如何将侦听器放在 web.xml java 中？
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How do I put listener in web.xml java?
提问by Karen Goh
I have created web application and I'd like to know where do I put my listener in the web.xml.
我已经创建了 Web 应用程序，我想知道将我的侦听器放在 web.xml 中的什么位置。
<servlet>
<servlet-name>ProcessReg</servlet-name>
<servlet-class>ProcessReg</servlet-class>
<init-param>
<param-name>pract123</param-name>
<param-value>jdbc:odbc:practODBC</param-value>
</init-param>
<listener>
<listener-class>config</listener-class>
</listener>
</servlet>
The error message I received is:
我收到的错误信息是：
cvc-complex-type.2.4.a: Invalid content was found starting with element 'listener'. One of '{"http://java.sun.com/xml/ns/javaee":init-param, "http://java.sun.com/xml/ns/javaee":load-on-startup, "http://java.sun.com/xml/ns/javaee":run-as, "http://java.sun.com/xml/ns/javaee":security-role-ref}' is expected. [17]
Here's my config file:
这是我的配置文件：
public class config implements ServletContextListener {
private static final String ATTRIBUTE_NAME = "config";
private DataSource dataSource;
@Override
public void contextInitialized(ServletContextEvent event) {
ServletContext servletContext = event.getServletContext();
String databaseName = servletContext.getInitParameter("pract123");
try {
dataSource = (DataSource) new InitialContext().lookup("java:/comp /env/jdbc/TestDB");
} catch (NamingException e) {
throw new RuntimeException("Config failed: datasource not found", e);
}}
@Override
public void contextDestroyed(ServletContextEvent event) {
// NOOP.
}
public DataSource getDataSource() {
return dataSource;
}
public static config getInstance(ServletContext servletContext) {
return (config) servletContext.getAttribute(ATTRIBUTE_NAME);
}
}
采纳答案by Suresh Atta
What you are doing is you are mixing up the <servlet>and <listener>tags .They should be seperate.
你正在做的是你混合了<servlet>和<listener>标签。它们应该是分开的。
That should be
那应该是
<servlet>
<servlet-name>ProcessReg</servlet-name>
<servlet-class>ProcessReg</servlet-class>
<init-param>
<param-name>pract123</param-name>
<param-value>jdbc:odbc:practODBC</param-value>
</init-param>
</servlet>
<listener>
<listener-class>fully.qaulified.path.ContextListener</listener-class>
</listener>
or
或者
<listener>
<listener-class>fully.qaulified.path.ContextListener</listener-class>
</listener>
<servlet>
<servlet-name>ProcessReg</servlet-name>
<servlet-class>ProcessReg</servlet-class>
<init-param>
<param-name>pract123</param-name>
<param-value>jdbc:odbc:practODBC</param-value>
</init-param>
</servlet>
And In your class
在你的课堂上
package fully.qaulified.path;
public class ContextListener implements ServletContextListener {
..
..
As a side note:
作为旁注：
In Java, class names start's with capital letters. public class configshould be
在 Java 中，类名以大写字母开头。 public class config应该
public class Config
回答by Philipp Sander
as you see in the content model of wep-app:
正如您在 wep-app 的内容模型中看到的：
Content Model : (((description*, display-name*, icon*)) | distributable | context-param | filter | filter-mapping |
listener | servlet | servlet-mapping | session-config | mime-mapping | welcome-file-list | error-page | jsp-config |
security-constraint | login-config | security-role | ((env-entry*, ejb-ref*, ejb-local-ref*, ((service-ref*)), resource-
ref*, resource-env-ref*, message-destination-ref*, persistence-context-ref*, persistence-unit-ref*, post-
construct*, pre-destroy*)) | message-destination | locale-encoding-mapping-list)*
it is a sibling of servletnot a child:
它servlet不是孩子的兄弟姐妹：
<servlet>
<servlet-name>ProcessReg</servlet-name>
<servlet-class>ProcessReg</servlet-class>
<init-param>
<param-name>pract123</param-name>
<param-value>jdbc:odbc:practODBC</param-value>
</init-param>
</servlet>
<listener>
<listener-class>config</listener-class>
</listener>
回答by Prasad Kharkar
Listener should not be within a servlettag and provide fully qualified name
侦听器不应位于servlet标记内并提供完全限定名称
<listener>
<listener-class>com.somePackage.ListenerClass</listener-class>
</listener>
回答by Deepak
Listener tag should be defined under web-app tag.
监听器标签应该在 web-app 标签下定义。
<web-app>
<display-name>MyListeningApplication</display-name>
<listener>
<listener-class>config</listener-class>
</listener>
<servlet-name>ProcessReg</servlet-name>
<servlet-class>ProcessReg</servlet-class>
<init-param>
<param-name>pract123</param-name>
<param-value>jdbc:odbc:practODBC</param-value>
</init-param>
</web-app>
回答by insung
web.xml would be
web.xml 将是
<listener>
<listener-class>com.my.ServletContextClass</listener-class>
</listener>
<context-param>
<param-name>pract123</param-name>
<param-value>jdbc:odbc:practODBC</param-value>
</context-param>
and com.my.ServletContextClass.java would be
和 com.my.ServletContextClass.java 将是
public class ServletContextClass implements ServletContextListener {
public void contextInitialized(ServletContextEvent sce) {
ServletContext context = sce.getServletContext();
String value = context.getInitParameter("pract123");
System.out.println("value: " + value);
}
}
enjoy coding :)
享受编码:)
