I am trying to retrieve the ProcessID of a Java app launched through a batch file. The problem is that the application does not start.
My code:
set cmd=java -jar XXXXXXX.jar XXXXXX.yml
for /f "tokens=2 delims==; " %%a in (' wmic process call create "%cmd%" ^| find "ProcessId" ') do set PID=%%a
start cmd /k echo %pid%
PAUSE
Batch files are pretty picky with whitespace.
Change this:
set cmd = java -jar XXXXXXX.jar XXXXXX.yml
To this:
set cmd=java -jar XXXXXXX.jar XXXXXX.yml
This works:
set cmd=notepad
for /f "tokens=2 delims==; " %%a in (' wmic process call create "%cmd%" ^| find "ProcessId" ') do set PID=%%a
start cmd /k echo %pid%
PAUSE
There may be something wrong with your jar file or perhaps java is not on your PATH. Try running your command directly at a command prompt to see if it works:
c:\>java -jar XXXXXXX.jar XXXXXX.yml
Also make sure that you use the full path to your jar file in your cmd otherwise the path will be relative to your system folder:
set cmd=java -jar c:\yourpath\XXXXXXX.jar XXXXXX.yml
I am not familiar with batch script, but I want to create a Windows Batch File to start a Java program. The problem is that it has to specific the path where JRE is installed. When you install both JRE7 and JRE8, the name of that JRE8 folder would call something like jre1.8.0_20 or jre1.8.0_40 with the version number in the back. When you have only JRE8 installed, the folder would call jre8. Is there an easier way to find where the most updated JRE installed and then execute it? Thanks.
start ..\..\Java\jre7\bin\javaw.exe -Xms512M -Xmx1024M -Djna.library.path=.\lib -cp example.jar; com.example.main
You should be able to get the location of javaw.exe by executing where java. This can be set as a variable inside a batch file like this:
# sets a variable called 'java' to the location of javaw.exe
for /f "delims=" %a in ('where javaw') do #set java=%a
# execute you jar file
%java% -jar <app.jar>
Noticed that the above only seems to work when running directly from the command line. Here is another example that should work in a batch file:
# run.bat
#echo off
setlocal enabledelayedexpansion
for /f %%a in ('where javaw') do (
set java=%%a
)
!java! -jar %1
The above batch file should be called with the name of the jar file:
run.bat app.jar
I think it's best to just user JAVA_HOME and/or JRE_HOME and let the user / sysadmin worry what's installed.
I have a set of files, labeled file1.java, file2.java, ...
I want to pipe the output of the files to .txt files, such as output1.txt, output2.txt, etc. corresponding to each java file.
I know that I can do this by doing:
javac file1.java
java file1 > output1.txt
However, is there a way to do this of all files in a given directory? (instead of doing it manually)
This is a good for loop example:
Command line usage:
for /f %f in ('dir /b c:\') do echo %f
Batch file usage:
for /f %%f in ('dir /b c:\') do echo %%f
if the directory contains files with space in the names, you need to change the delimiter the for /f command is using. for example, you can use the pipe char.
for /f "delims=|" %%f in ('dir /b c:\') do echo %%f
If the directory name itself has a space in the name, you can use the usebackq option on the for:
for /f "usebackq delims=|" %%f in (`dir /b "c:\program files"`) do echo %%f
And if you need to use output redirection or command piping, use the escape char (^):
for /f "usebackq delims=|" %%f in (dir /b "c:\program files" ^|
findstr /i microsoft) do echo %%f
It shows both command line + batch for loop examples. Then you will have the filenames and can just run some commands on those filenames like javac and java.
Just try to break it down to small tasks:
Get list of files (all or just .class)
Loop through each file and call javac and java on them
Google is your best friend - search for "DOS batch" + what you need
Your job is much easier if you restructure your output file names a bit: file1.java --> file1_output.txt
From the command line:
for %F in (*.java) do #javac "%F" >"%~nF_output.txt"
If in a batch file:
#echo off
for %%F in (*.java) do javac "%%F" >"%%~nF_output.txt"
If you keep your original naming convention, then you must parse out the number form each file name. Very doable, but it doesn't seem worthwhile in your case.
I have extracted an executable jar file containing xml,java class etc. Actually this executable jar file is a library with dependencies. i need to modify a single line of code in one of the class files in this library. I have successfully edited the class file, now i want to repack it to executable jar. how to do it.
Just zip the whole thing back, a jar is just a zip file.
Unzip the .jar
Modify the class
Rezip the whole structure into a .zip
Rename it to .jar
There you go.
You can automate it with the ant <jar> task for example.
As said by comments below you can use the jar tool that comes with the JDK installation.
jar -xvf your.jar to extract and jar -cvf your.jar inputfiles.
See the documentation.
What makes the JAR executable is that it contains in its structure a file located in META-INF/MANIFEST.MF that describe what is the entry point class, like:
Manifest-Version: 1.0
Main-Class: foo.bar.FooBar
using jar command you can do this
take a command prompt and got your root folder where all the classes. and use
following command
jar -cvf myjarname.jar *
* means all the files and folders in that location
Also, verify that you have the file association for executable jars. If you don't you can create this association with this batch file:
#ECHO off
SETLOCAL ENABLEDELAYEDEXPANSION
:: this script creates a file association for executable .jar files
ECHO Creating .jar file association...
ECHO JAVA_HOME is %JAVA_HOME%
IF NOT DEFINED JAVA_HOME GOTO :FAIL
REG ADD "HKCR\jarfile" /ve /t REG_SZ /d "Executable Jar File" /f
REG ADD "HKCR\jarfile\shell" /ve /f
REG ADD "HKCR\jarfile\shell\open" /ve /f
ECHO REG ADD "HKCR\jarfile\shell\open\command" /ve /t REG_SZ /d "\"%JAVA_HOME%\bin\javaw.exe\" -jar \"%%1\" %%*" /f
REG ADD "HKCR\jarfile\shell\open\command" /ve /t REG_SZ /d "\"%JAVA_HOME%\bin\javaw.exe\" -jar \"%%1\" %%**" /f
REG ADD "HKLM\jarfile" /ve /t REG_SZ /d "Executable Jar File" /f
REG ADD "HKLM\SOFTWARE\Classes\jarfile\shell" /ve /f
REG ADD "HKLM\SOFTWARE\Classes\jarfile\shell\open" /ve /f
REG ADD "HKLM\SOFTWARE\Classes\jarfile\shell\open\command" /ve /t REG_SZ /d "\"%JAVA_HOME%\bin\javaw.exe\" -jar \"%%1\" %%*" /f
ECHO Finished creating .jar file association for executable .jar files.
PAUSE
GOTO EOF
:FAIL
ECHO Script failed. JAVA_HOME not defined.
PAUSE
I need to know where JDK is located on my machine.
On running Java -version in cmd, it shows the version as '1.6.xx'.
To find the location of this SDK on my machine I tried using echo %JAVA_HOME% but it is only showing 'JAVA_HOME' (as there is no 'JAVA_PATH' var set in my environment variables).
If you are using Linux/Unix/Mac OS X:
Try this:
$ which java
Should output the exact location.
After that, you can set JAVA_HOME environment variable yourself.
In my computer (Mac OS X - Snow Leopard):
$ which java
/usr/bin/java
$ ls -l /usr/bin/java
lrwxr-xr-x 1 root wheel 74 Nov 7 07:59 /usr/bin/java -> /System/Library/Frameworks/JavaVM.framework/Versions/Current/Commands/java
If you are using Windows:
c:\> for %i in (java.exe) do #echo. %~$PATH:i
Windows > Start > cmd >
C:> for %i in (javac.exe) do #echo. %~$PATH:i
If you have a JDK installed, the Path is displayed,
for example: C:\Program Files\Java\jdk1.6.0_30\bin\javac.exe
In Windows at the command prompt
where javac
Command line:
Run where java on Command Prompt.
GUI:
On Windows 10 you can find out the path by going to Control Panel > Programs > Java. In the panel that shows up, you can find the path as demonstrated in the screenshot below. In the Java Control Panel, go to the 'Java' tab and then click the 'View' button under the description 'View and manage Java Runtime versions and settings for Java applications and applets.'
This should work on Windows 7 and possibly other recent versions of Windows.
In windows the default is: C:\Program Files\Java\jdk1.6.0_14 (where the numbers may differ, as they're the version).
Java installer puts several files into %WinDir%\System32 folder (java.exe, javaws.exe and some others). When you type java.exe in command line or create process without full path, Windows runs these as last resort if they are missing in %PATH% folders.
You can lookup all versions of Java installed in registry. Take a look at HKLM\SOFTWARE\JavaSoft\Java Runtime Environment and HKLM\SOFTWARE\Wow6432Node\JavaSoft\Java Runtime Environment for 32-bit java on 64 bit Windows.
This is how java itself finds out different versions installed. And this is why both 32-bit and 64-bit version can co-exist and works fine without interfering.
Plain and simple on Windows platforms:
where java
Under Windows, you can use
C:>dir /b /s java.exe
to print the full path of each and every "java.exe" on your C: drive, regardless of whether they are on your PATH environment variable.
The batch script below will print out the existing default JRE. It can be easily modified to find the JDK version installed by replacing the Java Runtime Environment with Java Development Kit.
#echo off
setlocal
::- Get the Java Version
set KEY="HKLM\SOFTWARE\JavaSoft\Java Runtime Environment"
set VALUE=CurrentVersion
reg query %KEY% /v %VALUE% 2>nul || (
echo JRE not installed
exit /b 1
)
set JRE_VERSION=
for /f "tokens=2,*" %%a in ('reg query %KEY% /v %VALUE% ^| findstr %VALUE%') do (
set JRE_VERSION=%%b
)
echo JRE VERSION: %JRE_VERSION%
::- Get the JavaHome
set KEY="HKLM\SOFTWARE\JavaSoft\Java Runtime Environment\%JRE_VERSION%"
set VALUE=JavaHome
reg query %KEY% /v %VALUE% 2>nul || (
echo JavaHome not installed
exit /b 1
)
set JAVAHOME=
for /f "tokens=2,*" %%a in ('reg query %KEY% /v %VALUE% ^| findstr %VALUE%') do (
set JAVAHOME=%%b
)
echo JavaHome: %JAVAHOME%
endlocal
In a Windows command prompt, just type:
set java_home
Or, if you don't like the command environment, you can check it from:
Start menu > Computer > System Properties > Advanced System Properties. Then open Advanced tab > Environment Variables and in system variable try to find JAVA_HOME.
Powershell one liner:
$p='HKLM:\SOFTWARE\JavaSoft\Java Development Kit'; $v=(gp $p).CurrentVersion; (gp $p/$v).JavaHome
In Windows PowerShell you can use the Get-Command function to see where Java is installed:
Get-Command -All java
Or
gcm -All java
The -All part makes sure to show all places it appears in the Path lookup. Below is example output.
PS C:> gcm -All java
CommandType Name Version Source
----------- ---- ------- ------
Application java.exe 8.0.202.8 C:\Program Files (x86)\Common Files\Oracle\Java\jav...
Application java.exe 8.0.131... C:\ProgramData\Oracle\Java\javapath\java.exe
Run this program from commandline:
// File: Main.java
public class Main {
public static void main(String[] args) {
System.out.println(System.getProperty("java.home"));
}
}
$ javac Main.java
$ java Main
More on Windows... variable java.home is not always the same location as the binary that is run.
As Denis The Menace says, the installer puts Java files into Program Files, but also java.exe into System32. With nothing Java related on the path java -version can still work. However when PeterMmm's program is run it reports the value of Program Files as java.home, this is not wrong (Java is installed there) but the actual binary being run is located in System32.
One way to hunt down the location of the java.exe binary, add the following line to PeterMmm's code to keep the program running a while longer:
try{Thread.sleep(60000);}catch(Exception e) {}
Compile and run it, then hunt down the location of the java.exe image. E.g. in Windows 7 open the task manager, find the java.exe entry, right click and select 'open file location', this opens the exact location of the Java binary. In this case it would be System32.
Have you tried looking at your %PATH% variable. That's what Windows uses to find any executable.
Just execute the set command in your command line. Then you see all the environments variables you have set.
Or if on Unix you can simplify it:
$ set | grep "JAVA_HOME"
This is OS specific. On Unix:
which java
will display the path to the executable. I don't know of a Windows equivalent, but there you typically have the bin folder of the JDK installation in the system PATH:
echo %PATH%
On macOS, run:
cd /tmp && echo 'public class Main {public static void main(String[] args) {System.out.println(System.getProperty("java.home"));}}' > Main.java && javac Main.java && java Main
On my machine, this prints:
/Library/Java/JavaVirtualMachines/jdk-9.0.4.jdk/Contents/Home
Note that running which java does not show the JDK location, because the java command is instead part of JavaVM.framework, which wraps the real JDK:
$ which java
/usr/bin/java
/private/tmp
$ ls -l /usr/bin/java
lrwxr-xr-x 1 root wheel 74 14 Nov 17:37 /usr/bin/java -> /System/Library/Frameworks/JavaVM.framework/Versions/Current/Commands/java
None of these answers are correct for Linux if you are looking for the home that includes the subdirs such as: bin, docs, include, jre, lib, etc.
On Ubuntu for openjdk1.8.0, this is in:
/usr/lib/jvm/java-1.8.0-openjdk-amd64
and you may prefer to use that for JAVA_HOME since you will be able to find headers if you build JNI source files. While it's true which java will provide the binary path, it is not the true JDK home.
I have improved munsingh's answer above by testing for the registry key in 64-bit and 32-bit registries, if needed:
::- Test for the registry location
SET VALUE=CurrentVersion
SET KEY_1="HKLM\SOFTWARE\JavaSoft\Java Development Kit"
SET KEY_2=HKLM\SOFTWARE\JavaSoft\JDK
SET REG_1=reg.exe
SET REG_2="C:\Windows\sysnative\reg.exe"
SET REG_3="C:\Windows\syswow64\reg.exe"
SET KEY=%KEY_1%
SET REG=%REG_1%
%REG% QUERY %KEY% /v %VALUE% 2>nul
IF %ERRORLEVEL% EQU 0 GOTO _set_value
SET KEY=%KEY_2%
SET REG=%REG_1%
%REG% QUERY %KEY% /v %VALUE% 2>nul
IF %ERRORLEVEL% EQU 0 GOTO _set_value
::- %REG_2% is for 64-bit installations, using "C:\Windows\sysnative"
SET KEY=%KEY_1%
SET REG=%REG_2%
%REG% QUERY %KEY% /v %VALUE% 2>nul
IF %ERRORLEVEL% EQU 0 GOTO _set_value
SET KEY=%KEY_2%
SET REG=%REG_2%
%REG% QUERY %KEY% /v %VALUE% 2>nul
IF %ERRORLEVEL% EQU 0 GOTO _set_value
::- %REG_3% is for 32-bit installations on a 64-bit system, using "C:\Windows\syswow64"
SET KEY=%KEY_1%
SET REG=%REG_3%
%REG% QUERY %KEY% /v %VALUE% 2>nul
IF %ERRORLEVEL% EQU 0 GOTO _set_value
SET KEY=%KEY_2%
SET REG=%REG_3%
%REG% QUERY %KEY% /v %VALUE% 2>nul
IF %ERRORLEVEL% EQU 0 GOTO _set_value
:_set_value
FOR /F "tokens=2,*" %%a IN ('%REG% QUERY %KEY% /v %VALUE%') DO (
SET JDK_VERSION=%%b
)
SET KEY=%KEY%\%JDK_VERSION%
SET VALUE=JavaHome
FOR /F "tokens=2,*" %%a IN ('%REG% QUERY %KEY% /v %VALUE%') DO (
SET JAVAHOME=%%b
)
ECHO "%JAVAHOME%"
::- SETX JAVA_HOME "%JAVAHOME%"
reference for access to the 64-bit registry
Maybe the above methods work... I tried some and didn't for me. What did was this :
Run this in terminal :
/usr/libexec/java_home
Simple method (Windows):
Open an application using java.
press ctrl + shift + esc
Right click on OpenJDK platform binary. Click open file location.
Then it will show java/javaw.exe then go to the top where it shows the folder and click on the jdk then right copy the path, boom. (Wont work for apps using bundled jre paths/runtimes, because it will show path to the bundled runtime)
in Windows cmd:
set "JAVA_HOME"
#!/bin/bash
if [[ $(which ${JAVA_HOME}/bin/java) ]]; then
exe="${JAVA_HOME}/bin/java"
elif [[ $(which java) ]]; then
exe="java"
else
echo "Java environment is not detected."
exit 1
fi
${exe} -version
For windows:
#echo off
if "%JAVA_HOME%" == "" goto nojavahome
echo Using JAVA_HOME : %JAVA_HOME%
"%JAVA_HOME%/bin/java.exe" -version
goto exit
:nojavahome
echo The JAVA_HOME environment variable is not defined correctly
echo This environment variable is needed to run this program.
goto exit
:exit
This link might help to explain how to find java executable from bash: http://srcode.org/2014/05/07/detect-java-executable/
Script for 32/64 bit Windows.
#echo off
setlocal enabledelayedexpansion
::- Get the Java Version
set KEY="HKLM\SOFTWARE\JavaSoft\Java Runtime Environment"
set KEY64="HKLM\SOFTWARE\WOW6432Node\JavaSoft\Java Runtime Environment"
set VALUE=CurrentVersion
reg query %KEY% /v %VALUE% 2>nul || (
set KEY=!KEY64!
reg query !KEY! /v %VALUE% 2>nul || (
echo JRE not installed
exit /b 1
)
)
set JRE_VERSION=
for /f "tokens=2,*" %%a in ('reg query %KEY% /v %VALUE% ^| findstr %VALUE%') do (
set JRE_VERSION=%%b
)
echo JRE VERSION: %JRE_VERSION%
::- Get the JavaHome
set KEY="HKLM\SOFTWARE\JavaSoft\Java Runtime Environment\%JRE_VERSION%"
set KEY64="HKLM\SOFTWARE\WOW6432Node\JavaSoft\Java Runtime Environment\%JRE_VERSION%"
set VALUE=JavaHome
reg query %KEY% /v %VALUE% 2>nul || (
set KEY=!KEY64!
reg query !KEY! /v %VALUE% 2>nul || (
echo JavaHome not installed
exit /b 1
)
)
set JAVAHOME=
for /f "tokens=2,*" %%a in ('reg query %KEY% /v %VALUE% ^| findstr %VALUE%') do (
set JAVAHOME=%%b
)
echo JavaHome: %JAVAHOME%
endlocal