I have installed ant-1.9.2 and jdk1.7.0 in windows 7 and tried the following ways to set the path
C:\>set ANT_HOME=C:\Users\sahaya_s\Downloads\apache-ant-1.9.2;
C:\>set JAVA_HOME=C:\Program Files\Java\jdk1.7.0_45;
C:\>set path=%ANT_HOME%\bin;%JAVA_HOME%\bin;
C:\>echo %path%
C:\Users\anto\Downloads\apache-ant-1.9.2;\bin;C:\Program Files\Java\jdk1.7.0_45;\bin;
here the path has divided into two. what is went wrong?
and tried by giving complete path
C:\>set path=C:\Users\sahaya_s\Downloads\apache-ant-1.9.2\bin;C:\Program Files\Java\jdk1.7.0_45\bin;
C:\>echo %path%
C:\Users\sahaya_s\Downloads\apache-ant-1.9.2\bin;C:\Program Files\Java\jdk1.7.0_45\bin;
C:\>java -version
java version "1.7.0_45"
Java(TM) SE Runtime Environment (build 1.7.0_45-b18)
Java HotSpot(TM) Client VM (build 24.45-b08, mixed mode, sharing)
C:\>ant -version
ANT_HOME is set incorrectly or ant could not be located. Please set ANT_HOME.
Still am getting problem with ANT_HOME. Could anyone help me with this?
You have unneeded ; causing your path to split. Use:
C:\>set ANT_HOME=C:\Users\sahaya_s\Downloads\apache-ant-1.9.2
Related
I was able to install Java and SonarQube, configured the wrapper.conf, and adjusted the port to 9000 to make it work 3 days ago.
Now, running StartSonar.bat gave this
Unable to execute Java command. Access is denied. (0x5) wrapper |
"C:\Program Files\Java\jdk-11.0.11\bin" -Dsonar.wrapped=true ...
C:\Program Files\Java\jdk-11.0.11\bin is what I have in wrapper.conf and it did work.
In a cmd window, C:\windows\system32>java -version
java version "11.0.11" 2021-04-20 LTS Java(TM) SE Runtime Environment
18.9 (build 11.0.11+9-LTS-194) Java HotSpot(TM) 64-Bit Server VM 18.9 (build 11.0.11+9-LTS-194, mixed mode)
navigate to the path above do java -version reports the same.
However I noticed my system Path has this, verified the one in javapath is the same version.
C:\Program Files\Common Files\Oracle\Java\javapath
C:\Program Files(x86)\Common Files\Oracle\Java\javapath
And java_home is empty: echo %java_home% got %java_home%
I'm totally confused now. What went wrong?
I am installing SUbversion Edge under CentOS 7.
The first step is
Set the JAVA_HOME environment variable, and point it to your Java 6 JRE
home. For example:
export JAVA_HOME=/usr/java/default
Test the variable:
$ $JAVA_HOME/bin/java -version
java version "1.6.0_20"
Java(TM) SE Runtime Environment (build 1.6.0_20-b02)
Java HotSpot(TM) Client VM (build 16.3-b01, mixed mode, sharing)
The problem is that /usr/java forlder not exists.
But /usr/bin/java is existing.
Have I use this folder instead?
So what do I have to do?
You should use correct Java home path in your system. If you installed Java in CentOS via YUM, then correct path should be
export JAVA_HOME=/usr/lib/jvm/jre
You should not use /usr/bin/java as JAVA_HOME, because it is not Java home path, it is symlink to java executable.
simple question. I got JDK 1.7 and 1.8 on my pc (Windows). I need both them. I got a "%JAVA_HOME%\bin;" entry in my PATH (JAVA_HOME=path_to_my_jdk_1_7). By command line I change the JAVA_HOME value like this:
set JAVA_HOME=path_to_my_jdk_1_8
Now if I type
C:\Users\Francesco>java -version
java version "1.8.0_20"
Java(TM) SE Runtime Environment (build 1.8.0_20-b26)
Java HotSpot(TM) 64-Bit Server VM (build 25.20-b23, mixed mode)
but when I type
C:\Users\Francesco>javac -version
javac 1.7.0_45
Why javac is still 1.7?
You need to run set PATH=%JAVA_HOME%\bin; again after setting the JAVA_HOME variable to 1.8 JAVA_HOME.
Please see these 3 commands and their outcomes in MacOS:
Korays-MacBook-Pro:~ koraytugay$ /usr/libexec/java_home
/Library/Java/JavaVirtualMachines/jdk1.8.0_11.jdk/Contents/Home
Korays-MacBook-Pro:~ koraytugay$ java -version
java version "1.7.0_55"
Java(TM) SE Runtime Environment (build 1.7.0_55-b13)
Java HotSpot(TM) 64-Bit Server VM (build 24.55-b03, mixed mode)
Korays-MacBook-Pro:~ koraytugay$ echo $JAVA_HOME
/Library/Java/JavaVirtualMachines/jdk1.7.0_55.jdk/Contents/Home
Why does the first one return 1.8? What exactly is /libexec/java_home ?
Koray,
As per the Oracle docs...
JAVA_HOME is just an environment variable used to trigger the 'java' found in your PATH to use a different JDK image. Unfortunately, not all 'java' startup scripts obey this env variable. It's also used by many java tool startup scripts to determine what 'java' to run, bypassing the 'java' found in the PATH setting. Setting this variable during a JDK build is a bad idea, don't do it.
I use Eclipse myself, and don't have a JAVA_HOME env variable defined, here is my output running the same commands.
brandon#brandons-mbp ~
$ /usr/libexec/java_home
/Library/Java/JavaVirtualMachines/jdk1.8.0_05.jdk/Contents/Home
brandon#brandons-mbp ~
$ java -version
java version "1.8.0_05"
Java(TM) SE Runtime Environment (build 1.8.0_05-b13)
Java HotSpot(TM) 64-Bit Server VM (build 25.5-b02, mixed mode)
brandon#brandons-mbp ~
$ echo $JAVA_HOME
Not sure why you have two different versions but did you try updating your JDK recently? Maybe when you upgraded it failed to launch a script to update your $JAVA_HOME.
Here's the instructions for installing JDK and setting JAVA_HOME on *nix based OSes.
Let me know if that helps I can help you dig deeper if need be.
I'm having a problem with determining what version of java am I using:)
cmd.exe is using java.exe from C:\WINDOWS\system32 as specified in Path environment variable.
C:\WINDOWS\system32>java.exe -version
java version "1.6.0_17"
Java(TM) SE Runtime Environment (build 1.6.0_17-b04)
Java HotSpot(TM) Client VM (build 14.3-b01, mixed mode)
But in windows when I right-click on that file and select Properties->Version->Full Version it says 1.6.0_22-b04.
Why?
This means you have two installations. Check your c:\program files\Java folder. Also check your PATH environment variable - it should point to the correct path.
Change to a directory which is not in your PATH environment variable and run java -version. Note that the current directory always is the first entry in PATH.