I'm running spring in debug mode, and after struggling to get it to recognize an external properties file I finally got it to do so
Loaded config file 'file:C:/Users/udyj/git/hermes-mq-tool.lib/rte/application-ALD-INT.properties'
However right after that line
Skipped (empty) config file 'file:C:/Users/udyj/git/hermes-mq-tool.lib/rte/application-ALD-INT.properties'
The thing is though, it isn't empty
Name=ALD-INT
Queues=
QueueManager=whatever
TAS_MQS_PASSWORD=pw
TAS_MQS_USER=name
I'm running spring this way,
pb = new ProcessBuilder().inheritIO().command(java, "-noverify", "-cp", classpath,
String.format("-Dspring.config.location=%s", "C:\\Users\\udyj\\git\\hermes-mq-tool.lib\\rte\\application-ALD-INT.properties"),
clazz.getCanonicalName()
);
What does the empty config file thing mean?
This is an open issue
The misleading logging is an unfortunate side-effect of it being
performed in a generic location that doesn't (and shouldn't)
understand the difference between .properties and .yaml configuration
files.
I want to set my relative file path in a properties file so my SaxReader can pick it up when it runs on Tomcat server. I know this should be easy but I've forgotten forgive me :)
I know Spring Boot has application.properties file but I don't see a way to hook in here.
Is there a way to set the relative path in a properties file that will get picked up by Spring Boot and the SaxReader will see it?
As it is I'm hard coding just the filename and putting the file in the resources folder that serves up the templates and static content such as css and js files. The filePath system.out gives: org.dom4j.DocumentException C:sts-bundle\sts-3.7.2.RELEASE\myFileName the toolsuite root location??? weird!!
Please tell me how to specify the relative path in a properties file.
Thanks!!
You can set the file path like any other string property in Spring Boot properties file and access it from inside the path.
E.g. I have the following set in my application.properties file:
download.directory=temp
and it is used as follows in java class:
#Value("${download.directory}")
private String downloadDirectory;
It is used to download the files, now, if I start the application with jar file present in let's say G:/applications folder then the files will be downloaded into G:/applications/temp.
So i can access properties file if its in WEB-INF/classes.
However if I keep the same file under TOMCAT/conf and updating catalina.properties to point to the path, I get an error like Name not bound.
I have almost tried everything...even tried with absolute path
if your file.properties file is in the WEB-INF/classes folder? Then:?
InputStream inputStream = this.getClass().getClassLoader().getResourceAsStream("/file.properties");
Only way I could solve it was
1. Create a env folder under tomcat_home and put the config.properties there
2. update catalina.properties and add the tomcat_home\env path in common loader
3. Comment out from spring security XML
Once I commented the above line It ran fine.
Not sure if this has an answer, but here goes. I'm using JUnit 4.8.1 to test my Spring 3.1.0.RELEASE project. I have two JAR files on my classpath. Within each, there are files of identical names -- /module/rootContext.xml .
In my testApplicationContext.xml file (my Spring context file for my JUnit tests), is it possible to include each of those? Right now, the only thing I know how to do is
<import resource="classpath:/module/rootContext.xml" />
but I don't know how to specify the exact JAR file where each file lives.
Because I'm dealing with code that's not my own, it is not an option to change the names of the XML files within the JARs.
If you want to include both files simultaneosly, you can do it as follows:
<import resource="classpath*:/module/rootContext.xml" />
See also:
4.7.2.2 The classpath*: prefix
I have a Spring MVC application and a problem with JUnit tests combined with the file applicationContext.xml.
In my JUnit test class I write:
final ApplicationContext context = new ClassPathXmlApplicationContext("applicationContext.xml");
service = (TestServiceImpl) context.getBean("testServiceImpl");
The error I get is that aplicationContect.xml can not be found:
org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from class path resource [applicationContext.xml]; nested exception is java.io.FileNotFoundException: class path resource [applicationContext.xml] cannot be opened because it does not exist
But it exists in the WEB-INF folder.
So, what's wrong here? Why does the file not exist for the JUnit test?
You should keep your Spring files in another folder, marked as "source" (just like "src" or "resources").
WEB-INF is not a source folder, therefore it will not be included in the classpath (i.e. JUnit will not look for anything there).
If you use maven, create a directory called resources in the main directory, and then copy your applicationContext.xml into it.
From your java code call:
ApplicationContext appCtx = new ClassPathXmlApplicationContext("applicationContext.xml");
I got the same error.
I solved it moving the file applicationContext.xmlin a
sub-folder of the srcfolder. e.g:
context = new ClassPathXmlApplicationContext("/com/ejemplo/dao/applicationContext.xml");
The ClassPathXmlApplicationContext isn't going to find the applicationContext.xml in your WEB-INF folder, it's not on the classpath. You could copy the application context into your classpath (could put it under src/test/resources and let Maven copy it over) when running the tests.
I also found this problem. What do did to solve this is to copy/paste this file everywhere and run, one file a time. Finally it compiled and ran successfully, and then delete the unnecessary ones. The correct place in my situation is:
This is under the /src/ path (I am using Intellij Idea as the IDE). The other java source files are under /src/com/package/ path
Hope it helpes.
This happens to me from time to time when using eclipse
For some reason (eclipse bug??) the "excluded" parameter gets a value *.* (build path for my resources folder)
Just change the exclusion to none (see red rectangle vs green rectangle)
I hope this helps someone in the future because it was very frustrating to find.
Click on the src/main/java folder and right click and create the xml file. If you create the application.xml file in a subpackage other than /, it will not work.
Know your structure is look like this
package/
| subpackage/
| Abc.java
| Test.java
/application.xml
enter image description here
I was struggling since a couple of hours for this issue because i was putting that file under resources folder but it didn't help me, finally i realized my mistake.
Put it directly under src/main/java.
For me, it worked by keeping file(applicationContext.xml) in the resources folder
ApplicationContext context = new ClassPathXmlApplicationContext("applicationContext.xml");
I placed the applicationContext.xml in the src/main/java folder and it worked
I solved it moving the file spring-context.xml in a src folder.
ApplicationContext context = new ClassPathXmlApplicationContext("spring-context.xml");
I fixed it by adding applicationContext.xml into jar/target/test-classes for Maven project. And use
XmlBeanFactory bf = new XmlBeanFactory( new ClassPathResource(
"/applicationContext.xml", getClass() ) )
instead of
XmlBeanFactory bf = new XmlBeanFactory( new ClassPathResource(
"/WEB-INF/applicationContext.xml", getClass() ) )
My solution:
If you have no folder WEB-INF please put the file applicationContext.xml into the folder source(src).
Then Java Project can read file applicationContext.xml -> getBean -> perform your business.
enter image description hereThe solution is to place the xml file in resources folder(src->main-> resources) and use this object creation new ClassPathXmlApplicationContext("applicationContext.xml");
Create a Directory at the bottom of main directory named resources. That solved my issue.
In Spring all source files are inside src/main/java.
Similarly, the resources are generally kept inside src/main/resources.
So keep your spring configuration file inside resources folder.
Make sure you have the ClassPath entry for your files inside src/main/resources as well.
In .classpath check for the following 2 lines. If they are missing add them.
<classpathentry path="src/main/java" kind="src"/>
<classpathentry path="src/main/resources" kind="src" />
So, if you have everything in place the below code should work.
ApplicationContext ctx = new ClassPathXmlApplicationContext("Spring-Module.xml");
Please do This code - it worked
AbstractApplicationContext context= new ClassPathXmlApplicationContext("spring-config.xml");
o/w: Delete main method class and recreate it while recreating please uncheck Inherited abstract method its worked
I'm using Netbeans, i solved my problem by putting the file in: Other Sources default package, then i called it in this way:
ApplicationContext context =new ClassPathXmlApplicationContext("bean.xml");
resources folder
While working with Maven got same issue then I put XML file into src/main/java path and it worked.
ApplicationContext context=new ClassPathXmlApplicationContext("spring.xml");
You actually need to understand the ApplicationContext. It is an interface and it will have different implementations based on configuration.
As you are using new ClassPathXmlApplicationContext("applicationContext.xml"); , kindly pay attention to initial right hand-side , it says ClassPathXmlApplicationContext , so the XML must be present in the class path.
So drag your applicationContext.xml wherever it is to the src folder.
Gist: new ClassPathXmlApplicationContext as the name
[ClassPathXml]will look for the xml file in the src folder of your
project, so drag your xml file there only.
ClassPathXmlApplicationContext—Loads a context definition from an XML
file located in the classpath, treating context definition files as classpath
resources.
FileSystemXmlApplicationContext—Loads a context definition from an XML
file in the file system.
XmlWebApplicationContext—Loads context definitions from an XML file contained
within a web application.
just change the containing package of your applicationContext.xml file.
applicationContext.xml must be in src package not in your project package.
e.g.
src(main package)
com.yourPackageName(package within src)
classes etc.
applicationContext.xml(within src but outside of yourPackage or we can say
parallel to yourPackage name)
your-module/src/applicationContext.xml
val context = ClassPathXmlApplicationContext("applicationContext.xml")
Check the directory path, the default path is /src/ and not /.
GL
I got the same issue while working on a maven project, so I recreate the configuration file spring.xml in src/main/java and it worked for me.
I solved it by moving the file applicationContext.xml in an src folder and main folder.
ApplicationContext context = new ClassPathXmlApplicationContext("spring-context.xml");