How to compile multiple java source files in command line

Question

I know running javac file1.java produces file1.class if file1.java is the only source file, then I can just say java file1 to run it.

However, if I have 2 source files, file1.java and file2.java, then how do I build the program?

Does that mean both the files have main method or there are 2 entrypoint of an application? — Pradeep Singh
– Pradeep Singh, Commented Jun 7, 2017 at 12:25

ryanprayogo · Accepted Answer · 2011-01-26 02:00:49Z

143

Try the following:

javac file1.java file2.java

answered Jan 26, 2011 at 2:00

ryanprayogo

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1 Comment

Alexander Mills Over a year ago

this is hardcoded, how do we use the results from $(find .) etc?

ajduke · Accepted Answer · 2011-01-26 07:59:47Z

101

or you can use the following to compile the all java source files in current directory..

javac *.java

answered Jan 26, 2011 at 7:59

ajduke

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3 Comments

Sagar Rout Over a year ago

No, it is not recursive.

user3735633 Over a year ago

* is the symbol for a wildcard. It means all things in this directory that end with ".java"

Rainb Over a year ago

this is not recursive but it is not bad

huch · Accepted Answer · 2017-06-07 12:17:10Z

Here is another example, for compiling a java file in a nested directory.

I was trying to build this from the command line. This is an example from 'gradle', which has dependency 'commons-collection.jar'. For more info, please see 'gradle: java quickstart' example. -- of course, you would use the 'gradle' tools to build it. But i thought to extend this example, for a nested java project, with a dependent jar.

Note: You need the 'gradle binary or source' distribution for this, example code is in: 'samples/java/quickstart'

% mkdir -p temp/classes
% curl --get \
    http://central.maven.org/maven2/commons-collections/commons-collections/3.2.2/commons-collections-3.2.2.jar \
        --output commons-collections-3.2.2.jar

% javac -g -classpath commons-collections-3.2.2.jar \
     -sourcepath src/main/java -d temp/classes \
      src/main/java/org/gradle/Person.java 

% jar cf my_example.jar -C temp/classes org/gradle/Person.class
% jar tvf my_example.jar
   0 Wed Jun 07 14:11:56 CEST 2017 META-INF/
  69 Wed Jun 07 14:11:56 CEST 2017 META-INF/MANIFEST.MF
 519 Wed Jun 07 13:58:06 CEST 2017 org/gradle/Person.class

xt j · Accepted Answer · 2019-01-16 09:18:52Z

2

1.use wildcard

2.use options

3.https://docs.oracle.com/javase/8/docs/technotes/tools/windows/javac.html

edited Jan 16, 2019 at 9:18

answered May 1, 2018 at 2:53

xt j

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2 Comments

Per Lundberg Over a year ago

How could this work? @file expects the file provided to contain the names of the files to compile.

xt j Over a year ago

sorry, i forget to check my answer. now, i paste official description.

Ryan Weiland · Accepted Answer · 2017-07-04 13:21:25Z

-7

OR you could just use javac file1.java and then also use javac file2.java afterwards.

answered Jul 4, 2017 at 13:21

Ryan Weiland

5

2 Comments

Ron Zhang Over a year ago

That does not answer the question. OP was asking if there is a way to compile multiple .java files at once.

Allohvk Over a year ago

what if there are 1000 files? javac *.java is best. Unfortunately it is not recursive

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