how to start a tomcat

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Hiii

I am new to advance java and trying to create servlets.I have downloaded TOMCAT 7.0.34 on windows 7(64bit) and jre6.

The problem is when i am working on command promt---- i have created JAVA environment variable(JAVA_HOME) and mentioned it on cmd promt like this:

C:\>set JAVA_HOME .

After this "c:\Program Files\java\jre6" comes itself in next line.
Then i have written next code on command promt is :
h:\>cd\my project\apache-tomcat-7.0.34\bin
.Then the code in next line is
H:\my project\apache-tomcat-7.0.34\bin>
For starting tomcat i am writing on cmd pmt is
H:\my project\apache-tomcat-7.0.34\bin> startup but it is not working

NOW want to start tomcat for which i am writing command startup for starting but it is not working.

Can anyone please suggest me how to start tomcat 7.0.34 on windows 7(64 bit)??

Posted 17-Jan-13 21:08pm

akminder2013

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phil.o · Answer 1 · 2013-01-17T21:38:00

Hi,

You should have a service in the service's console related to your apache/tomcat installation.
You have to see what is the name of the service, and if you want to launch it from command line, the syntax is :

net start tomcat

(Here I assume the name of the service is 'tomcat', but it can be different ; you have to check it first).

Hope this helps.

akminder2013 · Answer 2 · 2013-01-17T23:30:00

Solution 2

ya i will try it but i have noted one more thing that "bin" sub directory of Tomcat have only .sh files which is used for unix but i am using a window so i need .bat files but bin directory dont have these files.Is this a problem or not .

Pls explain

Posted 17-Jan-13 23:30pm