Java

20-Oct-10 1:29

now its clear Torsten.....thanks for ur suggestion too....

Classes and methods

vidhyaej19-Oct-10 21:28

19-Oct-10 21:28

hi....

Let me tell firstof all that i am a beginner in Java ,can anyone help me to understand the following piece of code...also to elaborate this class a bit so as to get an output...

class A
{
int a;
A()
{
this(10);
}
A(int x)
{
a=x;
}
}

Nagy Vilmos19-Oct-10 22:01

Nagy Vilmos

19-Oct-10 22:01

Firstly, can you add <pre> tags around your code so that can be read easily.

What you have is class A that has one member variable a and two constructors.
The default constructor A() simply delegates to the constructor taking an argument A(int x).

Example usages would be:

A a1 = new A();   // result is a1.a== 10
A a2 = new A(10); // same as above, a2.a == 10
A a3 = new A(42); // guess

Panic, Chaos, Destruction.
My work here is done.

or "Drink. Get drunk. Fall over." - P O'H

vidhyaej19-Oct-10 22:58

19-Oct-10 22:58

Thanks Vilmos....
But can u make the second line a bit more clear.....
ie:
"The default constructor A() simply delegates to the constructor taking an argument A(int x)."

Richard MacCutchan19-Oct-10 23:23

19-Oct-10 23:23

A constructor can be called just like a function so the line in the default constructor below:

A()
{
    this(10);
}

merely calls the constructor that takes an integer parameter and sets the member variable to the value 10. Take a look at The Java Tutorials[^] for more information.

Just say 'NO' to evaluated arguments for diadic functions! Ash

vidhyaej19-Oct-10 23:41

19-Oct-10 23:41

Thanks Richard...
Now its clear.....

TorstenH.19-Oct-10 23:27

19-Oct-10 23:27

class A // class defintion - this is the name of the object called later
{
 int a; // variable - this is a member, your object will have this. it is not initialized! nullpointer-exception might pop up.

 A() // 0-argument-constructor is called by "A myObject = new A();" and sets the member to a value of 10.
 {
  this(10);
 }
 A(int x) // 1-argument-constructor has a argument, wich lets you take control of the member. it is called by "A myObject = new A(5);" and in THIS CASE sets the value of x to 5.
 {        // "A myObject = new A(32457);" will set the value of x to 32457.
  a=x;
 }
 public int somemethod(int value){ // some method, it will return the result from x * value it is called by "myObject.somemethod(2);"
    return (x*value);
 }

Try to make your work easier by giving full classified names to objects, members and methods. and formate the code. make comments - feel free to express you.

regards,
Torsten

I never finish anyth...

vidhyaej19-Oct-10 23:43

19-Oct-10 23:43

Thanks,Torsten....

Its clear now....

Paint problem in java

tamash_ionut19-Oct-10 13:01

tamash_ionut

19-Oct-10 13:01

Hello,

I have the following problem :
- I have a class My frame that extends JFrame and
also I have a class MyComponent that extends JComponent
- the MyComponent class overrides the paint method and displays a picture on a certain position

The problem is that when I call in the MyFrame constructor the methods (inherited from JFrame )
add(new MyComponent( ...position parameters 1.. ))
and also
add(new MyComponent( ...position parameters 2.. ))
only the last one is displayed (if I add more only the last one is drawn).

I spoke to someone and told me that I could be that the frame draws the components on different layers but didn't knew how to solve it ... .I've googled it and still nothing .

If someone had this issue please reply.

Thanks in advance! Smile | :)

Nagy Vilmos19-Oct-10 22:03

Nagy Vilmos

19-Oct-10 22:03

In your overridden paint method, you need to repaint ALL the components not just the alst one added.

Panic, Chaos, Destruction.
My work here is done.

or "Drink. Get drunk. Fall over." - P O'H

TorstenH.19-Oct-10 23:12

19-Oct-10 23:12

You're calling the paint-method from JFrame - for shure the complete content of your JFrame gets (re)painted.
Call the paint-method of your component, that will do the trick.

Check this tutorial

Article "Painting in AWT and Swing" @ oracle.com

regards,
Torsten

I never finish anyth...

tamash_ionut20-Oct-10 10:45

tamash_ionut

20-Oct-10 10:45

Still nothing.

I'ved tried the following:

this are all in the class constructor that extends JFrame

1) 
   add(new MyComponent(0,0,50,50,12));
   add(new MyComponent(100,100,50,50,12));

2)
   this.getContentPane().add(new MyComponent(0,0,50,50,12));
   this.getContentPane().add(new MyComponent(100,100,50,50,12));

   this.repaint();

and only the second component was drawn in both cases. However if I omit comment the second component, the first one is drawn correctly.

My component extends JComponent and overrides paint method by painting a image at the specified coordinates. Confused | :confused:

TorstenH.21-Oct-10 23:00

Glassfish Deployment error

21-Oct-10 23:00

ups

we are in Swing ain't we?

You need to call the paintCompontent()-method. paint() won't do the trick.

article @ oracle.com about Painting in AWT and Swing

regards
Torsten

I never finish anyth...

JDragon31415919-Oct-10 11:09

JDragon314159

19-Oct-10 11:09

Hi:

I am deploying Glassfish 2.1.1 with J2EE 1.4. The project compiles-builds successfully. But it fails to deploy and open in the browser. The error from the log is:

com.sun.enterprise.config.ConfigException: I/O error while loading deployment descriptor for: hello2

caused by

Caused by: java.io.FileNotFoundException: C:/eepassion/UnZip/servletbasics/samples/hello2/build/web

I have looked in the web folder and there is an web folder. But do not know what I'm looking for. Your suggestion would be very helpful.

Lin

Re: Glassfish Deployment error

Richard MacCutchan19-Oct-10 23:17

Re: Glassfish Deployment error

19-Oct-10 23:17

The message seems fairly clear, Java is looking for a file or directory which does not exist, as described in the message. Or possibly there is a permission issue and the program cannot access that folder.

Just say 'NO' to evaluated arguments for diadic functions! Ash

TorstenH.19-Oct-10 23:31

Re: Glassfish Deployment error

19-Oct-10 23:31

Or possibly there is a permission issue and the program cannot access that folder.

often seen since Windows7 - check if your SDK/involved Programms is marked "run as Admin".

regards
Torsten

I never finish anyth...

Richard MacCutchan20-Oct-10 1:24

Processing JSP-Form using Servlet using AJAX Post method

20-Oct-10 1:24

Well I understand that, which is why I suggested it to the OP!

Just say 'NO' to evaluated arguments for diadic functions! Ash

khurram_shahzad17-Oct-10 7:17

khurram_shahzad

17-Oct-10 7:17

I am trying to process data by servlet sended by jsp form using AJAX..... with 'GET' method its working okey ... but using 'POST' method to do this creating problem that onreadystatechage condition not exectue at (xmlhttp.readyState==4 && xmlhttp.status==200)

xmlhttp.onreadystatechange=function() {
                    if (xmlhttp.readyState==4 || xmlhttp.status==200) {
                        alert("state=4 OR 200"); //for testing...
                        //document.getElementById("myDiv").innerHTML=xmlhttp.responseText;
                        document.getElementById("TextChatHistory").value = xmlhttp.responseText;
                    }
}

i tryed to solve this too much .... but i cant .... Any one plz help me ..... THANK ...

COMPLETE PROGRAM CODE IS GIVEN BELOW,

JSP_CODE

<%@page contentType="text/html" pageEncoding="UTF-8"%>
<%--@page import="java.util.Timer.*, java.util.TimerTask.*"
    Timer timer;
--%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
   "http://www.w3.org/TR/html4/loose.dtd">

<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>JSP Page</title>

        <script type="text/javascript" >
            function AjaxFunction() {
                xmlhttp = null;
                if(window.XMLHttpRequest) {
                    xmlhttp = new XMLHttpRequest();
                    //alert("alert_Browser");
                } else {
                    xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
                    //alert("alert_ActiveX");
                }
                xmlhttp.onreadystatechange=function() {
                    if (xmlhttp.readyState==4 || xmlhttp.status==200) {
                        alert("state=4 OR 200"); //for testing...
                        //document.getElementById("myDiv").innerHTML=xmlhttp.responseText;
                        document.getElementById("TextChatHistory").value = xmlhttp.responseText;
                    }
		}
                xmlhttp.open("POST", "TextChat", true);
                var Parameter_MsgSType = "msgText=" + document.getElementById("msgText").value;
                alert(Parameter_MsgSType);  //for testing...
                xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
                xmlhttp.setRequestHeader("Content-lenght", Parameter_MsgSType.lenght);
                xmlhttp.setRequestHeader("Connection", "close");
                xmlhttp.send(Parameter_MsgSType);
                alert("Ajax fuinction finished");   //for testing...
            }
        </script>

    </head>
    <body>
        <!--<h1>Hello JSP!</h1>-->
        <div id="myText"></div> <!--for testing...-->
        <form name="TextChat" method="post" action="" >
            <textarea cols="50" rows="10" readonly id="TextChatHistory" name="TextChatHistory" >TextChat History</textarea>
            <br />
            <textarea cols="50" rows="3" id="msgText" >Enter Text</textarea>
            <select name="SentToType">
                <option value="p">Presentor</option>
                <option value="h">Host</option>
                <option value="a">Attendee</option>
                <option value="l">All</option>
            </select>
            &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
            <input type="submit" value="SEND" onclick="AjaxFunction()" />
        </form>
    </body>
</html>

help required

Nima Dorji15-Oct-10 5:30

Nima Dorji

15-Oct-10 5:30

give me solution of Round robin scheduling code in java

Re: help required

Richard MacCutchan15-Oct-10 6:00

Swing equivalent in java mobile

15-Oct-10 6:00

Ask yourself this: Where would be a good place to search for something that fits my requirements?

This forum tends not to hand out ready made solutions on demand, try doing some research first, and come back here when you have a technical query.

Just say 'NO' to evaluated arguments for diadic functions! Ash

trioum14-Oct-10 8:03

trioum

14-Oct-10 8:03

what is the equivalent package on swing in j2me

Trioum

Re: Swing equivalent in java mobile

Nagy Vilmos14-Oct-10 21:51

Nagy Vilmos

14-Oct-10 21:51

J2ME has within it's framework its own set of controls that are desogned for generic mobile devices. If you are targetting a specific platform then you will also find that there are platform specific UI components. Check the JavaDoc for details.

Panic, Chaos, Destruction.
My work here is done.

or "Drink. Get drunk. Fall over." - P O'H

Problem in learning AJAX with PHP

khurram_shahzad13-Oct-10 11:04

khurram_shahzad

13-Oct-10 11:04

hi ...
i m trying to make this me first simple AJAX example using PHP that shows message from PHP on clicking Button in HTML page .... but this case some prolem ... that is when click on button to show message then nothing is happen .... is also check WAMP ... so any idea or suggestion !!!!

HTML CODE IS,

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>MyFirstAJAXProram</title>



</head>

<body>
<script type="text/javascript">

function JavaFunction() {
	var xmlhttp;
	<!--Getting XMLHttpRequest OR Active depends on browser support...-->
	if(window.XMLHttpRequest) {
		xmlhttp = new XMLHttpRequest();
	} else {
		xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
	}
	
	xmlhttp.onreadystatechange=function() {
		if(xmlhttp.readystate == 4) {
			document.timerForm.time.value = xmlhttp.responseText;
		}
	}
	xmlhttp.open("GET", "phpfile.php", true);
	xmlhttp.send(null);
}
</script>
<form name="timerForm">
<input type="button" onclick="JavaFunction()"; />
Time: <input type="text" id="timer" name="time" value="00:00:00" readonly="readonly" />
</form>
</body>
</html>

HERE's PHP CODE,

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<?
echo("KHURRAM SHAHZAD");
?>
</body>
</html>

Re: Problem in learning AJAX with PHP

Richard MacCutchan13-Oct-10 11:43