dhanil das

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Recent posts by dhanil das

This is ma first db2 code

using windows xp sp2



after running the code i got an error message

com.ibm.db2.jcc.am.ro: DB2 SQL Error: SQLCODE=-204, SQLSTATE=42704, SQLERRMC=ROO
T.EMPLOYEE, DRIVER=3.58.82
at com.ibm.db2.jcc.am.ed.a(ed.java:676)
at com.ibm.db2.jcc.am.ed.a(ed.java:60)
at com.ibm.db2.jcc.am.ed.a(ed.java:127)
at com.ibm.db2.jcc.am.tm.c(tm.java:2523)
at com.ibm.db2.jcc.am.tm.d(tm.java:2511)
at com.ibm.db2.jcc.am.tm.a(tm.java:1991)
at com.ibm.db2.jcc.t4.fb.g(fb.java:140)
at com.ibm.db2.jcc.t4.fb.a(fb.java:40)
at com.ibm.db2.jcc.t4.t.a(t.java:32)
at com.ibm.db2.jcc.t4.ub.i(ub.java:135)
at com.ibm.db2.jcc.am.tm.fb(tm.java:1962)
at com.ibm.db2.jcc.am.tm.a(tm.java:2981)
at com.ibm.db2.jcc.am.tm.a(tm.java:652)
at com.ibm.db2.jcc.am.tm.executeQuery(tm.java:636)
at Type2ExampleLegacy.main(Type2ExampleLegacy.java:27)

please give me a solution
i got the solution now

"classes" folder and "xml" file stored under "WEB-INF" which is under "JSPMultipleForms"

"Authentication.jsp" stored under "JSPMultipleForms" folder

use the url "http://localhost:8080/JSPMultipleForms/Authentication.jsp"

perfect it's working now .
8 years ago
JSP
hiii , this is ma jsp code , "Authentication.jsp"


This is ma java code "Authentication.java"



This is ma xml file "web.xml"



I have stored the "Authentication.jsp" and "web.xml" under "JSPMultipleForms" folder, which is under "webapps" folder
The class file stored in the "classes" folder which is under "JSPMultipleForms".

url used for runninng is "http://localhost:8080/JSPMultipleForms/Authentication.jsp" , when i press the "check button" i got these errors

HTTP Status 404 - /JSPMultipleForms/Authentication5

type Status report

message /JSPMultipleForms/Authentication5

description The requested resource (/JSPMultipleForms/Authentication5) is not available.

Apache Tomcat/6.0.20

Please help me how to solve this hurdle....
8 years ago
JSP
This is ma first JSP code , i got some errors during the program executes
here is the code


Error is :

type Exception report

message

description The server encountered an internal error () that prevented it from fulfilling this request.

exception

org.apache.jasper.JasperException: An exception occurred processing JSP page /BookEntryForm.jsp at line 25

22: if(request.getParameter("action") != null){
23: String bookname=request.getParameter("bookname");
24: String author=request.getParameter("author");
25: stmt.executeUpdate("insert into books_details(ID,BookName,Author) values('"+3+"','"+bookname+"','"+author+"')");
26:
27: rst=stmt.executeQuery("select * from books_details");
28: %>


Stacktrace:
org.apache.jasper.servlet.JspServletWrapper.handleJspException(JspServletWrapper.java:505)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:416)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:342)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:267)
javax.servlet.http.HttpServlet.service(HttpServlet.java:717)

root cause

java.lang.NullPointerException
org.apache.jsp.BookEntryForm_jsp._jspService(BookEntryForm_jsp.java:78)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:374)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:342)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:267)
javax.servlet.http.HttpServlet.service(HttpServlet.java:717)

Please help me
8 years ago
Thanks for your quick reply sir , here i got the solution to our problem ,here it is

set path=C:\WINDOWS\system32;C:\WINDOWS;C:\WINDOWS\System32\Wbem;C:\Program Files\MySQL\MySQL Server 5.0\bin;C:\Sun\SDK\bin;C:\Program Files\Java\jdk1.6.0_03\bin;;C:\Program Files\Java\jdk1.6.0_03\bin;


set classpath=;C:\Program Files\Java\jdk1.6.0_03\bin;;C:\ProgramFiles\Java\jdk1.6.0_03\bin;C:\mysql-connector-java-5.0.8\mysql-connector-java-5.0.8-bin.jar;


thank you sir

8 years ago
sir, I got some errors when i execute my first java program to connect the "MYSQL" database ,here is the code



it compiles successfully , but show this error during running ,

:>java JdbcConnection

java.lang.ClassNotFoundException: com.mysql.jdbc.Driver

please help me



sir I have got some error's when i run my first servlet program with Tomcat 6

Errors : HTTP Status 404 -
type Status report

message:

description: The requested resource () is not available.

my java code is :



XML code is :



class directory :C:\Program Files\Apache Software Foundation\Tomcat 6.0\webapps\myApp\WEB-INF\classes\TestingServlet.class

I have run the servlet with url :http://localhost:8080/myApp/servlet/Testing

8 years ago