JSP :: Dynamic Menu Duplicates Every Time A Link Is Clicked

Jan 2, 2015

I have following code to populate a dynamic menu from categories retrieved from database

<jsp:useBean id="category" class="category.CatBean" scope="application"></jsp:useBean>

<c:if test="${applicationScope.cartList == null}">
<c:set var="catList" value="${category.categoryList}" scope="application"/>
</c:if>
<div id="categoryBrowse"><label>Browse by Category</label></div>
<div id="cssmenu">

[Code] ....

It works fine. But every time I click on a link on the menu, it duplicates the whole menu.

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Dec 28, 2014

I have a question and wasn't able to find what I needed through the search. I am currently using Netbeans 8.0 and working on a desktop app. I have a main GUI file where I am going to add a menu item that can be clicked to open up an "About" Dialog box. The problem is, I don't know how to do this

The file are...

MainGui.java
About.java

How would I go about opening the About.java when clicking on a menu item? I tried using...

new About();
About.setVisible(true);

But this just gave a compilation error.

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Jul 7, 2014

i have a problem with navigating from a link in a primefaces tree menu to a specific page. The explanation is the following: I have a template with the following code :

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[code]....

i had to put a "*" in the from-view-id tag to be able to navigate to the pages specified with the navigation cases. If i put /templates/leftMenu.xhtml insted of * then the navigation doesn't work and i get the error "WARNING: JSF1064: Unable to find or serve resource, /deleteDepartment-page.xhtml."

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//construct and populate the Time menu
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[Code] ....

E:Calculator.java:64: error: cannot find symbol
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^
symbol: variable arg
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import java.awt.*;
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[code]....

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I want to read json file as follow;
 
{
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  },
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  "col1": "c10",
  "col2": "c11"
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Next to reading I need to assign to two dimensional array, but without giving "col1","col2","col3" a lot. 

The array is for example, Array[3][] = {{"c00"},{"c10","c11"},{"c20","c21","c22"}};

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May 14, 2015

I'm trying to count the number of elements in an ArrayList which also have duplicates. So for example, in an ArrayList of strings which contains cat, cat, dog, horse, zebra, zebra, the answer should be two.

If an element is found to be a duplicate, that element should then be exempt from the search so if that element is found again it should not increase the duplicate count.

Here is my code:

public int countDuplicates() {
int duplicates = 0;
// TODO: Write the code to get the number of duplicates in the list
for (int i = 0; i < list.size()-1;i++) {
boolean found = false;

[Code] ....

I know it's wrong because right now it's still increasing the duplicate count for elements that have already been detected as duplicates. How can I make it stop doing this?

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Apr 11, 2014

Write a method compressDuplicates that accepts a stack of integers as a parameter and that replaces each sequence of duplicates with a pair of values: a count of the number of duplicates, followed by the actual duplicated number. For example, suppose a variable called s stores the following sequence of values:

bottom [2, 2, 2, 2, 2, -5, -5, 3, 3, 3, 3, 4, 4, 1, 0, 17, 17] top

If we make the call of compressDuplicates(s);, after the call s should store the following values:

bottom [5, 2, 2, -5, 4, 3, 2, 4, 1, 1, 1, 0, 2, 17] top

This new stack indicates that the original had 5 occurrences of 2 at the bottom of the stack followed by 2 occurrences of -5 followed by 4 occurrences of 3, and so on. This process works best when there are many duplicates in a row. For example, if the stack instead had stored:

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Then the resulting stack after the call ends up being longer than the original:

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If the stack is empty, your method should not change it. You may use one queue as auxiliary storage to solve this problem. You may not use any other auxiliary data structures to solve this problem, although you can have as many simple variables as you like. You may not use recursion to solve this problem. For full credit your code must run in O(n) time where n is the number of elements of the original stack.

I wrote a code but still having a problem with it , am I allowed to use 3 while loops ?

public void compressDuplicates(Stack<Integer> s ){
Stack<Integer> backup= new Stack<Integer>();
int count = 1;
while(!s.isEmpty()){
int temp = s.pop();

[Code] .....
 
// example 1

Expected output : bottom [5, 2, 2, -5, 4, 3, 2, 4, 1, 1, 1, 0, 2, 17] top

My output: //bottom [17, 2, 2, 2, 2, 2, -5, -5, 3, 3, 3, 3, 4, 4, 1, 0, 17, 17] top

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Nov 5, 2014

I'm struggling with that piece of code, my intention is to check for the object I want to add before adding it, so there won't be any duplicate on my list. I'm not sure how could I do that, since I'm working with objects.

Person is a class with few parameters such as id, name, and few others.

I guess I should search for a person with the same id, since that has be unique, but can't get it right.

private ArrayList<person> model= new ArrayList<>();
//...
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Jan 23, 2013

It's supposed to count all of the duplicates in an array and print out how many occurrences of the value starting at whatever index, or if there are no duplicates state that. Basically:

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There are 3 more occurrences of value 2 starting at index 1

There are 2 more occurrences of value 2 starting at index 2....

This is what I've got so far:

Java Code:

public static void main(String[] args) {
int[] arr = {1, 2, 2, 3, 4, 2, 4, 3, 0, 5, 3, 2};
for(int i = 0; i<arr.length; i++){
int count = 0;
for(int j = i+1; j<arr.length; j++){
if((arr[j] == arr[i]) && (i!=j)){
count++;
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System.out.println(arr[i] + " starting at index " + i);
}
}
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I am stuck on this exercise and I don't know what exactly is wrong. I think it's something with the .remove and the for each loop, but I am not sure.

public class seven {
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list.add("aaa");
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[Code] ....

This is what i get

Exception in thread "main" java.util.ConcurrentModificationException
at java.util.ArrayList$Itr.checkForComodification(Unknown Source)
at java.util.ArrayList$Itr.next(Unknown Source)
at seven.removeDuplicates(seven.java:24)
at seven.main(seven.java:18)

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Apr 12, 2015

I have taken array int[] a = {33,33,5,5,9,8,9,9};

In this array so many values are duplicates means 33 comes twice & 5 also comes twice & 9 comes three times. But I want to count the first value which is duplicate means 33 is first value which comes twice so answer would be 2.

I try:

public class FindFirstDuplicate {
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int c=0;
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outerloop:
for(int i = 0; i < a.length; i++) {

[code]....

Output:

33
Count: 1

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This code is not best way to find the duplicate elements in a given array. Any alternative method for an optimized code.

Java Code:

import java.util.Arrays;
public class Find_Dupliicate_ArrayElement {
public static void main(String[] args) {
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[code]....

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So, with that being said, I would like to add one copy of the string, discard the duplicates, and do this thousands of times..

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DateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd");
jXDatePicker1.setFormats(dateFormat);
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groups| empId
-----------------
Group A| a1234 |
Group A| e3456 |
Group A| w3452 |
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Group A| j7689 |
[Code] ....

I want to format the .csv as follows:

--------------------------
empId | group 1 | group 2 |
--------------------------
a1234 | group A | group B |
---------------------------
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int[] numbers = new int[10];
Scanner input = new Scanner(System.in);
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Java Code:

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import java.util.Map;
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for example,

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90 is technically correct, the way the formula is setup, but I need it to display 0900 not 90.

Here is the code that I have. Be easy, I'm still learning, and this is just the file I created to get the formula to work. Also, the verbose in here is just for my own debugging to make sure values should be what I'm expecting them to be.

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[Code] ....

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code part to another class but nothing..this is what a second class should be doing

1 - Write a java class called Point to represent a 2-D point (With x and y)

- The constructor should take the x & y as double

- The class should have accessor / mutator methods for all coordinates

2 - Write a java class called Line to represent a line (with a starting point and an ending point)

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