First Java project distinguishing between different instances of class -


this first java application have built, have created class products, need declare 4 different instances of products can access each individual product in different parts of code. code have @ moment shown below can seem acces last product have created:

   products = new products(1,2.50,"coke",15);     products = new products(2,1.50,"crisps",5);     products = new products(3,2.00,"juice",2);     products = new products(4,2.50,"chocolate",0);     system.out.println(products.getproductid()+ " " + products.getproductname());

is there way 'where product id equal to' or assign individual name each instance of product? appolagies if question has been asked elsewhere had through related posts couldn't find on related java. code below shows products class:

public class products {   private int productid;  private double productprice;  private string productname;  private int productstock;   public products(int id, double price,string pname,int stock){   productid = id;   productprice = price;   productname = pname;   productstock = stock; }   public void setproductid(int id){   productid = id;  }  public void setproductprice(double price){   productprice = price;  }  public void setproductname(string pname){   productname = pname;  }  public void setproductstock(int stock){   productstock = stock;  }   public int getproductid(){   return productid;  }  public double getproductprice(){   return productprice;  }  public string getproductname(){   return productname;  }  public int getproductstock(){   return productstock;  }

thanks in advance help.

you should name variables different if aren't (you didn't show variable names ...) or make kind of collection (arraylist, list, etc.). eg.

products coke = new products(1,2.50,"coke",15); products crisps = new products(2,1.50,"crisps",5); products juice = new products(3,2.00,"juice",2); products chocolate = new products(4,2.50,"chocolate",0);  system.out.println(coke.getproductid()+ " " + juice.getproductname());

eg. 2

hashmap<integer, products> products = new hashmap<integer, products>(); products.put(products.size(), coke); products.put(products.size(), crisps); products.put(products.size(), juice); products.put(products.size(), chocolate);  system.out.println(products.get(1).getproductid() + " " + products.get(1).getproductname());

Comments

Post a Comment

Popular posts from this blog

php - failed to open stream: HTTP request failed! HTTP/1.0 400 Bad Request -

this code i'm using find latitude , longitude of place: function coords($address){ echo $url="http://maps.googleapis.com/maps/api/geocode/json?address=".$address; $json = file_get_contents(urlencode($url)); $data = json_decode($json, true); print_r($data['results'][0]['geometry']['location']['lat']); print_r($data['results'][0]['geometry']['location']['lng']); } however returns warning: failed open stream: http request failed! http/1.0 400 bad request if use above code in simple procedure works fine not in function. note: have tried curl_init()... method produced same result? instead of encoding whole url encode address part. following code work fine $add='jamshoro phase 2'; $url="http://maps.googleapis.com/maps/api/geocode/json?address=".urlencode($add); $json = file_get_contents($url); $data = json_decode($json, true); print_r($data['r
Read more

java - How to filter a backspace keyboard input -

i have code gives message when key number pressed: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochararray().length - 1]; if (!(ch >= '0' && ch <= '9')) { system.out.println("the char entered not number"); t.consume(); } } }); now if press on wrong button , press backspace remove it, error message. how can add backspace input if statement? i have found answer: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochar
Read more

java - Show Soft Keyboard when EditText Appears -

i have code in android app, alert dialog , keyboard, , when dialog shown want keyboard appear. not , not sure why? here code: final alertdialog.builder alert = new alertdialog.builder(this); alert.settitle("title"); alert.setmessage("message"); final edittext input = new edittext(this); input.setinputtype(inputtype.type_class_number); alert.setview(input); input.setonfocuschangelistener(new view.onfocuschangelistener() { @override public void onfocuschange(view v, boolean hasfocus) { if (hasfocus) { log.i(tag, "in focus"); inputmethodmanager inputmethodmanager = (inputmethodmanager) getsystemservice(context.input_method_service); inputmethodmanager.showsoftinput(input, inputmethodmanager.show_implicit); } } }); but keyboard not appear though edit text in focus, , tell appear why, doesn't come up? thanks in advance.
Read more