I can't seem to open up the Tomcat console despite having just been able to successfully do so. When I click on the "startup" Windows Batch File, the console opens then closes itself immediately afterwards. I've read that it has something to do with setting up the environment variables / paths correctly but I've looked at it and it seems fine; you can check the picture out.
NOTE: Over the course of step 6, the javac command did not work for me. To fix I added a second "C:\Program Files..jdk1.8.0_131\bin" into the Path variable (as you can see in the picture. This enabled me to use the javac command but I have no idea why it didn't detect the JAVA_HOME.
Trying to run startup via cmd console outputs the message: D:\myProject\tomcat\bin>startup
The JRE_HOME environment variable is not defined correctly
This environment variable is needed to run this program
I'm still a beginner with server management so I've been following this tutorial: https://www.ntu.edu.sg/home/ehchua/programming/howto/Tomcat_HowTo.html. I'm at the end of step 6c) where it tells me I should restart the server after compiling an servlet and creating an xml. But lo and behold, it won't open again after quitting...
You must not point JAVA_HOME to bin directory. Just point it to jdk directory, and add a %JAVA_HOME%\bin to your path variable.
I have read similar questions about this problem but it is not working anyways.
I've extracted maven 3.3.3 to D:\Maven (Java already installed and works fine) then added 2 system variables:
Variable
Value
M2
%M2_HOME%\bin
M2_HOME
D:\Maven
I've also appended next in the path:
%M2%
But it is still not working.
I can execute mvn -version from the D:\Maven\bin but can't do this from another directory.
Don't put extra space in your path after D:\Maven\bin; in D:\Maven\bin; D:\Program Files\NodeJS; Please don't make any space in path variable.
Normally, To set mvn in command line in windows :
Set JAVA_HOME property in system environment
eg. JAVA_HOME=C:\Program Files\Java\jdk1.8.0_65
Set M2_HOME property in system environment
eg. M2_HOME=D:\apache-maven-3.3.3
Finally add maven to PATH
;%M2_HOME%\bin; add this at the end of PATH property of system environment
Now Try mvn -version
Try to put full maven path upto bin like "C:\Users\ABC\Downloads\maven-2.0.5-bin\maven-2.0.5\bin" directly into path variable of os, then you will be able to understand where problem exists.
One mistake I did was, There will two environment variables section
One can be accessed via admin while the other one doesn't require any.
Please specify the paths in the second place
It will work for sure., also try echo %M2_HOME% to check whether you have configured correctly..before jumping to mvn install
Hope this helps for someone who has to use admin rights
I am trying to start oracle WebLogic Server 10.3.6 using startWebLogic.cmd located in domain but I get the error below;
\Java\jdk1.7.0_60 was unexpected at this time.
What am I doing wrong and how can I fix it?
I am assuming you have java installed in either C:\Program Files (x86)\Java\jdk1.6.0_23 or C:\Program Files\Java\jdk1.6.0_23 folder.
Path can't take a space in it, so you need to replace this PROGRA~2 instead of Program Files(86); it's the Windows short name for the same folder.
Similarly, use PROGRA~1 for the Program Files folder.
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I am installing Tomcat 7 in my PC it works in Eclipse but when I try to open in bin/ tomcat7.exe it just open and closes, (I am stopping the server in Eclipse). Also, in Eclipse I'am trying to open admin console like URL:(localhost:1234/admin) it shows 404 error. Can anyone who knows this problem explain a solution?
If you are using Windows, you can follow the following to run tomcat7.exe :
Open your Command prompt (cmd)
Set your JAVA_HOME or JRE_HOME. Eg. set JRE_HOME=C:\Program Files\Java\jre7
Go to the directory of your tomcat. Eg. cd C:\Program Files\apache-tomcat-7.0.35\bin
Run startup.bat file. Eg. startup.
This should start your tomcat server without using Eclipse.
First Set JAVA_HOME and CLASSPath path
For Windows Java Path Settings
Right click on Computer and open Properties.
In Windows Vista or Windows 7, go to Advanced System Settings.
Else go to next step. Go to Advanced Tab and click on Environment
Variables button.
Select CLASSPath under the list of System Variables, and press Edit and add If CLASSPath not available there click New button add
C:\Program Files\java\jre 1.x.x_xx\bin
after a semicolon.
Now click on New' button under system variables and enter
JAVA_HOME as variable name and path to jdk home directory (ex.
C:\Program Files\Java\jdk1.x.x_xx
After Completing Java Path Settings
Go to the directory of your Tomcat. Eg. cd C:\Program
Files\apache-tomcat-7.0.35\bin
Run startup.bat file
To start your application server in Windows :
Open command prompt .(Run as Administrator).
Go to the path (..\apache-tomcat-8.0.27\bin) and run startup.bat
A new command prompt opens up and should stay unclosed to tell your server is in running state.
If command prompt opens and closes automatically, then the reason can be as below.
tomcat is already in started state in Eclipse.
Check the connector port configured in server.xml file and check if that port is free in your local machine. [netstat command]
Change the connector port to some unused port, preferably in 80XX series.
Let me know if you are able to invoke localhost:(the http connector port) you have configured in server.xml through you web browser.
You don't start Tomcat by double clicking the exe, use the batch files instead.
startup.bat starts the server, shutdown.bat stops it again.
And ignore the advise to get the service installer. You don't need those unless you want the server to start on system boot and run in the background. Which you don't want unless you're going to configure a production server, which you aren't here.
The batch files are in the same directory on your computer as the exe.
you have to set JAVA_HOME and JRE_HOME in System environment.while setting the JAVA_HOME and JRE_HOME, you should assure that JRE path that you are setting is corresponding to the JDK whose path you have set in JAVA_HOME. some time what happens that there may be more than one JDK in the system so conflict occurs. better to have only one JDK and corresponding JRE.
This is probably due to a bad setup of an environment variable.
For example, if you setup incorrect JAVA_OPTS, tomcat startup will fail.
You can check your configuration by calling : catalina configtest
Here an example of the output for bad setup of JAVA_OPTS :
Error occurred during initialization of VM Could not reserve enough space for object heap
Error: Could not create the Java Virtual Machine.
Error: A fatal exception has occurred. Program will exit.
It works for me after updating the jdk path in JAVA_HOME compatible to your tomcat version in user Variables.
the JAVA_HOME and JRE_HOME environment variables set should not have semicolon or ; at the end...I was facing similar problem...removed semi-colon or ; from the end and tomcat server started from cmd screen after running startup command(of tomcat..bin folder).
I checked environment variables and compiler version everything was fine, deleting junk files from work folder worked for me.
Just download the tomcat 32-bit/64-bit Windows Service Installer (pgp, md5) as in this set up is given which will install tomcat on your PC and you can also start and stop tomcat using the tomcat service.
In my experience tomcat7 might not like it if JAVA_PATH or JDK_PATH contain bracket symbols like in "Program Files (x86)". Make sure you format path in your environmental variables correctly or move your JDK to another folder.
In my case, I had some reference to a jar file in catilana.bat, but the jar I had removed from my system. Tomcat started successfully once I removed that reference from catilana.bat
Check your JRE_HOME location in environment variables. if it didn't work then try
1). Open CMD in administration mode.
2). locate to your tomcat directory.
3). run startup.bat --->> It will show your problem.
Had the same problem with Apache Tomcat 9 version.
Solution is very simple.
Apache Tomcat 9 is supported by Java 8 and later versions. So, I had JRE_HOME with JRE7 path, I binded it to new JRE8 and everything works fine.
Different versions of Apache Tomcat are available for different versions of the Servlet and JSP specifications. The mapping between the specifications and the respective Apache Tomcat versions is:
why the error occurs when i am starting the jboss server 6.0 ?
'findstr' is not recognized as an internal or external command,
operable program or batch file.
It looks like your PATH environment variable hasn't been set up correctly. Does this link help?
http://community.jboss.org/wiki/FindstrCommandNotFound
EDIT: are you sure that the JBoss startup batch script (usually in %JBOSS_HOME%\bin\run.bat) is picking up the PATH correctly? It might be running as a different user with a different PATH. Edit this script and add the line echo %PATH% before the first line that contains findstr. What output does this give you?
The only other place findstr is used (in jboss-6.0.0.20100721-M4 anyway) is in the service.bat script in the same folder as run.bat . Again, you could try putting echo %PATH% before the line in this script that uses findstr if the previous step didn't help you.
EDIT 2: according to your comments, the echo %PATH% line I asked you to add gave the following output:
E:\jdk1.6\bin;E:\apache-ant-1.7.0\bin;E:\jboss-6.0\bin\run.bat
Clearly this doesn't contain C:\WINDOWS\system32, so JBoss definitely won't be able to find findstr. But I don't understand why the PATH is ending up like this. How are you starting JBoss - as a service or by running run.bat? Is JBoss being run under some user account which has been set up with a very restricted PATH? Do you have some other script which is manipulating the PATH before JBoss starts? Also, which version of Windows are you using?
Also, it's not immediately clear to me from your three comments
I'm already checked that the findstr application is already in that path C:\WINDOWS\system32\
E:\jboss-6.0\bin\run.bat
Am also set that in the system variables in Environmental Variables
whether C:\WINDOWS\system32 is in the PATH in Control Panel > System > Environment Variables. Is C:\WINDOWS\system32 in the PATH in the System Variables section within the Environment Variables dialog?
This doesn't seem to be an issue with JBoss. This seems to be more of an issue with the environment within which you are running it. I can quite imagine a lot of other programs would be unhappy with being run in a similar environment.