URgent !!!!!!!!! How do i add information to server.xml of tomcat

Hi ,
I want to add the conext information to my server.xml of tomcat for my hibernate configuration.....
the conext information is as follows ....
<Context path="/quickstart" docBase="quickstart">
<Resource name="jdbc/quickstart" scope="Shareable" type="javax.sql.DataSource"/>
<ResourceParams name="jdbc/quickstart">
<parameter>
<name>factory</name>
<value>org.apache.commons.dbcp.BasicDataSourceFactory</value>
</parameter>

<parameter>
<name>url</name>
<value>jdbc:postgresql://localhost/quickstart</value>
</parameter>
<parameter>
<name>driverClassName</name><value>org.postgresql.Driver</value>
</parameter>
<parameter>
<name>username</name>
<value>quickstart</value>
</parameter>
<parameter>
<name>password</name>
<value>secret</value>
</parameter>

<parameter>
<name>maxWait</name>
<value>3000</value>
</parameter>
<parameter>
<name>maxIdle</name>
<value>100</value>
</parameter>
<parameter>
<name>maxActive</name>
<value>10</value>
</parameter>
</ResourceParams>
</Context>
Where in my server.xml should i put
the server.xml looks like this :



<Server port="8005" shutdown="SHUTDOWN">

<Listener className="org.apache.catalina.core.AprLifecycleListener" />
<Listener className="org.apache.catalina.mbeans.ServerLifecycleListener" />
<Listener className="org.apache.catalina.mbeans.GlobalResourcesLifecycleListener" />
<Listener className="org.apache.catalina.storeconfig.StoreConfigLifecycleListener"/>

<GlobalNamingResources>

<Environment name="simpleValue" type="java.lang.Integer" value="30"/>

<Resource name="UserDatabase" auth="Container"
type="org.apache.catalina.UserDatabase"
description="User database that can be updated and saved"
factory="org.apache.catalina.users.MemoryUserDatabaseFactory"
pathname="conf/tomcat-users.xml" />
</GlobalNamingResources>


<Service name="Catalina">


<Connector
port="8080" maxHttpHeaderSize="8192"
maxThreads="150" minSpareThreads="25" maxSpareThreads="75"
enableLookups="false" redirectPort="8443" acceptCount="100"
connectionTimeout="20000" disableUploadTimeout="true" />

     



<Connector port="8009"
enableLookups="false" redirectPort="8443" protocol="AJP/1.3" />






<Engine name="Catalina" defaultHost="localhost">




<Realm className="org.apache.catalina.realm.UserDatabaseRealm"
resourceName="UserDatabase"/>







<Host name="localhost" appBase="webapps"
unpackWARs="true" autoDeploy="true"
xmlValidation="false" xmlNamespaceAware="false">









</Host>
</Engine>
</Service>
</Server>
Can Some one Help me pleaseeeeee

Please don't cross-post in multiple forums. I have answered
this in your other thread.

Similar Messages

URG !!!!!!! How do i add the conext information to my server.xml of tomcat

Hi ,
I want to add the conext information to my server.xml of tomcat for my hibernate configuration.....
the conext information is as follows ....
<Context path="/quickstart" docBase="quickstart">
<Resource name="jdbc/quickstart" scope="Shareable" type="javax.sql.DataSource"/>
<ResourceParams name="jdbc/quickstart">
<parameter>
<name>factory</name>
<value>org.apache.commons.dbcp.BasicDataSourceFactory</value>
</parameter>

<parameter>
<name>url</name>
<value>jdbc:postgresql://localhost/quickstart</value>
</parameter>
<parameter>
<name>driverClassName</name><value>org.postgresql.Driver</value>
</parameter>
<parameter>
<name>username</name>
<value>quickstart</value>
</parameter>
<parameter>
<name>password</name>
<value>secret</value>
</parameter>

<parameter>
<name>maxWait</name>
<value>3000</value>
</parameter>
<parameter>
<name>maxIdle</name>
<value>100</value>
</parameter>
<parameter>
<name>maxActive</name>
<value>10</value>
</parameter>
</ResourceParams>
</Context>
Where in my server.xml should i put

I don't think you should have to edit the server.xml
at all. your app can have its own context.xml that
obviates the need for editing server.xml
%I did try that in many permutations. For some reason, the server.xml worked for me in 5.5 but not the context.xml. (Granted, it was late on Friday and I might have missed a step). :^)
- Saish

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Hi Raktim:
I looked around on SDN and it seems that i have found what you have been looking for.
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Hi All,
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Thanks!
Pradeep
[email protected]

try following
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&MYFILE = GetFile("c:\temp\test.xml", "N", %FilePath_Absolute);
If &MYFILE.IsOpen Then
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End-If;content c:\temp\test.xml:
<?xml version="1.0"?>
<SOAP-ENV:Envelope xmlns:SOAP-ENC="http://schemas.xmlsoap.org/soap/encoding/" xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/">
<SOAP-ENV:Header xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/">
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</SOAP-ENV:Header>
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</SOAP-ENV:Body>
</SOAP-ENV:Envelope>

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Hi,
I will create a new web application, i.e named newApp. Then I create a file structure as follows:
- <server-root>/newApp
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Then I must tell the server that I have created a new web application. Then I must update my file server.xml, How shall I do this and where in the file shall I type in the new information?
I use windows XP Pro, and Tomcat 4.1.27.
My server.xml file looks like below:



<Server port="8005" shutdown="SHUTDOWN" debug="0">


<Listener className="org.apache.catalina.mbeans.ServerLifecycleListener"
debug="0"/>
<Listener className="org.apache.catalina.mbeans.GlobalResourcesLifecycleListener"
debug="0"/>

<GlobalNamingResources>

<Environment name="simpleValue" type="java.lang.Integer" value="30"/>

<Resource name="UserDatabase" auth="Container"
type="org.apache.catalina.UserDatabase"
description="User database that can be updated and saved">
</Resource>
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<value>org.apache.catalina.users.MemoryUserDatabaseFactory</value>
</parameter>
<parameter>
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<value>conf/tomcat-users.xml</value>
</parameter>
</ResourceParams>
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acceptCount="100" debug="0" connectionTimeout="20000"
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<Connector className="org.apache.coyote.tomcat4.CoyoteConnector"
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enableLookups="true" redirectPort="8443"
acceptCount="10" debug="0" connectionTimeout="0"
useURIValidationHack="false"
protocolHandlerClassName="org.apache.jk.server.JkCoyoteHandler"/>












<Engine name="Standalone" defaultHost="localhost" debug="0">



<Logger className="org.apache.catalina.logger.FileLogger"
prefix="catalina_log." suffix=".txt"
timestamp="true"/>


<Realm className="org.apache.catalina.realm.UserDatabaseRealm"
debug="0" resourceName="UserDatabase"/>







<Host name="localhost" debug="0" appBase="webapps"
unpackWARs="true" autoDeploy="true">





<Logger className="org.apache.catalina.logger.FileLogger"
directory="logs" prefix="localhost_log." suffix=".txt"
timestamp="true"/>

<DefaultContext reloadable="true"/>

<Context path="" docBase="ROOT" debug="0"/>

<Context path="/examples" docBase="examples" debug="0"
reloadable="true" crossContext="true">
<Logger className="org.apache.catalina.logger.FileLogger"
prefix="localhost_examples_log." suffix=".txt"
timestamp="true"/>
<Ejb name="ejb/EmplRecord" type="Entity"
home="com.wombat.empl.EmployeeRecordHome"
remote="com.wombat.empl.EmployeeRecord"/>




<Environment name="maxExemptions" type="java.lang.Integer"
value="15"/>
<Parameter name="context.param.name" value="context.param.value"
override="false"/>
<Resource name="jdbc/EmployeeAppDb" auth="SERVLET"
type="javax.sql.DataSource"/>
<ResourceParams name="jdbc/EmployeeAppDb">
<parameter><name>username</name><value>sa</value></parameter>
<parameter><name>password</name><value></value></parameter>
<parameter><name>driverClassName</name>
<value>org.hsql.jdbcDriver</value></parameter>
<parameter><name>url</name>
<value>jdbc:HypersonicSQL:database</value></parameter>
</ResourceParams>
<Resource name="mail/Session" auth="Container"
type="javax.mail.Session"/>
<ResourceParams name="mail/Session">
<parameter>
<name>mail.smtp.host</name>
<value>localhost</value>
</parameter>
</ResourceParams>
<ResourceLink name="linkToGlobalResource"
global="simpleValue"
type="java.lang.Integer"/>
</Context>
</Host>
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</Service>
</Server>

To use servlets u have indeed to update your web.xml...Well I'm not sure this is relevant to your case anyway.
You have to add a <servlet> element to this file.
Something like this:
<servlet>
<servlet-name>blabla</servlet-name>
<servlet-class>blablapackage.Blablaclass</servlet-class>
<init-param>...</init-param>
</servlet>
Now this may not solve your problem. Make sure you refer to your servlets using their full qualified names.btw, just to be sure, what is your definition of "servlet"? (i mean: any java class or only javax.servlet.Servlet)

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Hi does any body knows how can i obtain the information from a XML form file? I create some news with the xml forms and i need create a program to obtain some information from the XML form file like author, date of creation and so on but need obtain only specific information from the xml form.

Hi,
You can do one thing ,put in the fields u need in show form and it will show that fields that are required by you and not all fields.
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hello,
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When I try to startup the tomcat as root user every thing is ok, it
start (with port 8080)
I have create a new user with permissions to logs and work
directories, and try to startup the tomcat from command
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every thing is ok
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In Catalina.out
Jul 12, 2006 6:03:26 PM org.apache.catalina.startup.Catalina load
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Jul 12, 2006 6:03:26 PM org.apache.catalina.startup.Catalina load
WARNING: Can't load server.xml
Jul 12, 2006 6:03:26 PM org.apache.catalina.startup.Catalina start
INFO: Server startup in 0 ms
java.lang.reflect.InvocationTargetException
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        at
sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.jav
a:39)
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sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessor
Impl.java:25)
        at java.lang.reflect.Method.invoke(Method.java:585)
        at
org.apache.catalina.startup.Bootstrap.start(Bootstrap.java:271)
        at
org.apache.catalina.startup.Bootstrap.main(Bootstrap.java:409)
Caused by: java.lang.NullPointerException
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        at org.apache.catalina.startup.Catalina.start(Catalina.java:560) Can any one help,
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Click Start, click Run, type cmd in the
Open box, and then click OK.
Switch to the <var>Drive</var>:\WINDOWS\Microsoft.NET\Framework\v2.0.50727 folder. In this folder, <var>Drive</var> is the drive where Windows is installed. To do this, type the following line at the command prompt, and then press Enter:
cd <var>Drive</var>:\WINDOWS\Microsoft.NET\Framework\v2.0.50727
Add the appropriate permissions to the Network Service group for the Temporary ASP.NET Files folder. To do this, type the following line, and then press Enter:
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cd <var>Drive</var>:\WINDOWS\Microsoft.NET\Framework\v4
Add the appropriate permissions to the Network Service group for the Temporary ASP.NET Files folder. To do this, type the following line, and then press Enter:
aspnet_regiis -ir
http://social.technet.microsoft.com/Forums/sharepoint/en-US/bf6bb272-eb75-4435-8264-6980e9e7268f/sharepoint-server-2010-access-denied-error?forum=sharepointadminprevious
Is this a claims based web application? If yes, the super reader and super user need to be added with the in the format "i:0#.w|domain\username". Also, add them with full read and full control on the web application as well.
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// If is clicked
if (itemList
.button.hitTest(cursor)){
// Shows the current item
showCurrent(itemList)
// Shows the button clicked
showCurrent=function(current){
// Does this to all the buttons
for (i=0; i<total; i++){
// Does this to all the buttons exept the clicked
if (itemList
!=current){
// Close the other items if autoclose = true
if (autoClose){
// Close the other items
itemList.openBox=false
// fades the roll over effect of the other items
itemList
.over = false
//Does this to the clcked item only
} else {
// If it has sub items
if (total>0){
//Hides them if its open
if (itemList.openBox){
itemList
.openBox=false
//Shows them if its closed
} else {
itemList.openBox=true
// If it has no subitems goes to the link
} else {
getURL(xmlRoot.attributes.url, _self)

Please don't cross-post in multiple forums. I have answered
this in your other thread.

URgent !!!!!!!!! How do i add information to server.xml of tomcat

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