[英]Any source code example to load property file from src/main/resources in Java 9+ as module
I am migrating Java 8 application to JDK11 (early access). 我正在将Java 8应用程序迁移到JDK11(早期访问)。 New code suppose to be a module.
新代码应该是一个模块。 It read an error message from property file in Java8, but return 'null' in Java 11. Any help how to load/read resourceInputStream from /src/resources/messges/myerrors.properties
它从Java8中的属性文件中读取一条错误消息,但在Java 11中返回“ null”。有关如何从/src/resources/messges/myerrors.properties加载/读取resourceInputStream的任何帮助。
Any help appreciated. 任何帮助表示赞赏。
tree : 树 :
├── build.gradle
├── out
│ └── production
│ ├── classes
│ │ ├── com
│ │ │ └── acme
│ │ │ └── Main.class
│ │ └── module-info.class
│ └── resources
│ └── messages
│ └── mymessages.properties
├── settings.gradle
└── src
├── main
│ ├── java
│ │ ├── com
│ │ │ └── acme
│ │ │ └── Main.java
│ │ └── module-info.java
│ └── resources
│ └── messages
│ └── mymessages.properties
└── test
├── java
└── resources
Java : Java的 :
package com.acme;
import java.util.Locale;
import java.util.PropertyResourceBundle;
import java.util.ResourceBundle;
public class Main {
public static void main(String args[]){
Main m=new Main();
m.process();
}
void process(){
PropertyResourceBundle prb= (PropertyResourceBundle)ResourceBundle.getBundle("messages.mymessages",Locale.getDefault());
String key="key1";
String value= prb.getString(key);
System.out.println(key+":"+value);
}
}
Exception in thread "main" java.util.MissingResourceException: Can't find bundle for base name messages.mymessages, locale en_US
at java.base/java.util.ResourceBundle.throwMissingResourceException(ResourceBundle.java:2045)
at java.base/java.util.ResourceBundle.getBundleImpl(ResourceBundle.java:1679)
at java.base/java.util.ResourceBundle.getBundleImpl(ResourceBundle.java:1572)
at java.base/java.util.ResourceBundle.getBundleImpl(ResourceBundle.java:1546)
at java.base/java.util.ResourceBundle.getBundle(ResourceBundle.java:914)
at resource.main/com.acme.Main.process(Main.java:13)
at resource.main/com.acme.Main.main(Main.java:10)
For the code above solution is: "--class-path ./out/production/resources". 对于上面的代码,解决方案是:“-class-path ./out/production/resources”。 I figured out.
我想通了。 Also, another variation is this.
另外,这是另一种变化。 I have a set of JUnits.
我有一组JUnit。 During test code need to generate error message from /src/main/resources/messages/mymessages.properties.
在测试期间,代码需要从/src/main/resources/messages/mymessages.properties生成错误消息。 However, the test wasn't able to "see" resources from "main".
但是,该测试无法从“主”中“查看”资源。 After 1 day of trying different combinations solution was: "--module-path ...pathTotestClassesOrOtherProductionClasses;$buildDir/resources/main;$buildDir/resources/test;".
经过1天的尝试不同组合的解决方案是:“-module-path ... pathTotestClassesOrOtherProductionClasses; $ buildDir / resources / main; $ buildDir / resources / test;”。 So you need to add ".../resources/main" as module path
因此,您需要添加“ ... / resources / main”作为模块路径
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