I try to set the JAVA_HOME path as my Tomcat server is looking for it. I am trying to set it but it doesn't seem to work and causes an error when I do. I am trying to set the JAVA in the setclasspath.bat using
set JAVA_HOME="C:\Program Files (x86)\Java\jre7"
This is at the start of the setclasspath.bat
set JAVA_HOME="C:\Program Files (x86)\Java\jre7"
if not "%JAVA_HOME%" == "" goto gotJdkHome
if not "%JRE_HOME%" == "" goto gotJreHome
echo Neither the JAVA_HOME nor the JRE_HOME environment variable is defined
echo At least one of these environment variable is needed to run this program
goto exit
When I set this and run "startup.bat start" it displays
Files was unexpected at this time
Can you help me?
Note: I realise this is already quite an old question, but many of the answers posted here are either incomplete or inaccurate.. Hopefully this will help save a few headaches.
Firstly: Tomcat does not need a JDK to run, it will work just fine with a JRE, as long as it knows it's a JRE.
Secondly, the error from the original question is coming from an issue with syntax of the set JAVA_HOME=... command. Apache themselves could handle it better with stripping and adding " quote marks.
Also, I would highly recommend creating a setenv.bat file in the bin folder. It's absent by default, so if you don't already have one, create it and add your set JAVA_HOME=... or set JRE_HOME=... lines there.
Run with JRE
As per running.txt:
The JRE_HOME variable is used to specify location of a JRE. The
JAVA_HOME variable is used to specify location of a JDK.
Using JAVA_HOME provides access to certain additional startup options
that are not allowed when JRE_HOME is used.
If both JRE_HOME and JAVA_HOME are specified, JRE_HOME is used.
So, to startup this way, you'll need the following:
set "JAVA_HOME="
set "JRE_HOME=C:\Program Files (x86)\Java\jre7"
Clearing the JAVA_HOME variable is a failsafe, but it's not really required. As per the docs, Tomcat will try use the JRE variable first anyway.
Solution for Issue in Question
Take special note the position of the quotation marks. This way keep the entire string together in one variable, without including the quotation marks in the variable content itself.
For example:
set %TEST%="hello"
echo "%TEST%"
Will output ""hello"".
set "%TEST%=hello"
echo "%TEST%"
Will output "hello".
So, the startup batch file script was trying to use ""C:\Program Files (x86)\Java\jre7"", in which the first non-escaped space is between "Program" and "Files".
As already pointed out, removing the quotation marks (in this particular case at least) would work, but it's dangerous to rely on that. Rather play it safe from the start and wrap the variable name and value in quotation marks.
I had the same problem on Windows 7 with the following definition (I put it in a setenv.bat file in the jdk bin folder as specified by tomcat 7's RUNNING.txt.
set JAVA_HOME="C:\Program Files (x86)\Java\jre7"
I just tried removing the double quotes altogether:
set JAVA_HOME=C:\Program Files (x86)\Java\jre7
And tomcat then started.
First, install java and locate the instalation path.
Locate the tomcat installation path and find the startup.bat in \bin
Open startup.bat and add below(example) lines just after setlocal and before rem Guess CATALINA_HOME if not defined
:: JAVA
set JAVA_HOME=D:\thushara_data\Java\jdk1.8.0_73
set PATH=%JAVA_HOME%\bin;%PATH%
Now try to re-run startup.bat
Place the path in quotes:
set JAVA_HOME="C:\Program Files (x86)\Java\jre7"
The error is due to the fact that it's parsing the Files in Program Files as a separate parameter, which SET doesn't expect. Why SET isn't reading it properly I can't say without knowing more about what OS you're using, what command shell you're running Tomcat from, and so on.
If not able to set the variables manually:
to set JAVA_HOME and JRE_HOME
go to advance system setting and click on environment variables
under system variables click
new variable name : JAVA_HOME
variable value : [path of jdk]C:\Program Files\Java\jdk1.8.0_181 and click ok
similarly add JRE_HOME
click new variable name : JRE_HOME
variable value : [path of jre]C:\Program Files\Java\jre1.8.0_181
click Ok > Ok > Ok
You are pointing to a JRE. You need to point to a JDK, which given the location of your JRE might be something like C:\Program Files (x86)\Java\jdk1.6.0_26 or similar. Or, if you only have a JRE installed, you'll need to install a full JDK. Tomcat needs to be able to compile JSPs into .class files, which a JRE can't do.
The reason why your command failed is because it has white spaces between 'Program Files (x86)' and so it starts reading the command from 'Files (x86)'. So instead we put it in between double quotes such as the following.
try this..
set JAVA_HOME="C:\Program Files (x86)\Java\jre7"
Try
https://askubuntu.com/questions/446294/how-to-start-tomcat7-when-catalina-sh-does-not-work
It can give you some insight even though it's not Windows it still looks for those shell scripts except their extensions in Window is .bat not .sh
You Can Also Try This For jre1.8v:
Open cmd In Tomcat Server Folder And Enter Command
set "JAVA_HOME=C:\Program Files (x86)\Java\jre1.8.0_45"
Then Enter Command cd bin
Then To Start Tomcat Server Enter Command startup
To Stop Tomcat Server Enter Command shutdown
If It Fails To Start Although You Can Run Command: cd startup.bat
Related
I was trying to run ./gradlew bootRun in both Git Bash and PowerShell, and got this error: ERROR: JAVA_HOME is set to an invalid directory: C:\Program Files\Java\jre1.8.0_121
The gradlew file is inside the project directory that cloned from git. It is a practice project from Spring official guide. There's also a gradle.bat file with these lines:
:findJavaFromJavaHome
set JAVA_HOME=%JAVA_HOME:"=%
set JAVA_EXE=%JAVA_HOME%\bin\java.exe
I also tried to run Eclipse EE, which got no problem. The environment setting should be no problem.
In addition to Ernest's answer, the JAVA_HOME it wants is the actual directory, not the bin folder like the path wants for javac. Took me a while to work that out
Looks like your JAVA_HOME variable resolves to a JRE installation directory. But as you're talking about Gradle, you're surely trying to build/compile code.
If you have a JDK installed, change JAVA_HOME variable to point to it. The folder name should typically start with "jdk". Otherwise, you have to install a Java Development Kit.
You could also try exporting the variable manually before running gradle on the command line (SET JAVA_HOME=<path to the jdk directory>)
Please set the JAVA_HOME variable location to the path to JDK instead of JRE.
If that is correctly set, check the environment variables in Edit mode.
In my case, I found that a semicolon was appended to the JAVA_HOME environment variable at the end.
To check:
Open the environment variables list.
Select the variable and click Edit.
Remove the semicolon and Save.
Closed the settings and checked the path in cmd using echo %JAVA_HOME%, and ran gradle -v.
It ran perfectly.
Note: Check the variable value by clicking the Edit button as we cannot see the semicolon instead. Windows adds a semicolon so that the same variable can be used as the path to multiple locations.
Added picture for reference:
Found a solution I hadn't seen before in my googling.
Open up your gradlew.bat in some editor. On line 34-ish you will see this:
:findJavaFromJavaHome
set JAVA_HOME=%JAVA_HOME:"=%
set JAVA_EXE=%JAVA_HOME%\bin\java.exe
if exist "%JAVA_EXE%" goto init
echo.
echo ERROR: JAVA_HOME is set to an invalid directory: %JAVA_HOME%
echo.
echo Please set the JAVA_HOME variable in your environment to match the
echo location of your Java installation.
add this line to the error message: echo %JAVA_EXE%
so it looks like this
echo ERROR: JAVA_HOME is set to an invalid directory: %JAVA_HOME%
echo %JAVA_EXE%
echo.
echo Please set the JAVA_HOME variable in your environment to match the
echo location of your Java installation.
Then run your gradlew.bat file.
The error should now appear again, but with the added echo of the path to JAVA_EXE. In my case it was echoing \bin\java.exe and NOT the full path C:\...\Java\jre1.8.0_251\bin\java.exe
to fix this, I changed
set JAVA_EXE=%JAVA_HOME%\bin\java.exe
to
set JAVA_EXE=C:\...\Java\jdk1.8.0_261\bin\java.exe
The value is now hardcoded, true, but it worked. I don't know why gradle doesn't concat the variable name into the path.
I've found myself with the same issue, although my JAVA_HOME was set to the correct JDK path.
However Windows 10 appended a semicolon at the end of the path.
This way the JAVA_HOME variable would work anywhere but not with Gradle.
I'm posting this, because it might spare someone else the time I've been looking for it ;)
If you didn't have admin rights, override the JAVA_HOME in User Environment Variable. Avoid typo by selecting the Java home folder. You must include folder before the bin ( do not include /bin) for java home.
If none of the above works open the gradle.bat file in an editor and make sure that IT is correct , in my case gradle was the problem
I've tried just about everything but it will not recognize it.
I have the latest version of JDK installed, the path is located at "C:\Program Files\Java\jdk1.8.0_25". This is what I'm using for my "variable value" field when I add the new variable "JAVA_HOME"..
I also have a %JAVA_HOME%\bin; in my path variable. I've restarted multiple times, making small changes..
Help me :(
Keep in mind I don't have much of a clue of what I'm doing.. So I could be going about it completely wrong.. Running it in cmd
Change the path value %JAVA_HOME%\bin, instead of set this
C:\Program Files\Java\jdk1.8.0_25\bin
After setting that, open command prompt and check whether you have set JAVA_HOME correctly.
echo %JAVA_HOME%
java
echo %JAVA_HOME% will print the location where java installed and java will show the usage of java.
See here how to set JAVA_HOME
I want to use elastic search with my PHP application.
after running service install command , I am getting fallowing error.
C:\elasticsearch-0.90.10\bin>service install
JAVA_HOME points to an invalid Java installation (no java.exe found in "C:\Progr
am Files (x86)\Java\jdk1.7.0_25\bin"). Existing...
Open up and Look into the service.bat file:
It searches for:
%JAVA_HOME%\bin\java.exe
hence your %JAVA_HOME% should not include bin in it.
A workaround if you do not have privileges to set up the environmental variables:
open the service.bat file,
a) Remove the line:
if NOT DEFINED JAVA_HOME goto err
b) Replace %JAVA_HOME% with your java jdk path, something like: C:\Program Files\Java\jdk1.7.0_51
Note: This is just a workaround which works all the time.
from your error message we can see your JAVA_HOME points to
C:\Program Files (x86)\Java\jdk1.7.0_25\bin
while it should point to
C:\Program Files (x86)\Java\jdk1.7.0_25
I think the error message is fairly descriptive , you should look in this folder
is java.exe there ? If not then your JAVA_HOME environment variable to point to the correct installation
heres a page showing how to do that
https://confluence.atlassian.com/display/DOC/Setting+the+JAVA_HOME+Variable+in+Windows
1st hit on google.
Either JAVA_HOME points to the wrong location or you don't have Java installed. Check and see if there's an install of Java in the location specified.
Also note that the JAVA_HOME variable should point to the root of the Java install not the /bin folder. This is being added by the program you're executing.
To change the JAVA_HOME variable to test further use the following in the command prompt:
set JAVA_HOME=c:\xxxx
Once you've established the correct value to use this can be permanently changed in Control Panel>>System>>Advanced System Settings>>Advanced>>Environment Variables
set JAVA_HOME=C:\Program Files (x86)\Java
Above must fix the issue, I came across the same issue while installing the elastic search.
i gave a try by giving "bin" part of path but it did not work, so i just give the top folder "java" in path, it worked.
Just to add my own experience.
i was setting JAVA_HOME using tab autocompletion.
It ends up something like "C:\Program files...:" and it didn't work saying that it cannot find the java program. I solved typing the JAVA_HOME path in the set command without using tab completion. The problem was probably related to the " in the env variable
I am using Windows 7 and 8 in my PC .
I have installed jdk 7 and created class path for bin as it should be. But when I enter javac in the command prompt, it throws an error:
`javac is not recognizing as internal and external`.
I have searched for a solution in many websites but nothing could solve my problem.
Would you please help me and suggest where my mistake could be?
You need to set the environmental variables. Run sysdm.cpl from run dialog box to open advanced system properties.
In the environmental variables, add a new one like in this dialog.
Name : JAVA_HOME
VALUE: C:\Program Files\Java\jdk1.7.0_21\
Now edit the value of the PATH variable and add ;%JAVA_HOME%\bin\; at the end of it's value.
Now, restart your system and you can run java related commands from command prompt.
also set path in your *System variables:path=c:/..../jdk/bin/;c:/..../JRE/bin;.;
user Variables:CLASSPATH=c:/..../jdk/bin/;c:/..../JRE/bin;.;
Double check your system variable "Path" on System Properties appending %PATH_TO_JDK%\bin. Don't forget to restart the command prompt after.
It is path not classpath - -till your bin folder
and restart cmd
i've installed jboss-as-7.1.1.Final on my windows 7.
the installation was successfull, and i can see the server running on http:/localhost:8080/
But the problem is I am trying to add the users through add-user.bat file. It shows a error message like
C:\jboss-as-7.1.1.Final\bin>add-user.bat
The system cannot find the path specified.
Press any key to continue . . .
If I remove #echo off from the top of the script I see
> if "x" == "x" (set "JBOSS_MODULEPATH=C:\jboss-as-7.1.1.Final \modules" )
> "C:\java\jre7\bin\bin\java" -jar "C:\jboss-as-7.1.1.Fina l\jboss-modules.jar"
-mp "C:\jboss-as-7.1.1.Final\modules" org.jboss.as .domain-add-user
The system cannot find the path specified.
> if "x" == "x" pause
Press any key to continue . . .
I have set JBOSS_HOME and JAVA_HOME in environmental variables of my system.
Can any one please help to resolve this?
You've set your JAVA_HOME variable wrong. You've set it to
JAVA_HOME=C:\java\jre7\bin
whereas you want it without the bin directory, i.e. just
JAVA_HOME=C:\java\jre7
However it's my recollection that JBoss needs a JDK, not a JRE. If that works, great; if not, you'll need to download an install a JDK instead and when you've installed that, set JAVA_HOME to the root path of the JDK install, e.g.
JAVA_HOME=C:\java\jdk1.7.0_25
The one word solution:
set your java home value excluding bin folder like JAVA_HOME="C:\Program Files\java\jdk1.7"
Use the java home in path section of system variable like ...;%JAVA_HOME\bin%
Set the jboss home path like JBOSS_HOME="C:\jboss-as-7.1"
the reason behind this is your add-user.bat file need correct path for java home and jboss home.
You will probably need to add JAVA_HOME/bin to your PATH.
Add below lines for user variables:
variable = JAVA_HOME value= C:\Program Files\Java\jdk1.7.0_79
variable = JBOSS_HOME value= E:\anil\jboss-as-7.1.1.Final
variable=path value= C:\Windows\System32;%JAVA_HOME%\bin;%JBOSS_HOME%