Looks like Java appends to %path% its own paths. Nothing else.

Answer from Andremoniy on Stack Overflow
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Christianhujer
christianhujer.github.io › Echo-in-Java-and-many-other-languages
Echo in Java and many other languages – Christian Hujer – Agilist and Software Craftsman on Humanity, Management, Strategy and Technology.
1 christian@armor01:~$ echo 2 3 christian@armor01:~$ echo foo bar 4 foo bar 5 christian@armor01:~$ echo foo bar 6 foo bar 7 christian@armor01:~$ echo "foo bar" 8 foo bar 9 christian@armor01:~$ echo 1 2 3 10 1 2 3 11 christian@armor01:~$ echo "Hello, world!" 12 Hello, world! So, arguments are concatenated with a single whitespace. The last argument will be printed without a following whitespace. In any case, the last thing printed will be a newline. Let’s look at Java 8 first, because it provides the most convenient solution.
Top answer
1 of 2
5

If you're using Java 8, you can use StringJoiner.

/**
 * JEcho writes any command line argument to the standard output; each argument
 * is separated by a single whitespace and end with a newline (you can
 * specify '-n' to suppress the newline).
 *
 * This program doesn't interpret common backslash-escaped characters (for
 * exampe '\n' or '\c').
 */
public class JEcho {
    public static void main(String[] args) {
        boolean printNewline = true;
        int posArgs = 0;

        if (args.length > 0 && args[0].equals("-n")) {
                printNewline = false;
                posArgs = 1;
        }

        StringJoiner outputBuilder = new StringJoiner(" ");

        for (; posArgs < args.length; posArgs++) {
            outputBuilder.add(args[posArgs]);
        }

        String output = outputBuilder.toString();

        if (printNewline)
            System.out.println(output);
        else
            System.out.print(output);
    }
}
2 of 2
2

as a matter of an excersice i would advise you to keep your parameters as List<String>...

public static void main(String[] args) {
    boolean printNewline = true;
    List<String> parameters = new ArrayList<>(Arrays.asList(args);
    if (args.length > 0 && isNewLine(args[0]) ) {
        printNewline = false;
        parameters.remove(0); //remove the first one
    }

    String output = parameters.stream()
        .collect(Collectors.joining(" ", "", printNewLine ? System.lineSeparator() : "")

    System.out.print(output);
}

NOTE: the method isNewLine() is not shown here - but you should consider using such a method to prevent a missing argument... think of -n, -N, /n, /N, all these parameters can be handle in the isNewLine-method...

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Oracle
docs.oracle.com › cd › E19146-01 › 821-0791 › abwbh › index.html
echo Command (Sun Java System Web Server 7.0 Update 7 Developer's Guide)
Sun Java System Web Server 7.0 Update 7 Developer's Guide ... The echo command inserts the value of an environment variable. The var attribute specifies the environment variable to insert. If the variable is not found, “(none)” is inserted.
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Stack Overflow
stackoverflow.com › questions › 56325862 › how-to-run-echo-command-through-java
android - How to run “echo” command through Java? - Stack Overflow
You can do system call, Runtime.exec(). Or if you want to do it in pure Java you can reproduce the behavior of the command line. "echo" is just writing on the standard output, and the '>' operator is redirecting the standard output to a file ...
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University of Amsterdam
staff.fnwi.uva.nl › a.j.p.heck › Courses › JAVAcourse › ch1 › ss2_1.html
Getting Started: The Echo Application
By following the steps on this page, you can create and use a stand-alone Java application that echoes the command line argument twice.
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Javatpoint
javatpoint.com › echo-command
Echo Command in Linux/Unix with Examples - javatpoint
Echo Command in Linux/Unix with Examples with examples on files, directories, permission, backup, ls, man, pwd, cd, linux, linux introduction, chmod, man, shell, pipes, filters, regex, vi etc.
Find elsewhere
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YouTube
youtube.com › murexthecoder
Writing Java Programs using echo command - YouTube
Have you ever wrote any program using echo command , no IDE, no text editors strictly the echo command on your terminalwell in this video i write a short jav...
Published   November 13, 2022
Views   120
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Coderanch
coderanch.com › t › 529135 › java › Execute-echo-BASHPID-Linux-Java
Execute echo $BASHPID in Linux using Java (Java in General forum at Coderanch)
March 1, 2011 - Hi Erik. Take some time to read what the java.lang.Runtime documentation says on the various exec() methods. Note that exec() does not work like the command prompt.
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Ibmstreams
ibmstreams.github.io › streamsx.topology › doc › samples › javadoc › simple › Echo.html
Echo (Java Functional Samples)
This Java application builds a simple topology that echos its command line arguments to standard output. The application implements the typical pattern of code that declares a topology followed by submission of the topology to a Streams context com.ibm.streamsx.topology.context.StreamsContext.
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UBC Computer Science
cs.ubc.ca › ~tmm › courses › 111-10 › slides › lect8.code › Echo.java
Echo.java
import java.util.Scanner; public class Echo { public static void main (String[] args) { String message; Scanner scan = new Scanner (System.in); System.out.println ("Enter a line of text: "); message = scan.nextLine(); System.out.println ("You entered: \"" + message + "\""); } }
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Coderanch
coderanch.com › t › 549404 › java › echo-Environment-Variable
echo <Environment Variable> (Beginning Java forum at Coderanch)
$PATH is resolved by the shell, so you'll need to execute this command using a shell: sh -c $PATH. If you're using Windows, the command should be cmd /C echo %PATH%. Note that Windows uses %PATH% instead of $PATH. Edit: basically what N Sahni said There is just one flaw in that code: the error ...
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Oracle
docs.oracle.com › cd › E19146-01 › 820-1065 › 6ncotjb4j › index.html
echo Command (Sun Java System Web Server 7.0 Update 1 Developer's Guide)
Sun Java System Web Server 7.0 Update 1 Developer's Guide ... The echo command inserts the value of an environment variable. The var attribute specifies the environment variable to insert. If the variable is not found, “(none)” is inserted.
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Oracle
docs.oracle.com › javase › tutorial › essential › environment › cmdLineArgs.html
Command-Line Arguments (The Java™ Tutorials > Essential Java Classes > The Platform Environment)
For example, suppose a Java application called Sort sorts lines in a file. To sort the data in a file named friends.txt, a user would enter: ... When an application is launched, the runtime system passes the command-line arguments to the application's main method via an array of Strings. In the previous example, the command-line arguments passed to the Sort application in an array that contains a single String: "friends.txt". The Echo example displays each of its command-line arguments on a line by itself:
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Fiu
users.cis.fiu.edu › ~weiss › dsj4 › code › Echo.java
Echo.java
Here is the source code for Data Structures and Algorithm Analysis in Java (Fourth Edition), by Mark Allen Weiss. The materials here are copyrighted. I have successfully compiled and tested the programs with JDK 1.6.0 update 11. Organization of Files Most of the general, standalone, code is ...
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TheServerSide
theserverside.com › feature › How-to-set-JAVA_HOME-in-Windows-and-echo-the-result
How to set JAVA_HOME in Windows | TheServerSide
To echo JAVA_HOME in a DOS prompt, simply bookend the variable with percentage signs, like so: @REM How to echo JAVA_HOME in windows echo %JAVA_HOME% >> C:\_jdk25.0