Springload Hot-Deploy Java Application project __java

Thermal deployment is often a problem for operational personnel. Hot deployment In simple terms, is a partial or a file modified, without restarting the program, immediately effective.

The advantage is that a large amount of initialization data may be loaded when the program is initialized, and that restarting the program costs too much; If a program is a single module in a project, restarting can affect other applications.

For Web applications, most are deployed in servlet containers, such as jboss,weblogic,tomcat, which often provide a matching hot-deployment scenario.

But for application procedures, there are often no manufacturers to do this alone.

After my search, found Springload is a better solution. The simple test is as follows.

1, download Springload jar package:

Spring-loaded (https://github.com/spring-projects/spring-loaded)

Maven dependencies: (Not Configured)

<dependency>
			<groupId>org.springframework</groupId>
			<artifactid>springloaded </artifactId>
			<version>1.2.3.RELEASE</version>
		</dependency>

2, create the repo directory under the project root directory, put the Springloaded-1.2.3.release.jar below.

3. If it is eclipse, add the JVM startup parameters to the run configuration of the main class:

-javaagent:repo/springloaded-1.2.3.release.jar-noverify

Note the path.

If it's in Eclipse, remember to check for automatic compilation. Project automatically.

Once started, modifying the Java code will take effect immediately.

4, for Yu Yunwi, a lot of tools can not be used, such as eclipse, hot deployment is not so easy.

If you are using MAVEN management, you can configure environment variables:

Maven_opts

-javaagent:repo/springloaded-1.2.3.release.jar-noverify

Or add the maven startup parameter under%mava-home%/bin/mvn.bat:

Set Maven_opts=-javaagent:repo/springloaded-1.2.3.release.jar-noverify

Note that the path is the root directory of the project.

Give springload to start the project.

If you modify a Java class, how does it take effect immediately?

Compiles the file to the specified directory. such as compiling the Xxx.java into the dir directory.

javac-d dir  Xxx.java

As my current directory:

Notice that the package directory of the class is not added after-D.

Post the source code:

Package com.mobile263;

public class Calcutor {public
	
	static int-calc (int a,int b) {return
		a*b;
	}
	

Package com.mobile263;

Import Java.text.SimpleDateFormat;
Import java.util.Date;
Import Java.util.Timer;
Import Java.util.TimerTask;
public class Console {public

    static void Main (string[] args) {
    	Timer timer = new timer ();
    	TimerTask task = new TimerTask () {
		SimpleDateFormat sdf = new SimpleDateFormat ("Yyyy-mm-dd hh:mm:ss");
			@Override public
			Void Run () {
				int res = CALCUTOR.CALC (4, 5);
				System.out.println ("Exec Main Class ok! result= "+res +" @ "+sdf.format (new Date ());
			}
		;
		Timer.schedule (Task, 1000, 1000);
	}

At the cmd command line, go to the project root directory and execute the MAVEN command:

The result of a modified operation:

Exec Main Class ok! Result=9 @2015-05-20 11:10:57
Exec Main Class ok! Result=9 @2015-05-20 11:10:58
Exec Main Class ok! Result=9 @2015-05-20 11:10:59
Exec Main Class ok! Result=9 @2015-05-20 11:11:00
Exec Main Class ok! Result=9 @2015-05-20 11:11:01
Exec Main Class ok! Result=9 @2015-05-20 11:11:02
Exec Main Class ok! Result=20 @2015-05-20 11:11:03
Exec Main Class ok! Result=20 @2015-05-20 11:11:04
Exec Main Class ok! Result=20 @2015-05-20 11:11:05
Exec Main Class ok! Result=20 @2015-05-20 11:11:06
Exec Main Class ok! Result=20 @2015-05-20 11:11:07
Exec Main Class ok! Result=20 @2015-05-20 11:11:08
Exec Main Class ok! Result=20 @2015-05-20 11:11:09
Exec Main Class ok! Result=20 @2015-05-20 11:11:10
Exec Main Class ok! Result=20 @2015-05-20 11:11:11