I am programing in java and have also very little programming experience.
I am trying to make a program there you first write in a number of integers in a scanner. In the next window you are supposed to write only one integer and that integer the program will search for and tell if it is or isn't in the "Scanner numbers"
My problem is that when i for example write 1 2 3 and in the next window write 2 it doesn't recognize there is a 2 in the scanner but if I instead write a 1 it works nicely.
Heres the code:
public class Inlämningsuppgift_kap9 {
public static void main(String[] args) {
String s1 = JOptionPane.showInputDialog("Write any number of integers!");
Scanner sc1 = new Scanner(s1);
String s2 = JOptionPane.showInputDialog(
"Chose a integer that the program will search for!"
);
int a = Integer.parseInt(s2);
while(sc1.hasNextInt()){
if(a == sc1.nextInt()){
JOptionPane.showMessageDialog(null, "The integer can be found");
System.exit(0);
}
else {
JOptionPane.showMessageDialog(null, "The integer cannot be found");
System.exit(0);
}
}
}
}
Thanks for any help!
Not going to do your homework for you, but a hint there: it is not a good idea to mix Swing UIs and a scanner that reads from stdin.
In other words: either use a graphical UI for all input/output; or just read/write from/to stdin (using out.println and that scanner code).
And then: assuming that the user first enters a string such as "1 2 3 4 5"; you need some further processing. You have to split that string (for example on spaces); and then turn each of that substrings ... into a real number. As your next step is to ask the user to input a number. And when you want to know if one number is in a list of other numbers, then you need to turn your initial string into that list of numbers!
So, you want to study javadoc for:
String.split()
Integer.parseInt()
you should not parse the input. makes it much easier to do what you want.
String s1 = JOptionPane.showInputDialog("Write any number of integers!");
String s2 = JOptionPane.showInputDialog("Chose a integer that the program will search for!");
for(int i=0;i<s1.length();i++){
if(s1.charAt(i)==s2.charAt(0)){
JOptionPane.showMessageDialog(null, "The integer can be found");
System.exit(0);
}
}
JOptionPane.showMessageDialog(null, "The integer cannot be found");
System.exit(0);
i like this solution more since it has less code an still works fine
lets just anylyze your code and hope i'm not mistaken :) , the variables are initialized,then You start a loop with
while(sc1.hasNextInt())
it basically runs for x times where x is the number of inputs (sc1 variable's length) then you test if the number in current iteration is the number you look for. If it's you print success code and close the loop. if it isn't you print failure code and also close the loop. So when you provide 1 as input the execution is fine because 1 is the 1st element but 2 as second argument is never tested because of
System.exit(0);
in else's brackets
Related
The code below is for a simple calculator with the four basic mathematical operators. It is a working program, it works as expected. However, I have a few questions to understand and improve both my program as well as my understanding of Java. (I have used google but the amount of redundant info confuses me and haven't found any perfect answers on StackOverflow too, though there are dozens of related questions. Believe me, I did tried before posting here).
How can I make sure that the user input is exactly and strictly one char?
here in my program, it accepts more than one character (+-*) but operates on the first char (+) only. I want to make sure more than one character is not accepted as input.
After successful execution of the program, how can I somehow let the user repeat the main method? I mean, a user adds two numbers, gets his answer and he wants to do another calculation, maybe multiply two numbers this time. I can ask the user for yes or no to continue but how do I take him/her back to the beginning? (will a loop work? how?)
A the end of the program I used two methods to output a message. The system.out.print works fine but the JOptionPane method doesn't display the message and the program doesn't terminate (I have commented it out). I would like to understand why?
Is the default case required in the switch? And Am I following the correct code structure? (the arrangements and uses of curly braces)
NB: As I said this calculator works fine and can be used by newbies like myself to better understand the concept as I have commented on every detail. Please understand that I couldn't add everything in the question title due to limits...
package mycalculator;
import javax.swing.JOptionPane;
import java.util.*;
public class MyCalculator{
public static void main (String [] args){
// Let us code a simple calculator//
// Variable type declarations//
char OP;
int firstNum;
int secNum;
// Display an explanation of what this program does//
System.out.println("This is a simple calculator that will do basic
calculations such as :"
+ "\nAddition, Multiplication, Substraction and Division.");
// Create a scanner object to Read user Input.//
Scanner input = new Scanner(System.in);
// Ask user to input any positive number and read it//
System.out.println("Enter Any positive number followed by pressing
ENTER.");
firstNum = input.nextInt();
// Ask user to input/decide his choice operator and read it //
System.out.println("Enter a valid OPERATOR sign followed by pressing
ENTER.");
OP = input.next().charAt(0);
// Loop the below statement till one of the four (+,-,*,/) is entered.//
while(OP != '+' && OP != '-' && OP != '*' && OP != '/'){
System.out.println("Please Re-enter a valid Operator (+,-*,/)");
OP = input.next().charAt(0);}
// Ask user for any second number and read it//
System.out.println("Enter your Second number followed by an ENTER
stroke.");
secNum = input.nextInt();
// Various possible Resolution based on OP value.//
int RSum = firstNum+secNum;
int RSubs= firstNum-secNum;
int RPro = firstNum*secNum;
double DPro = firstNum/secNum;
// Conditional statements for Processing Results based on OP and display.//
switch(OP){
case '+': System.out.println("The Resulting sum is "+ RSum);
break;
case '-': System.out.println("The Resulting sum is "+ RSubs);
break;
case '*': System.out.println("The Resulting Product is "+ RPro);
break;
case '/': System.out.println("The Resulting Divisional product is "+
DPro);
break;
//Maybe the default case is actually not needed but is added for totality//
default : System.out.println("Try Again");
break;
}
// The following code maybe deleted, it is for experimental purpose//
// Just checking if additional statements executes after a program
completes//
System.out.println("Test Message ");
// JOptionPane.showMessageDialog(null, "The Process Ends Here!");
//The test message works fine//
//The JOptionPane statement don't work and program doesn't end. WHY?//
}
}
How can I make sure that the user input is exactly and strictly one
char? here in my program, it accepts more than one character (+-*) but
operates on the first char (+) only. I want to make sure more than one
character is not accepted as input.
If you use console application and Scanner, only thing that you can do is read a String and check its length. In case you use Swing, you could implement KeyPressListener and proceed exactly after user press a button (but not for console application).
After successful execution of the program, how can I somehow let the
user repeat the main method? I mean, a user adds two numbers, gets his
answer and he wants to do another calculation, maybe multiply two
numbers this time. I can ask the user for yes or no to continue but
how do I take him/her back to the beginning? (will a loop work? how?)
You can't repeat main method. In Java main method is been executing only once. To repeate your code, you could wrap whole main method content to the infinite loop or move the content to the separate method and call it from the loop in the main method.
A the end of the program I used two methods to output a message. The
system.out.print works fine but the JOptionPane method doesn't display
the message and the program doesn't terminate (I have commented it
out). I would like to understand why?
JOptionPane works only for graphic application (Swing/AWT). This is not available in console. You have only standard input and output there.
Is the default case required in the switch? And Am I following the
correct code structure? (the arrangements and uses of curly braces)
No, default case is optional by JVM syntax. I remember, that e.g. in C++ there was reccomendation to place it (event empty), to exclude side effects of compilators. I do not know, is there such reccomendation in Java, but when I use switch, I prefer to always add it to exclude logical problem (but this is definetly optional according to syntax case). You use switch correctly.
public static void main(String[] args) {
System.out.println("This is a simple calculator that will do basic calculations such as :"
+ "\nAddition (+)"
+ "\nMultiplication (*)"
+ "\nSubtraction (-)"
+ "\nDivision (/)");
System.out.println();
try (Scanner scan = new Scanner(System.in)) {
while (true) {
System.out.println("Enter Any positive number followed by pressing ENTER.");
int first = scan.nextInt();
System.out.println("Enter a valid OPERATOR (+,*,-,/) sign followed by pressing ENTER.");
String operator = scan.next();
while (operator.length() != 1 || !"+*-/".contains(operator)) {
System.out.println("Please Re-enter a valid Operator (+,-*,/)");
operator = scan.next();
}
scan.nextLine();
System.out.println("Enter your Second number followed by an ENTER stroke.");
int second = scan.nextInt();
if ("+".equals(operator))
System.out.println("The Resulting sum is " + (first + second));
else if ("*".equals(operator))
System.out.println("The Resulting mul is " + (first * second));
else if ("-".equals(operator))
System.out.println("The Resulting sub is " + (first - second));
else if ("/".equals(operator))
System.out.println("The Resulting div is " + ((double)first / second));
System.out.println();
System.out.println("Do you want to exit ('y' to exit)?");
if ("y".equals(scan.next()))
return;
System.out.println();
}
}
}
1) you can check size of string input.next() .If it is one then continue else again prompt for operator choice .
2)I would suggest better create a different method and put all logic in it and call it the number of time you want or call infinite number of times.
4)Should switch statements always contain a default clause?
I have just started my 2nd programming course at college and our first assignment is rather simple, intended to basically check our environment and to check we know how to submit assignments through the course website.
When I run the code we have been supplied with, it hangs where it is supposed to prompt for the user to enter a number, so that it can print it. I inserted a series of println statements to determine where it was hanging.
It prints TEST1, TEST2 and TEST3, but never makes it to TEST4. So there must be something wrong with the line:
number = input.nextInt();
But I can't for the life of me see what's wrong with that line. Any help would be greatly appreciated! :)
Anyway, here is the code
package rossassignment1;
import java.util.Scanner; // use the Scanner class located in the "java.util" directory
public class RossAssignment1 {
public static void main (String[] args) {
System.out.println("TEST 1");
int number;
System.out.println("TEST 2");
Scanner input = new Scanner(System.in);
System.out.println("TEST 3");
number = input.nextInt();
System.out.println("TEST 4"); // display the number with other messages
System.out.print("This program reads an integer from a keyboard,\n"
+ " and print it out on the display screen.\n"
+ "The number is: " + number + ".\n"
+ "make sure that you get the exact same output as the expected one!\n");
}
}
After the line input.nextInt() is run the program is waiting for the user (you, or whatever marking system is being used) to input an integer in the console. After you do so the program will continue to your TEST4 line.
If after entering an integer, the codes after that still don't show up, you can do this:
number = input.nextInt();
input.nextLine(); //Add this
It will clear the newline.
Alternatively, I prefer to do this:
number = Integer.parseInt(input.nextLine());
I'm very new to programming, especially Java. I need to create a program that counts how many orders each entry at a restaurant gets ordered. The restaurant carries 3 entries, hamburgers, salad, and special.
I need to set up my program so that the user inputs, say, "hamburger 3", it would keep track of the number and add it up at the end. If the user inputs "quit", the program would quit.
System.out.println("Enter the type (special, salad, or hamburger) of entrée followed by the number, or quit to exit the program.");
I'm thinking about using a while loop, setting it so if the user input != to "quit", then it would run.
What's difficult for me is I don't know how to make my program take into account the two different parts of the user input, "hamburger 3" and sum up the number part at the end.
At the end, I want it to say something like "You sold X hamburgers, Y salads, and Z specials today."
Help would be appreciated.
You'll probably want three int variables to use as a running tally of the number of orders been made:
public class Restaurant {
private int specials = 0;
private int salads = 0;
private int hamburger = 0;
You could then use a do-while loop to request information from the user...
String input = null;
do {
//...
} while ("quite".equalsIgnoreCase(input));
Now, you need some way to ask the user for input. You can use a java.util.Scanner easily enough for this. See the Scanning tutorial
Scanner scanner = new Scanner(System.in);
//...
do {
System.out.println("Enter the type (special, salad, or hamburger) of entrée followed by the number, or quit to exit the program.");
input = scanner.nextLine();
Now you have the input from the user, you need to make some decisions. You need to know if they entered valid input (an entree and an amount) as well as if they entered an available option...
// Break the input apart at the spaces...
String[] parts = input.split(" ");
// We only care if there are two parts...
if (parts.length == 2) {
// Process the parts...
} else if (parts.length == 0 || !"quite".equalsIgnoreCase(parts[0])) {
System.out.println("Your selection is invalid");
}
Okay, so we can now determine if the user input meets or first requirement or not ([text][space][text]), now we need to determine if the values are actually valid...
First, lets check the quantity...
if (parts.length == 2) {
// We user another Scanner, as this can determine if the String
// is an `int` value (or at least starts with one)
Scanner test = new Scanner(parts[1]);
if (test.hasInt()) {
int quantity = test.nextInt();
// continue processing...
} else {
System.out.println(parts[1] + " is not a valid quantity");
}
Now we want to check if the actually entered a valid entree...
if (test.hasInt()) {
int quantity = test.nextInt();
// We could use a case statement here, but for simplicity...
if ("special".equalsIgnoreCase(parts[0])) {
specials += quantity;
} else if ("salad".equalsIgnoreCase(parts[0])) {
salads += quantity;
} else if ("hamburger".equalsIgnoreCase(parts[0])) {
hamburger += quantity;
} else {
System.out.println(parts[0] + " is not a valid entree");
}
Take a look at The if-then and if-then-else Statements and The while and do-while Statements for more details.
You may also find Learning the Java Language of some help. Also, keep a copy of the JavaDocs at hand, it will make it eaiser to find references to the classes within the API
These two methods should be what you're looking for.
For splitting: String.split(String regex)
http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#split(java.lang.String)
For parsing String into an Interger: Integer.parseInt(String s)
http://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#parseInt(java.lang.String)
You can split your strings using input.split(" "). This method gives you two strings - two parts of the main string. The character you splitted with (" ") won't be found in the string anymore.
To then get an integer out of your string, you can use the static method Integer.parseInt(inputPartWithCount).
I hope this helps!
I want to accept user input as either an integer or a string in Java but I always get an error no matter what I do.
My code is very simple (I'm a beginner):
System.out.println(
"Enter the row number (or enter e to quit the application):"
);
Scanner rowInput = new Scanner(System.in);
int row1 = rowInput.nextInt();
I want the user also to be able to press "e" to exit.
I have tried several things:
1) To convert row1 to a String and and say:
if((String)(row1).equals("e"){
System.out.println("You have quit") }
2) To convert "e" to an integer and say:
if(row1 == Integer.ParseInt("e"){
System.out.println("You have quit") }
3) To do the switch statement but it said I had incompatible types (String and an int).
My errors usually say: Exception in thread "main" java.util.InputMismatchException
Is there somebody who could help me?
Many thanks in advance!
You can try something like this
Scanner rowInput = new Scanner(System.in);
String inputStr = rowInput.nextLine();
try{
int row1 = Integer.parseInt(inputStr);
} catch (NumberFormatException e) //If exception occurred it means user has entered 'e'
{
if ("e".equals(inputStr)){
System.out.println("quiting application");
}
}
you should read the input from your scanner as String type and the try to convert that String input into integer using Integer.parseInt().
If its successful in parsing, it means user has input an Integer.
And If it fails, it will throw an exception NumberFormatException which will tell you that its not an Integer. So you can go ahead and check it for e if it is, do whatever you want as per your requirement.
You can't use nextInt() if you aren't sure that you will have a number coming in. Also, you can't parse e as an integer, because it is not an integer, it's either char or string (I'll suggest string in this case).
That means, your code should look like:
System.out.println("Enter the row number (or enter e to quit the application):");
Scanner rowInput = new Scanner(System.in);
string row1 = rowInput.next();
Then you have the data in a string. You can simply check if the string is e:
if (row1.Equals("e")) ...
It is also a good idea to check if the input is actually an integer then. Good way to check it is described here:
How to check if a String is numeric in Java
1) you should say Integer.parseInt() instead of Integer.ParseInt()
2) if you pass "e" to Integer.parseInt() you'll get java.lang.NumberFormatException
3) get your input as a string this way (cause it's safer)*:
Scanner rowInput = new Scanner(System.in);
String row = rowInput.next();
if (row.equals("e")) {
//do whatever
}
else if(row.matches("\\d+$")) {
//do whatever
}
*In your approach if user enters a non-integer input you'll encounter java.util.InputMismatchException
I have tried several things ....
Attempting to solve this problem by "trying things" is the wrong approach.
The correct approach is to understand what is going on, and then modify your solution to take this into account.
The problem you have is that you have a line of input that could be either a number or the special value e which means quit. (And in fact, it could be a couple of other things too ... but I will come to that.)
When you attempt to either:
call Scanner.nextInt() when the next input is not an integer, OR
call Integer.parseInt(...) on a String that is not an integer
you will get an exception. So what do you do?
One way is to change what you are doing so that you don't make those calls in those situations. For example:
call Scanner.hasInt() to see if the next token is an integer before calling nextInt(), or
test for the "e" case before you attempt to convert the String to an integer.
Another way to deal with this is to catch the exception. For example, you could do something like this:
String s = ...
try {
number = Integer.parseInt(s);
} catch (NumberFormatException ex) {
if (s.equals("e")) {
...
}
}
Either way will work, but I'm going to leave you to work out the details.
But the real point I'm trying to make is that the right way to solve this is to understand what is happening in your original version / versions, and based on your understanding, modify what you were doing. If you just randomly "try" alternatives that you found with Google, you won't progress to the point of being a productive programmer.
I said there were other cases. They are:
The user types something that is neither a number of your special e character. It could be anything. An empty line, hi mum ...
The user types the "end of file" character; e.g. CTRL-D on Linux or CTRL-Z on windows.
If you want your program to be robust you need to deal with these cases too.
Try this:
public class Demo{
public static void main(String[] args) {
Scanner s=new Scanner(System.in);
String choice = null;
System.out.println("Enter e to exit or some other key to stay:");
choice=s.next();
if(choice.equals("e"))
{
System.out.println("quits");
System.exit(0);
}
else
{
System.out.println("stays");
}
}
}
You can't convert a string into integer, use char ASCII codes instead.
This is an assignment i have to complete.
Can someone lead me in the right direction?
The program compiles but wont run correctly.
The error is InputMissmatch exception.
The error you are getting means that you are trying to use some kind of data as another one, in your case, you are probably trying to use a String as a float.
When using any of the next methods in the Scanner class you should first be sure that there's an appropiate input from the user.
In order to do so, you need to use the has methods.
Your problem is that you are not checking wether the input is a correct float or not before using your Scanner.nextFloat();
You should do something like this:
if (hope.hasNextFloat())
{
// Code to execute when you have a proper float,
// which you can retrieve with hope.nextFloat()
}
else
{
// Code to execute when the user input is not a float
// Here you should treat it properly, maybe asking for new input
}
That should be enough to point you in the right direction.
Also, check the Scanner api documentation for further details.
EDIT
Also, you are asking the user to input characters (or strings): "A", "B", etc..., but you are trying to compare them with a float. That's wrong, you should compare them with a string or character, like this:
if (hope.hasNextString())
{
if (hope.nextString().equals("A"))
{
// Code for option "A"
}
else if (hope.nextString().equals("B"))
{
// Code for option "B"
}
else ...
}
You could use a switch there, but it seems that you are not yet very fammiliar with java, so I'll leave it for another time.
Your problem is that you are entering a letter into a float field.
In your program you're asking the user to enter a float:
A = hope.nextFloat();
But if you enter the letter "A", you're going to get an exception because "A" is not a float, it's a string.
A simpler way to solve your problem is instead of having all the choices fields, you just read the input the user enters from the scanner like:
String choice = hope.next();
Next in the if statement, you check if the value from the string choice is equal to a specific letter, for example
if (choice.equals("A")) {
number4 = (number1 + number2 + number3);
System.out.printf("Your results are:" + (number4));
}
And you can do the same thing for the other choices you have.