Develop Reference

Compiling a java project from linux command line

I am having trouble understanding how to actually use the command line to run a big java project ( by big I mean with multiple files and folder).
Imagine I have a project containing :
~/src/main/fr/file1.java
~/src/main/fr/file2.java
~/src/test/test1.java
All my life people have done the makefile for me. I just code the java src with vim and compile and run with make. Now there is no makefile ! I compile with maven (that I am still understanding how it works.). After compiling with maven (I just run maven compile). I then have a new folder named target.
~/target/main/fr/file1.class
~/target/main/fr/file2.class
~/target/test/test1.class
Now how can I run test1 ? I tried using java -classpath =... test1 but I always get errors ...
If can someone help me (or just give me some resources so I can finally understand basic project structuring and scripting) it will be amazing. Thank you !

Here is a minimal working example to compile java source files from multiple locations and pack them in a single runnable jar.
$ tree
.
├── makefile
├── src_folder_1
│   └── Main.java
└── src_folder_2
└── Person.java
The contents of java files appear at the end for completeness. Here is the makefile.
BIN_DIR = bin
WITH_DEBUG_INFO = -g
.DEFAULT_GOAL:
app.jar:
compile_phase.done: ${SRC_FILES}
#mkdir -p ${BIN_DIR}
#javac ${WITH_DEBUG_INFO} ${SRC_FILES} -d ${BIN_DIR}
#echo "DONE" >> $#
manifest.mf:
#echo "Class-Path: ${BIN_DIR}/*.class\nMain-Class: Main\n" > $#
app.jar: manifest.mf compile_phase.done
#jar cfm $# $< -C ${BIN_DIR} .
#rm -rf ${BIN_DIR}
#rm $^
See that all java source files are compiled within the same rule, and the corresponding *.class files are put in a dedicated bin directory. To emphasize that everything needed to run the jar is inside it I completely removed the bin directory and the manifest.mf. Now you can run the program with
$ java -jar app.jar oren
Hello oren
Here are the java files for completeness:
$ cat src_folder_1/Main.java
class Main {
static public void main(String[] args) {
Person p = new Person(args[0]);
System.out.format("Hello %s\n",p.toString()); }}
$ cat src_folder_2/Person.java
class Person {
private String name;
public Person(String n) { this.name = n;}
public String toString() { return name; }}

Error parsing file arguments error on jar with entry point creation

I've create simple java project with single class - Main with main method printing Hello.
package com.foo;
public class Main {
public static void main(String[] args) {
System.out.println("Hello!");
}
}
Code was compiled to bin directory. I'm trying to create jar using command
jar -cfe project.jar com.foo.Main -C bin\
with no results, always returning Error parsing file arguments error.
I also tried many different variations, like
jar --create --file project.jar --main-class com.foo.Main -C bin
but none of it worked. I'm using Java 16

Try running this. It should create the 'project.jar'
jar --create --file project.jar --main-class com.foo.Main -C bin .
-C flag Temporarily changes directories (cd dir) during execution of the jar command while processing the following inputfiles argument.
This command changes to bin directory and adds to project.jar all files within the bin directory (without creating a bin directory in the jar file).
For more information check the -C part in the OPTIONS section :
https://docs.oracle.com/javase/7/docs/technotes/tools/solaris/jar.html#options

You need both a "=" after main-class and a "." at the end to selct al lthe files
jar --create --file project.jar --main-class=com.foo.Main -C bin .
or
jar cfe project.jar com.foo.Main -C bin .

How can i remove part of a directory in batch?

Here's the code
SET PATH="C:\Program Files\Java\jdk1.7.0_40\bin"
dir *.java /b /s >> ./sources_list.txt
javac -cp ".;lib/*" #sources_list.txt -d compiled
dir compiled\*.class /b /s >> .\classes_list.txt
jar cfm app.jar MANIFEST.MF #classes_list.txt
del sources_list.txt
del classes_list.txt
PAUSE
so this is for compiling my java code (1st 3 lines) which works, then to copy the compiled classes into my jar using jar
now my problem is on line 4, finding the compiled classes and printing the path to the classes_list.txt, that works however it returns the full C:\somethin\compiled\something.class
i need it to return only the
compiled\subfolders\something.class
how can i edit that to return the path i need?

jar command can take path to classes and create directory. You need to pass only the directory not all classes.
jar cfm app.jar MANIFEST.MF compiled
should work.

Batch file to execute java program

I have multiple .java files inside a folder (e.g temp/code/project)
I want a batch file that will compile and run these java files.
The batch file should create class files inside the same structure where java files are located starting with the classes folder (i.e classes/temp/code/project)
I don't know how to write a batch file. Can any body help me with this? Thanks in advance.

Theoretically you should write the following:
#echo off
::compile classes
javac -cp YOUR_CLASSPATH com/yourcompany/YourClass1.java
javac -cp YOUR_CLASSPATH com/yourcompany/YourClass2.java
javac -cp YOUR_CLASSPATH com/yourcompany/YourClass3.java
javac -cp YOUR_CLASSPATH com/yourcompany/YourClassLauncher.java
:: create jar
jar cvfM Manifest.txt myjar.jar *.class
echo.
echo Hit any key to launch project.
pause
java -jar myjar.jar
pause
If you have one class that depends on all other classes in your project it is enough to run javac with this class only: compiler will compile everything.
# Manifest.txt
Manifest-Version: 1.0
Class-Path: .;MyUtils.jar
Created-By: 1.6.0
Main-Class: com.yourcompany.YourClassLauncher
This is only example and a good exercise. In real life people use special build tools like good old Ant, Maven or newer Graidle or Buildr. I'd recommend you to take one of them.

Something along the lines of
cd C:\temp\code\project
javac -classpath . -d C:\classes\temp\code\project\ *.java
You may not need to include the -classpath tag (I always do so that I don't have to worry about maintaining my CLASSPATH variable. The -d tag specifies a particular directory to place generated class files.

How can I compile and run a Java class in a different directory?

I'm writing a makefile that compiles a .java file in a different directory, and then I want to run it, without changing directories. I want to do something along the lines of:
$(SQM_JAVA_TOOL_DONE) : $(SQM_JAVA_TOOL)
$(shell cd /home_dir)
javac myjavafile.java
java myjavafile
where the Java file is /home/myjavafile.java, and the makefile isn't running from /home.
How can I do this?

I might be misunderstanding the question, but you can compile with
javac /home/MyJavaFile.java
This will create MyJavaFile.class in /home
You can then run it by including /home on the classpath. e.g.
java -cp /home MyJavaFile
If you want to generate the class file in a different directory then you can use the -d option to javac.

Use the -d command line parameter with javac to tell it what directory you'd like to store the compiled class files in. Then, to run the program, simply include this directory in the classpath:
javac -d some/directory myjavafile.java
java -cp some/directory myjavafile

Just to add to the existing answers, you may want the --source-path flag:
--source-path <path>, -sourcepath <path>
Specify where to find input source files
I believe this effectively sets the package root javac will compile from (i.e. <path> will be stripped from the expected package name of the files). It's still necessary to enumerate the files to compile, and this should still be relative to the current working directory, not the path passed to --source-path.
For example, to compile and run from a project's root where source is stored in src/ and you want it build in bin/:
$ javac --source-path src -d bin src/mypackage/*.java
$ java -cp bin mypackage.Main
This works even from directories elsewhere in the filesystem, e.g.:
$ javac --source-path /some/absolute/path/src -d /some/absolute/path/bin /some/absolute/path/
$ java -cp /some/absolute/path/bin mypackage.Main

I am using VS Code and installed java and code runner extensions. When I created new java project using the extension, it was creating the .class file in src instead of bin. To solve the issue I opened settings.json file from File > Preferences > Settings and searched for "settings" (or "code-runner"). Then I added following lines in that file.
"code-runner.executorMap": {
"java": "cd \"$workspaceRoot\\\" && javac --source-path src -d bin src\\$fileName && java -cp bin $fileNameWithoutExt",
}
If you don`t want to see the command that runs before code file then add these lines instead:
"code-runner.clearPreviousOutput": true,
"code-runner.showExecutionMessage": false,
"code-runner.executorMap": {
"java": "there is && clear added in the execution paramater"
"java": "cd \"$workspaceRoot\\\" && javac --source-path src -d bin src\\$fileName && clear && java -cp bin $fileNameWithoutExt",
}
I hope this finds someone with similar issue.