Language: Java, IDE: eclipse mars
The program is supposed to prompt the user (using JOptionPane) for a positive value. I'm trying to catch the invalid entries. My while statement catches the negative numbers but not the strings. When a negative number is entered, the prompt is shown again, but when a string value is entered, the exception is caught and the program moves on (when it should re prompt the user).
Once a positive value has been entered, the program assigns it to a value in another class. (We're learning the MVC OOP design pattern).
Double.isNaN(Double.parseDouble(h)) ---> can anyone help me find what am I missing?
// prompt to get bandwidth from user
// check for validity
// if invalid, prompt again
try{
h = JOptionPane.showInputDialog("Enter bandwidth as a positive number");
// loop until parsed string is a valid double
while (Double.isNaN(Double.parseDouble(h)) || Double.parseDouble(h) <=0) {
h = JOptionPane.showInputDialog("Enter bandwidth as a positive number");
}
// h has been set to valid double, set it to bandwidth
model.setBandwidth(Double.parseDouble(h));
}catch(NumberFormatException|NullPointerException NFE){
System.err.println("Caught exception: " + NFE.getMessage());
}
This is because of how parseDouble() works.
Throws:
NumberFormatException - if the string does not contain a parsable double.
(See here)
So if the String is not a double parseDouble() will not return NaN but throw an exception, which means your catch clause will be called.
To solve this problem maybe use recursively algorithm which will call your method again if an exception is thrown.
As 4castle already stated, you need to move your try/catch block inside your while loop.
However, for validating user input you can basically stick to the following pattern:
public Foo getUserInput() {
Foo result;
do {
try {
String s = requestUserInput(); // something like Scanner.nextLine()
result = parseUserInput(s); // something like Double.parseDouble(String)
}
catch(Exception exc) {
// maybe you want to tell the user what's happened here, too
continue;
}
}
while(!isValid(result)); // something like (0 < result)
return result;
}
Related
I am loath to reinvent this bicycle and am hoping to be shown the tried and true way to handle this problem.
I'm collecting numeric values from users via some String interface (text input for instance).
I want to make sure that regardless of what type of memory space I'm using for collecting this info, I don't allow the user to enter a number that exceeds this value.
My intuition tells me that the only way to do this is to actually measure the string length of the max value... such as...
if( (userInput + "").length() > (Integer.MAX_VALUE + "").length()){
//user has entered too many digits for an Integer to hold.
}
But this looks ugly to me and I'm guessing there's a cleaner way to handle this.
When you get the userInput for the first time you should verify that what the user enters is valid and if it is, then Integer.parseInt() will work. If it's not valid, i.e. a value greater than Integer.MAX_VALUE, it will throw an exception.
The behavior you're describing leads to using the catch as flow control which is not a good design...
BAD:
try{
Integer.parseInt(max);
//do something with the integer
}catch (NumberFormatException e)
{
//user has entered too many digits for an Integer to hold.
userInput = Integer.MAX_VALUE + "";
}
The constructor of Integer would detect that for you, by throwing a NumberFormatException if the user input is either out of range or not really an integer. See the following test program example:
public class UserInputBigInteger {
public static void main(String[] args) {
String[] inputStrings = {
String.valueOf(Integer.MAX_VALUE)
, String.valueOf(Integer.MAX_VALUE)+"0" // x10
, String.valueOf(Integer.MAX_VALUE)+"a" // not an integer
};
for (String inputString : inputStrings) {
try {
Integer inputInteger = new Integer(inputString);
final int MAX = Integer.MAX_VALUE;
System.out.format("userInput %s is within range %,d%n"
, inputString, MAX);
} catch (NumberFormatException ex) {
System.out.format("userInput does not appear to be valid interger: %s%n"
, ex.getMessage()); }
}
}
}
The output would be:
userInput 2147483647 is within range 2,147,483,647
userInput does not appear to be an interger: For input string: "21474836470"
userInput does not appear to be an interger: For input string: "2147483647a"
You could also try this to get the bit information:
BigInteger userInputCheck=new BigInteger(userInput);
if(userInputCheck.bitLength()>31){
//error : cannot convert the input to primitive int type
}
EDIT
If you are using Apache Commons there is a method createNumber# in the NumberUtils utility class.
From the documentation:
...it starts trying to create successively larger types from the type specified
until one is found that can represent the value...
Source Code for the above method
First post so my apologies if this was done incorrectly (and am also relatively new to programming, so any extraneous tips are also appreciated).
So I have written up a basic calculator program in Java. It works well currently, but I am having a particular issue with NumberFormatException. Here's the code:
private static double spaceTestAndConvert(String numInput){
Scanner input= new Scanner(System.in);
if (numInput.equalsIgnoreCase("quit")){
System.exit(1);
}
else if(numInput.equalsIgnoreCase("C/E")){
Restart();
}
try{
return Double.parseDouble(numInput.trim());
}
catch(NumberFormatException nfe){
numInput = "";
System.out.println("Please enter only one number without any spaces or letters: ");
numInput = input.nextLine();
spaceTestAndConvert(numInput.trim());
return Double.parseDouble(numInput.trim());
}
}
The issue is that after trying to force an error by entering in several inputs which would cause NumberFormatException and then entering in a valid input, the program will crash from a NumberFormatException citing the previous invalid input.
I.E. -
"1 2 3"
loops back
"1 2 q 3"
loops back
"12q3 3 sqw 1"
loops back
"12"
crash - Exception in thread "main" java.lang.NumberFormatException: For input string: "12q3 3 sqw 1"
It only occurs after several occurrences of the exception. I'm curious why it is doing this. Any advice on how to fix this or explanation of what is happening? If you need any other part of the code, please let me know! Thanks!
I don't follow everything that you're saying, but these 2 lines (from within your catch block) look problematic...
spaceTestAndConvert(numInput.trim());
return Double.parseDouble(numInput.trim());
You are calling the spaceTestAndConvert function recursively, but throwing away the value. I don't understand why you would call it and not be interested in the value.
The second line is also a mess. You so carefully surround the first call to Double.parseDouble() with try/catch, but then you call it again within your catch block. If the second Double.parseDouble() generates a NumberFormatException, it will not be caught.
removing the return in catch will solve your problem. because if you have return on it, you are going to return an invalid Number format since you are in a catch. What you want to do is to return a value when it is now valid, you are now actually doing it inside the try. Don't force your program to return the value with error (since it is in a catch) because it will really give you an error.
returning to previous method after you had the right value (because of recursion) will still have the stack of error value aside from success value you gained from the end part because they are different variables.
private static double spaceTestAndConvert(String numInput){
Scanner input= new Scanner(System.in);
if (numInput.equalsIgnoreCase("quit")){
System.exit(1);
}
else if(numInput.equalsIgnoreCase("C/E")){
Restart();
}
try{
return Double.parseDouble(numInput.trim());
}
catch(NumberFormatException nfe){
numInput = "";
System.out.println("Please enter only one number without any spaces or letters: ");
numInput = input.nextLine();
spaceTestAndConvert(numInput.trim());
}
}
I'm looking for an exception on how to catch an invalid String that is user input. I have the following code for a exception on an integer input:
try {
price = Integer.parseInt(priceField.getText());
}
catch (NumberFormatException exception) {
System.out.println("price error");
priceField.setText("");
break;
But I'm not aware of a specific exception for strings, the input is a simple JTextBox so the only incorrect input I can think of is if the user enters nothing into the box, which is what I'm looking to catch.
if (textField.getText().isEmpty())
is all you need.
Or maybe
if (textField.getText().trim().isEmpty())
if you also want to test for blank inputs, containing only white spaces/tabs.
You generally don't use exceptions to test values. Testing if a string represents an integer is an exception to the rule, because there is no available isInt() method in String.
You can do like
if (priceField.getText().isEmpty())
throw new Exception("priceField is not entered.");
You could check if priceField contains a string by using this:
JTextField priceField;
int price;
try {
// Check whether priceField.getText()'s length equals 0
if(priceField.getText().getLength()==0) {
throw new Exception();
}
// If not, check if it is a number and if so set price
price = Integer.parseInt(priceField.getText());
} catch(Exception e) {
// Either priceField's value's length equals 0 or
// priceField's value is not a number
// Output error, reset priceField and break the code
System.err.println("Price error, is the field a number and not empty?");
priceField.setText("");
break;
}
When the if-statement is true (If the length of priceField.getText() is 0) an exception gets thrown, which will trigger the catch-block, give an error, reset priceField and break the code.
If the if-statement is false though (If the length of priceField.getText() is greater or lower than 0) it will check if priceField.getText() is a number and if so it sets price to that value. If it not a number, a NumberFormatException gets thrown, which will trigger the catch-block etc.
Let me know if it works.
Happy coding :) -Charlie
if you want your exception to be thrown during the normal operation of the Java Virtual Machine, then you can use this
if (priceField.getText().isEmpty())
throw new RunTimeException("priceField is not entered.");
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.
I have a method that a wrote. This method just scans for a user entered integer input. If the user enters a character value it will throw an input mismatch exception, which is handled in my Try-Catch statement. The problem is that, if the user inputs anything that is not a number, and then an exception is thrown, I need the method to loop back around to ask the user for input again. To my understanding, a Try catch statement automatically loops back to the Try block if an error is caught. Is this not correct? Please advise.
Here is my method (it's pretty simple):
public static int getMask() {
//Prompt user to enter integer mask
Scanner keyboard = new Scanner(System.in);
int output = 0;
try{
System.out.print("Enter the encryption mask: ");
output = keyboard.nextInt();
}
catch(Exception e){
System.out.println("Please enter a number, mask must be numeric");
}
return output;
}//end of getMask method
Here is how the method is implemented into my program:
//get integer mask from user input
int mask = getMask();
System.out.println("TEMP mask Value is: " + mask);
/***********************************/
Here is my updated code. It creates an infinate loop that I can't escape. I don't understand why I am struggling with this so much. Please help.
public static int getMask() {
//Prompt user to enter integer mask
Scanner keyboard = new Scanner(System.in);
int output = 0;
boolean validInput = true;
do{
try {
System.out.print("Enter the encryption mask: ");
output = keyboard.nextInt();
validInput = true;
}
catch(InputMismatchException e){
System.out.println("Please enter a number, mask must be numeric");
validInput = false;
}
}while(!(validInput));
return output;
/********************/FINAL_ANSWER
I was able to get it finally. I think I just need to study boolean logic more. Sometimes it makes my head spin. Implementing the loop with an integer test worked fine. My own user error I suppose. Here is my final code working correctly with better exception handling. Let me know in the comments if you have any criticisms.
//get integer mask from user input
int repeat = 1;
int mask = 0;
do{
try{
mask = getMask();
repeat = 1;
}
catch(InputMismatchException e){
repeat = 0;
}
}while(repeat==0);
To my understanding, a Try catch statement automatically loops back to the Try block if an error is caught. Is this not correct?
No this is not correct, and I'm curious as to how you arrived at that understanding.
You have a few options. For example (this will not work as-is but let's talk about error handling first, then read below):
// Code for illustrative purposes but see comments on nextInt() below; this
// is not a working example as-is.
int output = 0;
while (true) {
try{
System.out.print("Enter the encryption mask: ");
output = keyboard.nextInt();
break;
}
catch(Exception e){
System.out.println("Please enter a number, mask must be numeric");
}
}
Among others; your choice of option usually depends on your preferred tastes (e.g. fge's answer is the same idea but slightly different), but in most cases is a direct reflection of what you are trying to do: "Keep asking until the user enters a valid number."
Note also, like fge mentioned, you should generally catch the tightest exception possible that you are prepared to handle. nextInt() throws a few different exceptions but your interest is specifically in an InputMismatchException. You are not prepared to handle, e.g., an IllegalStateException -- not to mention that it will make debugging/testing difficult if unexpected exceptions are thrown but your program pretends they are simply related to invalid input (and thus never notifies you that a different problem occurred).
Now, that said, Scanner.nextInt() has another issue here, where the token is left on the stream if it cannot be parsed as an integer. This will leave you stuck in a loop if you don't take that token off the stream. To that end you actually want to use either next() or nextLine(), so that the token is always consumed no matter what; then you can parse with Integer.parseInt(), e.g.:
int output = 0;
while (true) {
try{
System.out.print("Enter the encryption mask: ");
String response = keyboard.next(); // or nextLine(), depending on requirements
output = Integer.parseInt(response);
break;
}
catch(NumberFormatException e){ // <- note specific exception type
System.out.println("Please enter a number, mask must be numeric");
}
}
Note that this still directly reflects what you want to do: "Keep asking until the user enters a valid number, but consume the input no matter what they enter."
To my understanding, a Try catch statement automatically loops back to the Try block if an error is caught. Is this not correct?
It is indeed not correct. A try block will be executed only once.
You can use this to "work around" it (although JasonC's answer is more solid -- go with that):
boolean validInput = false;
while (!validInput) {
try {
System.out.print("Enter the encryption mask: ");
output = keyboard.nextInt();
validInput = true;
}
catch(Exception e) {
keyboard.nextLine(); // swallow token!
System.out.println("Please enter a number, mask must be numeric");
}
}
return output;
Further note: you should NOT be catching Exception but a more specific exception class.
As stated in the comments, try-catch -blocks don't loop. Use a for or while if you want looping.