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Old 05-26-2011, 02:48 PM   #1
weixx827
 
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Default In the Windows NT environment to create JSP

Windows98, Windows NT 4.0, Windows 2000 skilled, windows 2000 server and other
jdk1_2_2-001-win.exe
tomcat.zip (three.1)

a software obtain
JDK

Tomcat


2nd, the software program installation
( a ) JDK
1, double-click jdk1_2_2-001-win.exe file to set up , utilize the default installation configuration , JDK 's The default set up directory is C: jdk1.two.2, JRE 's default installation directory is C: Program FilesJavaSoftJRE1.two;
two, restart the personal computer ;
three, update the following surroundings variables : the C: jdk1.two.two? in the directory is appended to the Route in ; to .; C: jdk1.2.2lib ools.jar; C: jdk1.2.2libdt.jar additional towards the CLASSPATH. Windwos modify the Autoexec.bat , NT Update process: Control Panel - A directory , I unzip it to D:, it will instantly develop a tomcat directory , to ensure that the D generate for much more than a directory D: / tomcat, I utilised the $ TOMCAT identification ;
three, Tomcat running on port No. The default is 8080 , you can modify , observe that they have a separate Tomcat HTTP server, it need to use a port range hasn't been used , with $ TOMCAT_PORT emblem, modified as follows :
one) Open the $ TOMCATconfserver.xml;
2) modify the ContextManager:
4, double-click the directory $ TOMCAT startup.bat to begin Tomcat;
5, enter in the browser page says that Edition three.0 of the Tomcat set up was productive ;
six, click into the Servlet interface Servlet Instance , must be capable to execute Servlet.
7, click the JSP Example JSP interface to enter , ought to be capable to execute JSP , and when not, modify the $ TOMCATconfstart.bat: add SET JAVA_HOME = C: jdk1.2.2 can .
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