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how many time the characters of string are to be found in another string

my professor gave me an exercise to find how many time the characters of string called "filter" are to be found in a second string called "query".
before I begin I am java noob and English isnt my native language.
example:
String filter="kjasd";
String query="kjg4t";
Output:2
getting how many times a char has been found in another string isnt my problem but the problem that the professor gave us some rules to stick with:
class filter. The class must be the following public
Provide interfaces:
public Filter (String letters) (→ Constructor of class)
The string representing the filter should be stored in the letters string
public boolean contains (char character)
Returns true if the passed character is contained in the query string, otherwise false
-public String toString ()
Returns an appropriate string representation of the class (just to be clear I have no clue about what does he means with this one!)
To actually determine the occurrences of the filter in the query, another class QueryResolver is to be created.
The class should be able to be used as follows:
QueryResolver resolver = new QueryResolver();
int count = resolver.where(query).matches(filter).count();
the filter and the query are given by the user.
(i couldnt understand this one! )The methods "where" and "matches" configure the "QueryResolver" to include a subsequent call of "count" the calculation based on the previously passed variables
"query" and "filter" performs.
The count method should use the filter's previously-created method.
The modifier static is not allowed to use!
I dunno if he means that we cant use static {} or we cant use public (static) boolean contains (char character){}
we are not allowed to use void
so the problems that encountered me
- I can not pass a char to the method contains as long as it is not static.
error "Non-static variable can not be referenced from a static context"
i did not understand what i should do with the method toStirng!
what I've done so far:
Approach Nr 1:
so I just wrote everything in the main method to check whether the principle of my code works or not and then I wanted to create that whole with constructor and other methods but unfortunately I did not succeed.
Approach Nr 2:
then I tried to write the code in small mthoden as in the exercise but I did not succeed !.
in both aprroaches i violated the exercise rules but i cant seem to be able to do it alone thats why i posted the question here.
FIRST APPROACH:
public class filter{
public filter(String letters) {
//constructor of the class
String filter;
int count;
}
public boolean contains (char character){
/*Subprogram without static!
*the problem that I can't pass any char to this method if it wasn't static
*and I will get the following error"Non-static variable cannot be referenced from a static context"
*I understand why I'm getting the error but I don't know how to get around it X( */
return true ;
}
public String toString (){
/*he told us to include it in the program but honestly, I don't know what shall I write in it -_-
*I make it to null because you have to return something and I don't know what to do yet
*so, for now, I let it null. */
return null;
}
public static void main(String[] args) {
Scanner in =new Scanner (System.in);
System.out.println("please enter the query string! ");
String query= in.next();
System.out.println("please enter the filter stirng!");
String filter= in.next();
System.out.println("the query string is : [" + query+ "]");
System.out.println("the filter string is : [" + filter+ "]");
int count=0;
// I initialized it temporarily because I wanted to print it!
//later I need to use it with the boolean contains as a public method
boolean contains=false;
//to convert each the query and the filter strings to chars
char [] tempArray=query.toCharArray();
char [] tempArray1=filter.toCharArray();
//to iterate for each char in the query string!
for (int i = 0; i < tempArray.length; i++) {
char cc = tempArray[i];
//to iterate for each char in the filter string!
for (int j = 0; j < tempArray1.length; j++) {
// if the value in the filter string matches the value in the temp array then increment the counter by one!
if(tempArray1[j] == cc){
count++;
contains=true;
}
}
}
System.out.println("the characters of the String ["+filter+"] has been found in the forworded string ["+query+"] exactly "+count+" times!" );
System.out.println("the boolean value : "+ contains);
in.close();
}
}
SECOND APPROACH
- But here too I violated the rules of the task quite brutally :(
- First, I used void and did not use the tostring method.
- Second, I did not use a constructor.
- I did not add comments because that's just the same principal as my first attempt.
public class filter2 {
public static void main(String[] args) {
Scanner in = new Scanner (System.in);
System.out.println("enter the filter string:");
String filterStr=in.next();
System.out.println("enter the query string:");
String querystr =in.next();
Filter(filterStr, querystr);
in.close();
}
public static void Filter(String filterstr , String querystr){
char [] tempArray1 = filterstr.toCharArray();
contains(tempArray1, querystr);
}
public static void contains(char[]tempArray1, String querystr){
boolean isThere= false ;
int counter=0;
char [] tempArray = querystr.toCharArray();
for (int i = 0; i < tempArray.length; i++) {
char cc = tempArray[i];
for (int j = 0; j < tempArray1.length; j++) {
if(tempArray1[j] == cc){
counter++;
isThere=true;
}
}
}
System.out.println("the letters of the filter string has been found in the query string exactly "+counter+" times!\nthus the boolean value is "+isThere);
}
/*
* sadly enough i still have no clue what is meant with this one nor whatshall i do
* public String toString (){
* return null;
* }
*
*/
}
Few hints and advice would be very useful to me but please demonstrate your suggestions in code because sometimes it can be difficult for me to understand what you mean by the given advice. ;)
Thank you in advance.
(sorry for the gramatical and the type mistakes; english is not my native language)

As already mentioned, it is important to learn to solve those problems yourself. The homework is not for punishment, but to teach you how to learn new stuff on your own, which is an important trait of a computer scientist.
Nonetheless, because it seems like you really made some effort to solve it yourself already, here is my solution, followed by some explanation.
General concepts
The first thing that I feel like you didn't understand is the concept of classes and objects. A class is like a 'blueprint' of an object, and the object is once you instanciated it.
Compared with something like a car, the class would be the description how to build a car, and the object would be a car.
You describe what a class is with public class Car { ... }, and instanciate an object of it with Car myCar = new Car();.
A class can have methods(=functions) and member variables(=data).
I just repeat those concepts because the code that you wrote looks like you didn't fully understand that concept yet. Please ask some other student who understood it to help you with that.
The Filter class
public class Filter{
String letters;
public Filter(String letters) {
this.letters = letters;
}
public boolean contains (char character){
for(int i = 0; i < letters.length(); i++) {
if(letters.charAt(i) == character)
return true;
}
return false;
}
public String toString (){
return "Filter(" + letters + ")";
}
}
Ok, let's brake that down.
public class Filter{
...
}
I guess you already got that part. This is where you describe your class structure.
String letters;
This is a class member variable. It is unique for every object that you create of that class. Again, for details, ask other students that understood it.
public Filter(String letters) {
this.letters = letters;
}
This is the constructor. When you create your object, this is the function that gets called.
In this case, all it does is to take an argument letters and stores it in the class-variable letters. Because they have the same name, you need to explicitely tell java that the left one is the class variable. You do this by adding this..
public boolean contains (char character){
for(int i = 0; i < letters.length(); i++) {
if(letters.charAt(i) == character)
return true;
}
return false;
}
This takes a character and looks whether it is contained in this.letters or not.
Because there is no name collision here, you can ommit the this..
If I understood right, the missing static here was one of your problems. If you have static, the function is class-bound and not object-bound, meaning you can call it without having an object. Again, it is important that you understand the difference, and if you don't, ask someone. (To be precise, ask the difference between class, object, static and non-static) It would take too long to explain that in detail here.
But in a nutshell, if the function is not static, it needs to be called on an object to work. Look further down in the other class for details how that looks like.
public String toString (){
return "Filter(" + letters + ")";
}
This function is also non-static. It is used whenever the object needs to be converted to a String, like in a System.out.println() call. Again, it is important here that you understand the difference between class and object.
The QueryResolver class
public class QueryResolver {
Filter filter;
String query;
public QueryResolver where(String queryStr) {
this.query = queryStr;
return this;
}
public QueryResolver matches(String filterStr) {
this.filter = new Filter(filterStr);
return this;
}
public int count() {
int result = 0;
for(int i = 0; i < query.length(); i++) {
if(filter.contains(query.charAt(i))){
result++;
}
}
return result;
}
}
Again, let's break that down.
public class QueryResolver {
...
}
Our class body.
Note that we don't have a constructor here. It is advisable to have one, but in this case it would be an empty function with no arguments that does nothing, so we can just leave it and the compiler will auto-generate it.
public QueryResolver where(String queryStr) {
this.query = queryStr;
return this;
}
This is an interesting function. It returns a this pointer. Therefore you can use the result of the function to do another call, allowing you to 'chain' multiple function calls together, like resolver.where(query).matches(filter).count().
To understand how that works requires you to understand both the class-object difference and what exactly the this pointer does.
The short version is that the this pointer is the pointer to the object that our function currently lives in.
public QueryResolver matches(String filterStr) {
this.filter = new Filter(filterStr);
return this;
}
This is almost the same as the where function.
The interesting part is the new Filter(...). This creates the previously discussed Filter-object from the class description and puts it in the QueryResolver object's this.filter variable.
public int count() {
int result = 0;
for(int i = 0; i < query.length(); i++) {
if(filter.contains(query.charAt(i))){
result++;
}
}
return result;
}
Iterates through the object's query variable and checks for every letter if it is contained in filter. It keeps count of how many times this happens and returns the count.
This function requires that filter and query are set. Therefore it is important that before someone calls count(), they previously call where(..) and matches(..).
In our case, all of that happens in one line, resolver.where(query).matches(filter).count().
The main function
I wrote two different main functions. You want to test your code as much as possible during development, therefore the first one I wrote was a fixed one, where you don't have to enter something manually, just click run and it works:
public static void main(String[] args) {
String filter="kjasd";
String query="kjg4t";
QueryResolver resolver = new QueryResolver();
int count = resolver.where(query).matches(filter).count();
System.out.println(count);
}
Once you understand the class-object difference, this should be straight forward.
But to repeat:
QueryResolver resolver = new QueryResolver();
This creates your QueryResolver object and stores it in the variable resolver.
int count = resolver.where(query).matches(filter).count();
Then, this line uses the resolver object to first call where, matches, and finally count. Again, this chaining only works because we return this in the where and matches functions.
Now finally the interactive version that you created:
public static void main(String[] args) {
Scanner in =new Scanner(System.in);
System.out.println("please enter the query string! ");
String query= in.next();
System.out.println("please enter the filter stirng!");
String filter= in.next();
System.out.println("the query string is : [" + query+ "]");
System.out.println("the filter string is : [" + filter+ "]");
QueryResolver resolver = new QueryResolver();
int count = resolver.where(query).matches(filter).count();
System.out.println("the characters of the String ["+filter+"] has been found in the forworded string ["+query+"] exactly "+count+" times!" );
in.close();
}

Questions about code error detected by eclipse and creating objects to run the methods

I am a newbie in Java. I am trying to create a Class named Database that will read a text file to create an array. There is no main method in this code since I have another java file that acts as a main application, and it has the main method.
Here is my code:
public class Database {
public static String[][] items = new String[20][3];
public static String[][] fillArray(String myFile)
{
TextFileInput in = new TextFileInput(myFile);
for(int i=0; i<20; i++)
{
String line =in.readLine();
StringTokenizer token = new StringTokenizer(line, ",");
for(int j=0; j<3; j++)
{
items[i][j] = token.nextToken();
}
}
in.close();
return items;
}
public static String getName(String code)
{
for(int i=0; i<20; i++)
{
if (code.equals(items[i][0]))
return items[i][1];
}
}
public static String getPrice(String code)
{
for(int i=0; i<20; i++)
{
if(code.equals(items[i][0]))
return items[i][2];
}
}
}
Question 1: Eclipse shows errors by indicating on both methods (getName and getPrice). It says: " This method must return a result of type String". Can anyone please explain and fix this?
Question 2: This Database class includes methods and an array created by reading in the text files. And I have another java file which is the main application file that includes the main method. The main application file is the file where I would like to create object and call the methods in. I understand the concept, but I keep getting errors when I try to create a database object and call the methods on this object. Can anyone please show me an example by using this class?

Answer 1: Your return statement in getName & getPrice method is inside if block which might be executed or not based upon the condition satisfied in if hence compiler will give error.
You need to have return statement before the method returns.
Answer 2: Since all the methods in your database class are static you don't need to create object, you can invoke it directly using classname e.g.Database.getName("code")

Eclipse shows errors by indicating on both methods (getName and
getPrice). It says: " This method must return a result of type
String". Can anyone please explain and fix this?
The method signature states that the return type is String:
public static String getName(String code)
So you must return it. For example:
public static String getName(String code) {
String result = null;
// Perform work and update value in result.
return result;
}
Can anyone please show me an example by using this class?
You need to instantiate the class before invoking any instance method but not for static methods. For example:
String result = Database.getName("code"); //pass some string as input argument and get result.
On a side note why have you declared all methods static? You need to understand why you really need static methods.

Java and Eclipse cannot know if code will ever be equal to items[i][0]. That's why, there might not be a return value for getName.
public static String getName(String code)
{
for(int i=0; i<20; i++)
{
if (code.equals(items[i][0]))
return items[i][1];
}
return "";// or null
}

With This method must return a result of type String the eclipse is saying your method return type is String so you must return a String in any case but your return statement is inside if() what if the condition inside if is never true then there is no return statement so add a default return statement to your method where you are return type String.
public static String getPrice(String code) {
for (int i = 0; i < 20; i++) {
if (code.equals(items[i][0])) return items[i][2];
}
return null;
}

In your method getName and getPrice you have put the if condition, and method only return when your condition in true that is why you have the compilation issue.
To fix that add the else condition in which you should return or add return after completion of if block.

Your method declarations are
public static String getName(){...}
public static String getPrice(){...}
let's analyse this, static keyword means it's a static method.
String means this method returns a object type of String.
However you return a String array object, not a String object.
Also return is inside a if statement and it the if condition not met, it won't return anything, but you must always return a object type of String.
therefore add a return after the if statement, incase if the code never enters your if block
if(YOUR CONDITION) {
//do stuff
//return stuff
}
//if never enters if block returns null
return null;
Your deletions must be
public static String[][] getName(){...}
public static String[][] getPrice(){...}

Get the Objects out of a list Java

Im doing a little text adventure. Im current working on a inventory. I tried to do my inventory with a list:
ArrayList<Object> inventar = new ArrayList<Object>();
I initialize Objects as Items:
Item holzschwert = new Item("Holzschwert", 1, 5);
inv.addToInventar(holzschwert);
Here's my add Method:
public void addToInventar(Object ...item){
inventar.add(item);
}
But everytime i want to print out my inventory (here's my toString() function) i still get the Hash code:
#Override
public String toString() {
return "Inventar [ Inventar = " + inventar + " ]";
}
and my print method
Object test = inv.getInventar();
String ausgabe = inv.toString();
System.out.println(ausgabe);
and the console log:
Inventar [ Inventar = [[Ljava.lang.Object;#281c35ec] ]
Thank you for your help :)

The mistake is
public void addToInventar(Object ...item){
inventar.add(item);
}
Here item is an Object[], because you used varargs (...). What you are seeing is the result of calling toString on an array.
You want
public void addToInventar(Object item){
inventar.add(item);
}
or possibly
public void addToInventar(Object... items){
for (Object item : items)
inventar.add(item);
}
EDIT
In addition to this answer, I should point out that you should probably not be using List<Object> anyway. List<Item> would be better as mistakes like this are more likely to be spotted by the compiler. The signature addToInventar(Object... items) should really be avoided like the plague, as it will accept literally any sequence of arguments of any length. All reference types are Objects and all primitives can be autoboxed.

Override the Item class toString() method.
class Item{
#override
public String toString(){
//Give your implementation
}
}

#Override
public String toString() {
StringBuffer stuff = new StringBuffer();
for (Object item : inventar) {
stuff.append(item.getName()); // you need a getter for the item name
stuff.append(" ");
}
return stuff.toString();
}
Both your inventory and items are Objects. If you just print the objects and not the fields you will only get their memory address as output:
Ljava.lang.Object;#281c35ec
not the properties you've set up for them.

If you do not care about the exact format of the output, you can do something like this in your Item class:
#Override
public String toString() {
return ToStringBuilder.reflectionToString(this);
}
ToStringBuilder is part of the Apache Commons Lang library. If you want it specifically formatted a certain way, then go with #Code Whisperer and write your own implementation.

Replacing a string in an array, where the part of the string is an int

I have code where I have a constructor for an object with mixed data types:
public GuidedTour(String tourID, String tourDescription, double tourFee,
String tourDate, int tourSize, String tourLeader)
I need to be able to allow the user to update the tourSize.
I can search and find particular tours, but am having trouble working out how to replace a substring with an inputted int value int newTourSize.
So the code I am trying to make the replacement is:
String tourDetails;
String newDetails = tourDetails.replace(tourSize, newTourSize);
It won't let me do it, as I am using int values. I have trying parsing the int into a String.
I am stuck on how to do this, beyond putting all the values back in, the original values with the new value, I am sure it would be better to find and replace that part of the substring and write this new string into the array.
I'm floundering a bit (but have improved much since my first java questions here!)
Any help?

It seems like your class looks like this:
public class GuidedTour {
private String tourDetails;
public GuidedTour(String tourID, int tourSize) { // other arguments omitted for brevity
this.tourDetails = tourID + tourSize;
}
public String getTourDetails() {
return tourDetails;
}
}
And that you're now wondering how to change the tourSize into the tourDetails.
Well, it's easy. Don't forget about the parts of the tour details. Replace your class by something like
public class GuidedTour {
private String tourID;
private int tourSize;
public GuidedTour(String tourID, int tourSize) { // other arguments omitted for brevity
this.tourID = tourID;
this.tourSize = tourSize;
}
public String getTourDetails() {
return tourID + tourSize;
}
public void setTourSize(int newTourSize) {
this.tourSize = newTourSize;
}
}
Now you don't need to do any String replacement anymore. You just set the new tour size, and ask the GuidedTour instance to recreate the tourDetails String by reassembling all the parts.

If you just want to convert a primitive typed value to a String, do that:
String.valueOf(tourSize)
The String class contains several static methods for creating a String object from various inputs.

An addition of empty string "" using String Concatenation Operator "+" to any primitive type will result in a string. It is a equivalent operation to String.valueOf(data). For example:
String strSize = tourSize+"";

polymorphic binary search tree

I have 2 classes, NonEmptyTree and EmptyTree that implement the Tree interface. toString() method in NonEmptyTree class should return a string: key=>value key=>value key>value etc...
I do not know how to remove last space at end of result.
EDIT: I cannot use any methods in the String class, and I cannot compare anything to null.
public String toString() {
String result = "";
result+=(this.left.toString());
result+=this.key;
result+="=>";
result+=this.value;
result+=" ";
result+=this.right.toString();
return result;
}
I've tried having a variable for the class that indicates if a NonEmptyTree instance is the largest in the current tree, but console displayed same string.
for example, the string would look like this:
"7=>seven 10=>ten 12=>twelve 15=>fifteen 16=>sixteen 17=>seventeen 20=>twenty 30=>thirty "
any help would be appreciated. thanks

Use String.trim():
return result.trim()
BTW. if you do a lot of String additions it's better to use StringBuilder and its method append() instead of adding multiple Strings with +.

public String toString() {
StringBuilder result = new StringBuilder();
result.append(this.left.toString());
result.append(this.key);
result.append("=>");
result.append(this.value);
result.append(" ");
result.append(this.right.toString());
return result.toString().trim();
}

Can you simply add an 'if'? See below:
public String toString()
{
String result = "";
result+=(this.left.toString());
result+=this.key; result+="=>";
result+=this.value;
if (this.right!=null) // Just add this, so it doesn't add an extra space if no right result exists
{
result+=" ";
result+=this.right.toString();
}
return result;
}