its only my second program with java and im running into some issues.
I'm trying to get input from a user, either yes or no, then based on that go to an if else statemene. Heres what I have so far
String answer= UI.askString("Do you want to continue?");
if(answer=="yes"){
UI.println("Lets go");
}
else if(answer == "no"){
UI.println("Thank you. Goodbye");
}
else{
UI.println("Please enter yes or no");
}
Im thinking perhaps its better to use booleans for this?
Any help is gladly appreciated!
(also if you're wondering, its a custom import hence the weird syntax in some lines)
Cheers.
When you compare two Strings in Java with the == operator, they are compared to see if they are the same object, rather than whether they contain the same text. So, you could type "yes", and when you use if (answer == "yes") the comparison fails, because the object you got back from UI.askString is a different object, stored at a different place in memory, than the String the compiler generated from the literal "yes".
To compare the value of the two Strings you need to write answer.equals("yes"), or "yes".equals(answer). Either one will work, and will call the equals method of the String class, which will compare the actual text.
The latter syntax, "yes".equals(answer), is often recommended because it will not cause a NullPointerException, even if the variable answer is set to null. This is because the equals method handles null and simply returns false. If, on the other hand, you used the answer.equals("yes") form, and answer was null, you would be trying to invoke a method on null and an exception would be thrown.
what you are looking for is a dialog box. Here is oracle examples, with code. It is more than I can write here. There are ton of yes, no boxes and detection's of user input with them.
http://docs.oracle.com/javase/tutorial/uiswing/components/dialog.html
Quick answer:
int dialogResult = JOptionPane.showConfirmDialog (null, "Would You Like to Save your Previous Note First?","Warning",dialogButton);
if(dialogResult == JOptionPane.YES_OPTION){ ... }
Other choices ...
YES_OPTION, NO_OPTION, CANCEL_OPTION, OK_OPTION, and CLOSED_OPTION
For a command line program you need...
import java.util.Scanner;
The code will look like ...
Scanner in = new Scanner(System.in);
String line = in.nextLine();
//ask them to write yes, no, whatever
if(line.equal("yes"){ }
else if (line.eqals("no") {}
else {}
using MikeG010590's answer, you can try:
Scanner in = new Scanner(System.in);
String line;
System.out.println("you want to continue?");
Boolean exit = null;
do {
line = in.nextLine();
switch (line) {
case "yes":
exit = false;
break;
case "no":
exit = true;
break;
default:
System.out.println("Please enter yes or no");
break;
}
}
while (exit == null);
System.out.println(exit ? "Thank you. Goodbye" : "Lets go");
Related
The idea is that the while loop should loop through the code if the result is "wrong password",
until the correct password is entered and breaks the loop when it matches the login method return value.
while (true){
System.out.println("\nLogin: ");
System.out.println("\nEnter a username: ");
String loginUsername = sc.nextLine();
System.out.println("Enter a password: ");
String loginPassword = sc.nextLine();
System.out.println(login(loginUsername,loginPassword));
if (login(loginUsername,loginPassword).equalsIgnoreCase("Successfully logged in!")) {
break;
}
}
this code is the return statements from the login method
if (check == true){
return ("\nSuccessfully logged in!");
} else {
return ("\nWrong username/password");
}
But "Successfully logged in!" is not the same string as "\nSuccessfully logged in!".
More importantly... Why use strings for this at all? If you want to know whether something is true or false, there's a perfectly good data type to convey that information. Return that type instead:
return check;
Rename the method to something more meaningful than login, and use its result in a semantically clear condition:
if (isLoginSuccessful(loginUsername,loginPassword)) {
break;
}
This puts the semantics of what you're doing in the code itself, rather than in magic strings that you need to copy/paste everywhere and manually keep track of. Which, already in this one tiny example, you've lost track of by making the strings different. (With the added benefit that using booleans for conditional logic probably performs a little better than string comparison.)
I'm trying to learn java but I'm stuck trying to do a single program which concerns Do While Statement with two conditions. Specifically, I want a method to run until the user write "yes" or "no". Well, down there is my thing, what is wrong with it?
String answerString;
Scanner user_input = new Scanner(System.in);
System.out.println("Do you want a cookie? ");
do{
answerString = user_input.next();
if(answerString.equalsIgnoreCase("yes")){
System.out.println("You want a cookie.");
}else if(answerString.equalsIgnoreCase("no")){
System.out.println("You don't want a cookie.");
}else{
System.out.println("Answer by saying 'yes' or 'no'");
}while(user_input == 'yes' || user_input == 'no');
}
}}
I'd do something similar to Tim's answer. But to do things the way you were trying to do them, you have a lot of problems that need to be fixed:
(1) String literals in Java are surrounded by double quote marks, not single quote marks.
(2) user_input is a Scanner. You can't compare a scanner to a string. You can only compare a String to another String. So you should be using answerString in your comparison, not user_input.
(3) Never use == to compare strings. StackOverflow has 953,235 Java questions, and approximately 826,102 of those involve someone trying to use == to compare strings. (OK, that's a slight exaggeration.) Use the equals method: string1.equals(string2).
(4) When you write a do-while loop, the syntax is do, followed by {, followed by the code in the loop, followed by }, followed by while(condition);. It looks like you put the last } in the wrong place. The } just before the while belongs to the else, so that doesn't count; you need another } before while, not after it.
(5) I think you were trying to write a loop that keeps going if the input isn't yes or no. Instead, you did the opposite: you wrote a loop that keeps going as long as the input is yes or no. Your while condition should look something like
while (!(answerString.equals("yes") || answerString.equals("no")));
[Actually, it should be equalsIgnoreCase to be consistent with the rest of the code.] ! means "not" here, and note that I had to put the whole expression in parentheses after the !, otherwise the ! would have applied only to the first part of the expression. If you're trying to write a loop that does "Loop until blah-blah-blah", you have to write it as "Loop while ! (blah-blah-blah)".
I might opt for a do loop which will continue to take in command line user input until he enters a "yes" or "no" answer, at which point the loop breaks.
do {
answerString = user_input.next();
if ("yes".equalsIgnoreCase(answerString)) {
System.out.println("You want a cookie.");
break;
} else if ("no".equalsIgnoreCase(answerString)) {
System.out.println("You don't want a cookie.");
break;
} else {
System.out.println("Answer by saying 'yes' or 'no'");
}
} while(true);
I am doing a coding project for a basic non recursive, non GUI form of Minesweeper. One of the requirements is that input commands must be strictly formatted like so:
For marking commands (reveal, guess, mark):
["reveal"/"r"] [int] [int]
For help and quit:
just ["help"/"h"] or ["quit"/"q"]
Any inputs outside of these restrictions must be considered ill-formatted. My code for reveal looks something like this:
case "reveal":
case "r":
roundsCompleted ++;
if(input.hasNextInt()){
par1 = input.nextInt();
}
else{
commandIn = null;
}
if(input.hasNextInt()){
par2 = input.nextInt();
correctInput = true;
}
else{
commandIn = null;
}
if(correctInput && isInBounds(par1, par2)){
reveal(par1, par2);
where this is all inside a switch statement of course. The commandIn = null statements are designed to throw the default case which prints "command not recognized". I realize part of my issue here is that these are in two separate if-else statements. Another problem is that input.hasNextInt() doesn't seem to evaluating to false when there is not an int after the first one.
The essence of this problem is in completely restricting these commands to the formats I listed above. Can anyone give me some insight into this issue? Thanks.
I'd use regex to first see if something is either a good input or not just cause it'd be easier
String input = "r 3 4";
if (input.matches("^(help|h|quit|q|((r|reveal) \\d \\d))$"))
//switch case
System.out.println("match");
else
//null string
System.out.println("no match");
then after you've got a match you can use your switch case like what you're doing, if it's a "reveal" or "r", I would just use split() and turn it into an array to get the different x and y coordinates
The code doesn't exit after I type "stop" - for some reason. Why?
Step-by-step debugging shows that after I enter "stop" it's value consists of exactly 's','t','o','p' without any line breaks, etc. - however, the code still goesn't exit. Could anyone tell why, please?
import java.util.Scanner;
public class Application {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
// asking username
System.out.print("Username: ");
String username = input.nextLine();
String inpText;
do {
System.out.print(username + "$: ");
inpText = input.nextLine();
System.out.print("\n");
// analyzing
switch (inpText) {
case "start":
System.out.println("> Machine started!");
break;
case "stop":
System.out.println("> Stopped!");
break;
default:
System.out.println("> Command not recognized");
}
} while (inpText != "stop");
System.out.println("Bye!..");
}
}
To compare Strings use .equals() and not ==, unless you really know what you are doing.
inpText != "stop" //Not recommended
!"stop".equals(inpText) //recommended
You cannot use a String in a switch unless you are using jdk 1.7+.
Cannot switch on a value of type String for source level below 1.7.
Only convertible int values or enum variables are permitted
You are comparing pointers not strings with this piece of code:
while (inpText != "stop");
Should be something like this:
while (!"stop".equals(inpText));
change while (inpText != "stop"); to while (!(inpText.equals("stop")));
If your JDK is 1.6 or lower you can't switch() on a String
P.S.
switching on a String is probably not the best solution Yeah in java 1.6 you can only switch int, boolean, double, long, and float I believe.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How do I compare strings in Java?
So, I have a question regarding the testing operators and strings. I'm trying to write a program that will take in user input of either "boy" or "girl". When the user inputs "boy", the output should be "You are a boy." When the user inputs "girl", the output should be "You are a girl."
However, when I compile and run the program, no matter what I input, the output is always "You are a girl."
Is this because the strings have no actual value like integers and therefore, testing operators cannot be used to compare them?
Also, is there anything like the assert function from python in java?
import java.util.Scanner;
class apples{
public static void main(String args[]){
System.out.println("Are you a boy or a girl?");
Scanner lalala = new Scanner(System.in);
String text = lalala.nextLine();
if (text == "boy"){
System.out.println("You are a boy.");
}
else{
System.out.println("You are a girl.");
}
}
}
Thanks a lot~
Use
text.equals("boy")
instead of
if (text == "boy")
The reason is, In Java, == always just compares two references (for non-primitives, that is) - i.e. it tests whether the two operands refer to the same object.
However, the equals method can be overridden - so two distinct objects can still be equal.
But the better option is to use equalsIgnoreCase, the reason is that user may enter boy or BOY or Boy or any other combination. equalsIgnoreCase method just ignores the case and compares two strings.
text.equalsIgnoreCase("boy")
Side note. Input other than Boy will get you to print Girl. Put one more extra if condition. Also before going for comparision, being on safer side trim the string for blank spaces.
Use
text.equals("boy")
or
text.equalIgnoreCase("boy")
if its not case sensitive.
instead of
if (text == "boy")
Your coade will be something like this
import java.util.Scanner;
class apples{
public static void main(String args[]){
System.out.println("Are you a boy or a girl?");
Scanner lalala = new Scanner(System.in);
String text = lalala.nextLine();
text = text.trim();
if (text.equalIgnoreCase("Boy")){
System.out.println("You are a boy.");
}
else if(text.equalIgnoreCase("Girl")){
System.out.println("You are a girl.");
} else {
System.out.println("Invalid Gender");
}
}
}
use text.equalsIgnoreCase("boy")
if you think about case also it will check content of string is same.
== is used to check for object equality.
text.equals("boy") is ok if you not consider about case.
Please use equals method for string like : if (text.equals("boy")){
== is not for string contents equality check in java
== tests if object identity is the same. When you have two string objects containing the same value this object identity won't be equal. Use the equals function to test for logical equivalence.
Your if else is wrong. Use only if here. Otherwise,
If I entered "some text", the output will be "You are a girl"
Like this
if (text.equalsIgnoreCase("boy")) {
System.out.println("You are a boy.");
}
else if (text.equalsIgnoreCase("girl")) {
System.out.println("You are a girl.");
} else {
System.out.println("You are neither a boy nor a girl");
}
It would be best to use the .equals method instead of == boy
For your other question regarding assert:
Java has an assert function. This is the sample code.
assert(x <= 50): "Not Accepted";
This code checks if integer x is greater than 50. You will see the output at the console.