Does Configuration.addResource() method load resource file like ClassLoader of java or it just encapsulates ClassLoader class.Because I find it can not use String like "../resource.xml" as argument of addResource() to load resource file out of classpath, this property is just the same as ClassLoader.
Thx!
Browsing the Javadocs and source code for Configuration, Strings are assumed to be classpaths (line 1162), rather than relative to the file system - you should use URLs to reference files on the local file system as follows:
conf.addResource(new File("../resource.xml").toURI().toURL());
Related
In spring boot application how do I give an external windows path using #Value Spring annotation and Resource
The below example works fine that look into resources folder but I want to give the path outside of application like c:\data\sample2.csv
#Value("classPath:/sample2.csv")
private Resource inputResource;
...
#Bean
public FlatFileItemReader<Employee> reader() {
FlatFileItemReader<Employee> itemReader = new FlatFileItemReader<Employee>();
itemReader.setLineMapper(lineMapper());
itemReader.setLinesToSkip(1);
itemReader.setResource(inputResource);
and if I want to get the value from properties file in annotaion, whats the format to put the path in windows?
i tried these, none of them worked:
in code
#Value("${inputfile}")
in properties file:
inputfile="C:\Users\termine\dev\sample2.csv"
inputfile="\\C:\\Users\\termine\\dev\\sample2.csv"
inputfile="C:/Users/termine/dev/sample2.csv"
inputfile="file:\\C:\Users\termine\dev\sample2.csv"
inputfile="file://C://Users//termine///dev//sample2.csv"
When you use classpath spring will try to search with the classpath even if you provide the outside file path.
so instead of using classpath: you can use file:
Ex.
#Value("file:/sample2.csv") //provide full file path if any
Use the key spring.config.location in properties to set the config location. Spring-boot will by default load properties from the locations, with precedence like below :
A /config subdir of the current directory.
The current directory
A classpath /config package
The classpath root
and apart from this when you start the jar or in application.properties you can provide the location of the config file like :
$ java -jar myproject.jar --spring.config.location=classpath:/default.properties,classpath:/override.properties
You can serve static files from the local disk, by making the resource(s) "sample2.csv" as a static resource. An easy way to do this is by adding spring.resources.static-locations configuration to your applicaiton.properties file. Example:
spring.resources.static-locations=file:///C:/Temp/whatever/path/sample2.csv",classpath:/static-files, classpath:/more-static-resource
When I did this in one of the projects, I was able to access the file form the browser using localhost:8080/sample2.csv.
I have a Spring Boot application, the code need to access a file under resources/templates folder.
here is my application.properties file:
pont.email.template.location=templates/mailTemplate.html
This is the java file where I use the variable:
#Value("${pont.email.template.location}")
private String templateLocation;
----------------
BufferedReader reader = new BufferedReader(new FileReader(templateLocation));
The problem is not get the varibale, it returns correctly, The problem is that the application do not found any file for this path.
I always get
java.io.FileNotFoundException: templates/mailTemplate.html (No such file or directory)
I have checked that the file is in the path..
what is wrong in my code?
Help please, thanks.
You cannot read a File from inside a JAR. This fails due to the fact that the File has to point to an actual file resource on the file system and not something inside a JAR.
Let Spring do the heavy lifting and use the Resource abstraction to hide the nasty internals. So instead of using a String use a Resource and prefix the value of the property with classpath: to make sure it is loaded from the classpath. Then use an InputStreamReader instead of FileReader to obtain the information you need.
#Value("${pont.email.template.location}")
private Resource templateLocation;
----------------
BufferedReader reader = new BufferedReader(new InputStreamReader(templateLocation.getInputStream()));
In your application.properties prefix with classpath:.
pont.email.template.location=classpath:templates/mailTemplate.html
Now it should work regardless of the environment you are running in.
I have a very simple spring test app. But I get exception even though everything seems to be on order. I might be missing something. Please check the pic to see the project structure and web.xml file contains as well as exception:-
efinitionStoreException: IOException parsing XML document from class path resource [WEB-INF/servlet-context.xml]; nested exception is java.io.FileNotFoundException: class path resource [WEB-INF/servlet-context.xml] cannot be opened because it does not exist
at org.springframework.beans.factory.xml.XmlBeanDefinitionReader.loadBeanDefinitions(XmlBeanDefinitionReader.java:341)
There are two kinds of resources in a servlet environment:
servlet resources - files uner the root of the web application (loaded via ServletContext)
classpath resources - resources on web application's classpath (loaded via ClassLoader)
When Spring is supposed to load its configuration it needs to know which mechanism to use.
classpath:foo/bar.xml - will load as classpath resource
checking WEB-INF/classes, contents of WEB-INF/lib/*.jar and other shared servlet container's classpath locations
when using maven and its project structure, all files from src/main/resources will be placed on classpath
foo/bar.xml - will load as servlet resource
when using maven and its project structure the src/main/webapp folder is the root of your application
TL;DR As I wrote in the comment, either remove classpath: prefix when referencing XML file or move your XML file to src/main/resources and remove the WEB-INF part.
I am using Spring with Maven and I would like to load a file contained in the maven folder src/main/resources inside a spring service annotated with #Service contained in a JAR file (used by another WAR file as a Maven dependency). This does not seem to work for many different reasons (read: for many different exceptions according to many different attempts).
I know that Maven, when building, puts the stuff in src/main/resources at the root of the JAR file. I've opened the JAR file and I've verified that.
I would prefer a File object (something I can use with utility libraries like Apache Commons IO), but also "something" else that's working is fine.
I would prefere going for stuff within the Java class annotated with #Service (i.e. no application context XML file, with stuff inside that I recall within the service etc.)
The file contains stuff that I eventually use as a List<String> where every item is a line in the original file. I need to keep this file as it is, I can not use a database or another solution different than a file.
This is not a .properties file and is not intended to be.
The following are the not working attempts (throwing NullPointerException, FileNotFoundException, IllegalArgumentException: URI is not hierarchical and so on...).
Among the attempts I have tried also adding/removing the prefix src/main/resources while trying to figure out what was wrong, but then I figured out that Maven puts all the stuff at the root of the JAR file as stated before.
#Value("classpath:src/main/resources/myFile.txt")
private Resource myFileResource;
private File myFile =
new File(this.getClass().getResource("/myFile.txt").toURI());
private File myFile =
new FileSystemResource("src/main/resources/myFile.txt").getFile();
private BufferedReader bufferedReaderMyFile =
new BufferedReader(
new InputStreamReader(getClass().getResourceAsStream("myFile.txt")));
What's wrong with what I am doing?
File inside jar file is no longer a File, so treat it as Resource and read it via Stream
InputStream inputStream = this.getClass().getResourceAsStream("/myFile.txt");
assumming you have actually packed that myFile.txt inside jar file and it is available at the root of JAR file
Resource resource = new ClassPathResource("classpath:src/main/resources/template/datafields.properties");
Properties props = PropertiesLoaderUtils.loadProperties(resource);
Your problem is that your file is actually not in the application classpath. Looking at your folder paths I am assuming that you have a maven project structure and your properties file is present within resources directory. When your project is compiled, everything inside the resources directory is at the root of the classpath along with your compiled java classes. So you should instead use
Resource resource = new ClassPathResource("template/datafields.properties");
Classpath resource loads resources from the application classpath, so you need to be aware what all directories/jar files are in your classpath and their directory structure to successfully load resources.