Getting error: variable private acces in Component

Getting error: variable private acces in Component

By : Icon-pc Icon-compute
Date : November 22 2020, 12:01 PM
help you fix your problem Your class Gamey extend the JFrame class which in-turn extend the Component class. In the method startGame(), you have used a field called name in this statement.
code :
label1 = new JLabel(name + "Snow glows white on the mountain tonight");
name has a private access in Component.
public class Gamey extends JFrame {
    // Other fields
    private String name;
    // Getter & setter for name


    public static void main(String[] args) {
        Gamey game = new Gamey();
        game.setName(JOptionPane.showInputDialog(null, "enter name: ")); // Set the name with the value from the input dialog

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php call private variable inside another private variable error

php call private variable inside another private variable error

By : reptar
Date : March 29 2020, 07:55 AM
it helps some times Class members have to be initialized with static values. You can't use a function result in the initialization, so
code :
    'password' => sha1('salt'.$this->password)
PHP: Get Acces To Private Variable In Class

PHP: Get Acces To Private Variable In Class

By : Jgarner
Date : March 29 2020, 07:55 AM
wish of those help Look at this approach.
first: create Entity that stores and retrieves data inside of private $attributes array, and with magic __set(), __get() You can also do like: $object->variable = 123
code :
    class Entity {
        private $attributes;

        public function __construct($attributes = []) {

        public function setAttribute($key, $value) {
            $this->attributes[$key] = $value;
            return $this;

        public function setAttributes($attributes = []) {
            foreach($attributes AS $key => $value) {
              $this->setAttribute($key, $value);

        public function getAttribute($key, $fallback = null) {
            return (isset($this->attributes[$key]))? 
                   $this->attributes[$key] : $fallback;

        public function __get($key) {
            return $this->getAttribute($key);

        public function __set($key, $value) {
            $this->setAttribute($key, $value);

    class Human extends Entity {
        public function __construct($attributes = []) {

        public function hasValidAge() {
            return ($this->getAttribute('age') > 12)? true : false;

    $boy = new Human(['name' => 'Mark', 'age' => 14]);
    if($boy->hasValidAge()) {
        echo "Welcome ".$boy->name."!";

Acces and place to session variable inside the sql query (NullReferenceException) error

Acces and place to session variable inside the sql query (NullReferenceException) error

By : Toufik Mujawar
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further In a nutshell, Session["myConf"] is a NULL value, so you need to guard against that by performing a null check on that session value:
code :
private void BindGrid()
    if (Session["myConf"] == null) 
        return; // Or whatever you want to do.

    Object conference = Session["myConf"];
using a Linked List with a Stack: private acces in node error

using a Linked List with a Stack: private acces in node error

By : Ketki Raje
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , I inserted comments in your code about the issues I saw. In the Node class there was only a single one, in setNext():
code :
public void setNext(Node n){
    // assign the parameter to the field
    this.next = n;
public class StackLL{
    private Node head; //Whatever is on top of the stack
    private int n; //Suze of the stack
    // the field next is not used; just delete it
//  private Node next;

    public StackLL(){
        head = null;
        n = 0;

    public boolean isEmpty(){
        return n == 0;

    public void push(){
        Node temp = head;
        // The Node constructor takes two arguments
        head = new Node("Jim", 541365250);
        // x is not defined; delete this statement
//      head.x = x;
        // Node.next is not visible (it’s private), so use the setter 

    public Node pop(){
        // if you want to return a Node, just return head
        Node x = head;
        // next is private, use the getter
        head = head.getNext();
        return x;

    public Node top(){
        // if you want to return a Node, just return head
        return head;

    // printStack method for StackLL
    public void printStack() {
        Node temp = head;
        while (temp != null) {
        temp = temp.getNext();

How to fix "*private variable* is a private member of '*class name*' error

How to fix "*private variable* is a private member of '*class name*' error

By : user2629251
Date : March 29 2020, 07:55 AM
help you fix your problem I'm writing code that uses friend functions but I am not sure why I get the error "is a private member of" in the function "sum" since I declared the function as a friend in the header file. , You want something like this:
code :
rational sum (const rational& r1, const rational& r2)
    // formula: a/b + c/d = ( a*d + b*c)/(b*d)

    int numerator = ((r2.denominator * r1.numerator) + (r1.denominator * r2.numerator));

    int denominator = (r1.denominator * r2.denominator);
    return rational(numerator, denominator);
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