So I was trying to set my JAVA_HOME variable for a program and though I was sure I set the right directory the program was printing:
The system cannot find the path specified.
I did some testing and now I'm just baffled.
Other info:
C:\Program Files\Java\jre7\bin>java.exe -version
java version "1.7.0_60"
Java(TM) SE Runtime Environment (build 1.7.0_60-b19)
Java HotSpot(TM) 64-Bit Server VM (build 24.60-b09, mixed mode)
Windows version
OS Name: Microsoft Windows 8.1 Pro
OS Version: 6.3.9600 N/A Build 9600
The cd command handles paths with spaces but dir requires that you quote the path if it contains spaces.
dir "%JAVA_HOME%"
will do what you expect.
The issue is that cd expects only one path, so even if the path contains a space it's still treated as a single string. dir can handle multiple paths, delimited by spaces, so when %JAVA_HOME% gets expanded you get
dir C:\Program Files\Java\jre7\bin
which contains TWO paths (as seen by the dir command), neither of which exists:
C:\Program
Files\Java\jre7\bin
Related
There are a few of these questions. But my PATH variable is correct
;%JAVA_HOME%\bin;
My JAVA_HOME is C:\Program Files\Java\jdk1.8.0_131
When I type java -version, I get
java version "1.8.0_131"
Java(TM) SE Runtime Environment (build 1.8.0_131-b11)
Java HotSpot(TM) 64-Bit Server VM (build 25.131-b11, mixed mode)
echo %JAVA_HOME% and echo %PATH% return what they are supposed to. What can be the reason wsimport doesn't work?
My full PATH is
C:\Informatica\9.5.1\clients\DT\bin;C:\Informatica\9.5.1\clients\tools\datadirect;C:\Informatica\9.5.1\clients\tools\odbcdrv;C:\oracle\client_1\bin;C:\ProgramData\Oracle\Java\javapath;%SystemRoot%\system32;%SystemRoot%;%SystemRoot%\System32\Wbem;%systemroot%\System32\WindowsPowerShell\v1.0\;%systemroot%\System32\WindowsPowerShell\v1.0\;C:\Program Files (x86)\Skype\Phone\;C:\Program Files\TortoiseSVN\bin;C:\Program Files\PuTTY\;C:\Program Files (x86)\PuTTY\;%JAVA_HOME%\bin;%ZOOKEEPER_HOME%\bin;C:\Program Files\Git\cmd;C:\Program Files\apache-maven-3.5.0\bin;
%JAVA_HOME%\bin won't work in %PATH%. It should be full path without variables. There is no variable expansion when OS search for executable.
Your java executable found in some system folder.
I suppose %SystemRoot% is some sort of special case.
Solution
Replace %JAVA_HOME%\bin by real path.
Firstly, try to check javac -version. Maybe you provide some wrong path for JDK and system recognize only JRE path.
What is echo %PATH% output? Is there some different path for java before ;%JAVA_HOME%\bin?
I recently installed Java but when I open the command line and I enter:
java -version
I get back:
C:\Users\Tim>java -version
java version "1.8.0_60"
Java(TM) SE Runtime Environment (build 1.8.0_60-b27)
Java HotSpot(TM) 64-Bit Server VM (build 25.60-b23, mixed mode)
But when I enter:
javac -version
I get back:
C:\Users\Tim>javac -version
'javac' is not recognized as an internal or external command,
operable program or batch file.
I added the following line to the end of my path in environment variables:
;C:\Program Files\Java\jdk1.8.0_60\bin;
I installed this version:
jdk-8u60-windows-x64.exe
I use Windows 10. I have sought out multiple sources for help but none of them seem to be working. I think I've installed the right package as the name is JDK, it is for Windows on a 64 machine and java is found, but the compiler is not. I'm trying to train on Java now, but I'm stuck here and can't go much further until I iron this out. I appreciate any help. Thank you.
You missed some steps. Go to variable environment and do the following:
System Variables:
Edit path variable adding the following:
;C:\ProgramFiles\Java\Java_version\bin
Create a new variable. Its name will be classpath and its value will be:
;C\ProgramFiles\Java\Java_Version\src.zip
User Variables :
Create a variable called path, its value will be:
C:\ProgramFiles\Java\Java_verson\bin
Create a variable called classpath , its value will be:
C\ProgramFiles\Java\Java_Version\src.zip
I had that problem.. the solution is to remove all spaces from the java path. So instead of having it located at:
JAVA_HOME = C:\Program Files\Java\jdk1.8.0_60
move the jdk folder to,
for example: JAVA_HOME = C:\Java\jdk1.8.0_60
Then add the following line to the end of path in environment variables:
C:\Java\jdk1.8.0_60\bin;
it worked for me :)
I'm currently setting up a Tomcat Server under Ubuntu 14.10.
in ${CATALINA_HOME}/RUNNING.txt it says, that either JAVA_HOME or JRE_HOME need to be set. I filled the file ${CATALINA_HOME}/bin/setenv.sh with:
JAVA_HOME=/usr/lib/jvm/java-7-openjdk-amd64
since
sudo dpkg -L 'openjdk-7-jdk'
indicated that this is the place where my JDK is installed. To test the location, i also tried:
JAVA_HOME=/usr/lib/jvm/some-none-existing-place
which gave no error when starting tomcat using ${CATALINA_HOME}/bin/startup.sh (although the path does obviously not exist)
So, when is this variable used by Tomcat and how can I check if I set the correct path (for Tomcat)?
Type in terminal,
echo $JAVA_HOME
It will display JAVA_HOME variable path.
IF nothing appears then you can set it manually by,
export JAVA_HOME=/usr/lib/jvm/java-7-openjdk-amd64
This will differ according to your JDK type & version
Edit: Also you just follow the below to ensure you set in tomcat configuration file.
Hit Ctrl+Alt+T to open terminal and run:
sudo gedit /etc/default/tomcat7
Find commented JAVA_HOME, uncomment and change the directory to your JDK environment. Restart now and check for the error.
Open a command prompt, type :
java -version
java version "1.8.0_60" Java(TM)
SE Runtime Environment (build > 1.8.0_60-b27)
Java HotSpot(TM) 64-Bit Server VM (build 25.60-b23, mixed mode)
javac -version
javac 1.8.0_60
echo %JAVA_HOME%
C:\Program Files\Java\jdk1.8.0_60
So, what is your problem:
a) Tomcat does not start, or
b) it starts, but you do not know what version of Java it is using?
Check that your setenv.sh file is readable. If it is not readable, it will be ignored.
If JRE_HOME variable is set then JAVA_HOME will be ignored.
If neither one is set, then there are some defaults that are tried, including a call to which java. (The magic is in ${CATALINA_HOME}/bin/setclasspath.sh).
The ${CATALINA_HOME}/bin/version.sh script will print the versions of Tomcat and Java.
Tomcat scripts print the actual value of JRE_HOME when Tomcat starts if you have a terminal. (if [ $have_tty -eq 1 ]; block in catalina.sh).
In recent versions of Tomcat a org.apache.catalina.startup.VersionLoggerListener can be configured in server.xml and will print the path to java and other settings.
http://tomcat.apache.org/tomcat-8.0-doc/config/listeners.html
I want to install Apache ant on windows. Here is some information below:
http://ant.apache.org/manual/install.html
Error installing Ant: ANT_HOME is set incorrectly
I followed its descriptions and step by step but it still cannot work.
I think the entire environment variables I set are correct. May please someone helps me? And some copies from cmd.
C:\>echo %ANT_HOME%
C:\Program Files\apache-ant-1.9.4
C:\>echo %JAVA_HOME%
C:\Program Files\Java\jdk1.6.0_34
C:\>java -version
java version "1.6.0_34"
Java(TM) SE Runtime Environment (build 1.6.0_34-b04)
Java HotSpot(TM) Client VM (build 20.9-b04, mixed mode, sharing)
C:\>ant
is not recognized as an internal or external command,operable program or batch file.
My Ant version is 1.9.4
I think the path is not set properly.
set JAVA_HOME=C:\Program Files\Java\jdk1.7.0
set ANT_HOME=D:\software\apache-ant-1.9.4
set PATH=%PATH%;%JAVA_HOME%\bin;%ANT_HOME%\bin
Edit: fixed PATH definition
Change your path From:
C:\Program Files\apache-ant-1.9.4
C:\Program Files\Java\jdk1.6.0_34
To:
C:\Progra~1\apache-ant-1.9.4
C:\Progra~1\Java\jdk1.6.0_34
When setting the paths.
Change you path , not install on C:\Program Files\
Add Path, like:
set PATH=%PATH%;%JAVA_HOME%\bin;%ANT_HOME%\bin
I have two javas installed on my pc. jdk1.7.0_45 (x64) and jdk1.6.0_45(x86). I want that default java on my machine would be jdk1.6.0_45(x86). I added JAVA_HOME with value C:\Program Files (x86)\Java\jdk1.6.0_45 and added C:\Program Files (x86)\Java\jdk1.6.0_45\bin to system path, but still when i type in command prompt "java -version" it says:
java version "1.7.0_45"
Java(TM) SE Runtime Environment (build 1.7.0_45-b18)
Java HotSpot(TM) 64-Bit Server VM (build 24.45-b08, mixed mode)
I have no idea why the things are like that, because my OS should see jdk1.7.0_45 (x64) version. Why is that so?
I use windows 8
edit:
after setting java home to jdk1.7.0_45 (x64) folder and updating path accordingly javac works, but then I set java home and path to jdk1.6.0_45(x86) javac does not work.
You need to add the %JAVA_HOME%\bin to the very beginning of your path. This is because Java is installed in your %SystemRoot%\system32, too. If you add the %JAVA_HOME%\bin to the beginning of your path the Java installed in your JAVA_HOME will be used.
JAVA_HOME does not help when we run java from command line, you should fix PATH system var
I added JAVA_HOME with value C:\Program Files (x86)\Java\jdk1.6.0_45
In Windows, you should excape spaces in the path: JAVA_HOME="C:\Program Files (x86)\Java\jdk1.6.0_45". But the best way is to set Java to a folder that doesn't have spaces in its path. For example, I use C:\Java\Java6
and added C:\Program Files (x86)\Java\jdk1.6.0_45\bin to system path
Probably you added after path to the Java 7. Make sure that Java 7 doesn't exist in the system path.