City.mmdb").getPath();
File database = new File(path);
try
{
DatabaseReader reader = new DatabaseReader.Builder(database).build(); //<-this line get a FileNotFoundException
}
catch (IOException e)
{
LOG.info(e.toString());
}
My project package zip file is "classes-WOdCPQCHjW-hRNtrfrnZMw.zip"
(it contains class files and GeoLite2-City.mmdb)
The path value is "file:/dataflow/packages/staging/classes-WOdCPQCHjW-hRNtrfrnZMw.zip!/GeoLite2-City.mmdb", however it cannot be opened.
and This is the options.
--runner=BlockingDataflowPipelineRunner
--project=peak-myproject
--stagingLocation=gs://mybucket/staging
--input=gs://mybucket_log/log.68599ca3.gz
The Goal is transform the log file on GCS, and insert the transformed data to BigQuery.
When i ran locally, it was success importing to Bigquery.
i think there is a difference local PC and GCE to get the resource path.

I think the issue might be that DatabaseReader does not support paths to resources located inside a .zip or .jar file.
If that's the case, then your program worked with DirectPipelineRunner not because it's direct, but because the resource was simply located on the local filesystem rather than within the .zip file (as your comment says, the path was C:/Users/Jennie/workspace/DataflowJavaSDK-master/eclipse/starter/target/classe‌​s/GeoLite2-City.mmdb, while in the other case it was file:/dataflow/packages/staging/classes-WOdCPQCHjW-hRNtrfrnZMw.zip!/GeoLite2-City.mmdb)
I searched the web for what DatabaseReader class you might be talking about, and seems like it is https://github.com/maxmind/GeoIP2-java/blob/master/src/main/java/com/maxmind/geoip2/DatabaseReader.java .
In that case, there's a good chance that your code will work with the following minor change:
try
{
InputStream stream = myClass.class.getResourceAsStream("/GeoLite2-City.mmdb");
DatabaseReader reader = new DatabaseReader.Builder(stream).build();
}
catch (IOException e)
{
...
}

Play 2.2.3 can not see file by Play.application().resource

Play 2.2.3. Windows 7.
public static Result menu() throws IOException {
String path = Play.application().resource("resources/menu.json").toString();
String content = Files.toString(new File(path), Charsets.UTF_8);
return ok(content).as("JSON");
}
Got an error:
... scala-2.10\classes\resources\menu.json The filename, directory
name, or volume label syntax is incorrect
Checking that path in file-system, I'm able to find my file there:
..\target\scala-2.10\classes\resources\menu.json
I'm able to find it there. Why play can't?
--
UPDATE:
I've just figured out I can not create files on C:\ root folder on my machine. That maybe the issue. But on other hand I'm not accessing root folder, and ad trying to get read only access. And I do have write access to that file on that path anyway.

As actually you want to use your menu.json file as a static asset you can put it i.e. into public/resources/menu.json file and then read it with simple:
<script>
$.get('#routes.Assets.at("resources/menu.json")');
</script>
or just directly by request:
http://localhost:9000/assets/resources/menu.json
To do what you want via controller you need to read the InputStream by classpath (remember that finally it will be archived into jar file!) but it need to be placed in conf folder i.e.: conf/resources/menu.json then from controller:
public static Result menuViaControllerJson() {
InputStream is = Play.application().classloader().getResourceAsStream("resources/menu.json");
return (is != null)
? ok(is)
: notFound();
}
Anyway you will get exactly the same result as for common Assets.at, so consider if it's worth of effort.
Edit: If you want to use this file as custom config just use HOCON syntax file i.e.: conf/ses.conf:
foo = "bar"
and in controller:
import com.typesafe.config.Config;
import com.typesafe.config.ConfigFactory;
....
Config cfg = ConfigFactory.parseResources(Play.application().classloader(), "ses.conf");
debug("My 'foo' is configured as: " + cfg.getString("foo"));

How I can specify directories in war file?

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)