sai murali

+ Follow
since Oct 03, 2001
Merit badge: grant badges
For More
Cows and Likes
Total received
In last 30 days
Total given
Total received
Received in last 30 days
Total given
Given in last 30 days
Forums and Threads
Scavenger Hunt
expand Ranch Hand Scavenger Hunt
expand Greenhorn Scavenger Hunt

Recent posts by sai murali

Find one excellent book @
17 years ago
Please try the following:
HttpUtils httpUtil = new HttpUtils();
String strTempURL = (httpUtil.getRequestURL(request)).toString();
by performing substring() on strTempURL you can get JSP file name.
20 years ago
When I was getting exception, I had "struts.jar" only in "jre/lib/ext" not in both "WEB-INF/lib" and ""jre/lib/ext"".
20 years ago
Actually, I read following lines from the article:
You can also copy the struts.jar into the jre\lib\ext directory since you will need it to do compiles of programs using Struts. You should also add the servlet.jar file to that directory if you haven't already done it. You can find the servlet.jar file in the c:\tomcat\common\lib directory.
So, I thought I can retain struts.jar in jre\lib\ext even after compiling the files.
Now, I took off all struts.jar instances from throughout my system except from \application\WEB-INF\lib\, now it works fine.
Thankyou so much I am out of loop now.
20 years ago
I have not changed anything from the code downloaded from the link. I just followed the steps for installing struts and deploying the sample files which were given in article.
I am using Tomcat 4.1. I have "struts.jar" in
C:\j2sdk1.4.0_01\jre\lib\ext\ instead of having it in struts/WEB-INF/lib (application) folder.
I am still getting the same:
javax.servlet.ServletException: Missing message for key title.login Exception.
Please help to come out of this loop.
20 years ago
If you can post your sample code what you are using in your web page, it will be more easy to understand the problem.
20 years ago
I have read the article from:
and installed struts as specified in the article. I tried the sample application from the article, but I was getting following exception:

org.apache.jasper.JasperException: Missing message for key title.login
Please help me out.
20 years ago
change your build.xml as follows:
<?xml version="1.0"?>
<project name="test" default="main" basedir=".">
<property name="src" value="." />
<property name="build" value="." />
<path id="project-classpath">
<pathelement location="."/>
<pathelement location="${build}"/>
<target name="init">
<target name="compile" depends="init">
<javac srcdir="${src}" destdir="${build}" />

<target name="run">
<java classname="test" fork="yes" >
<classpath refid="project-classpath" />

<target name="main" depends="compile,run" />
and try again.
20 years ago
Thankyou Valentin,
Your answer cleared my doubt to some extent.
s.concat( s2 );
It will not change the original content of s, will result a new String variable.
So s value will be still "Hello", but the previous
s value is "HelloJava".
Answer is false(b).
Roopa and Ragu,
I got what you were telling, but we can
construct a String using constructor also as:
String s1=new String("java");
But only Strings declaration allows to create
a String object as:
String s1="java";
I just want to know is there any other specific
reason for allowing like this.
thank you....

a is correct because we can REQUEST to invoke garbage
collection explicitly by invoking either Runtime.gc().
b is correct because finalize method will be called just before the object is collected. This is just a notification mechanism
to tell that object is about to be collected. So, this will
be called only before collection.
c is correct, because finalize method will always be
an override of a superclass method. So finalize method should
invoke superclass's finalize method.
This process is having a significance, i.e., before making
an object available for garbage collection we should make
sure that all it is not having any references,
and d is not correct because garbage collection behavior will never be predictable.
[This message has been edited by sai murali (edited October 03, 2001).]
Hi All,
We can create a String class as:
String s1="sample";
But we can't create a StringBuffer as:
StringBuffer s1="sample";
Can you please explain the reason for this
other than syntax.