I am very new to programming and am quite young.
I have no friends/family who can help me so I am seeking help on the internet. There is problem with my code as it isn't working as I intend it.
Instead of printing out what the variable "TheChoice", it just ends.
This isn't all the code and I have consised it so that it will be easier to read and maybe more people will be able to quickly answer.
However, this is definately the part of my code which I have messed up).
public String Choices(String value1, String value2)
{
Scanner x = new Scanner(System.in);
if(x.next() == value1){return value1;}
if(x.next() == value2){return value2;}
}
// Separate class...
ChoiceClass Object1 = new ChoiceClass();
String TheChoice = Object1.Choices("Ham", "Cheese");
System.out.println(TheChoice);
There's a number of issues with your code.
Firstly, you compare Strings with == instead of equals (there's
a ton of literature about String comparison and Object equality
in Java, I suggest you take a look here and here.
Secondly, you don't always return a value in your choices method. Your method must return a String (even in its default value, as null), but you're not considering user inputs other than given arguments.
Also your Scanner next wouldn't work as you're calling it twice,
when you only want to call it once.
Finally, you should take a look at Java naming conventions: method
names are camelBack. See here for code conventions.
Here's a snippet that'll probably help you out:
public static String choices(String value1, String value2) {
Scanner x = new Scanner(System.in);
System.out.println("Type your choice and ENTER...");
String input = x.nextLine();
if (input.equals(value1)) {
return value1;
}
else if (input.equals(value2)) {
return value2;
}
// handling other user inputs
else {
return "nothing";
}
}
public static void main(String[] args) {
// usage of method. Note that you could also refactor with varargs, as in:
// String... values in method signature.
// This way you could iterate over values and check an arbitrary number of values,
// instead of only 2.
System.out.println(choices("foo", "bar"));
}
Output:
Type your choice and ENTER...
... then either "foo" or "bar" or "nothing" according to input.
You can use something similar to this:-
public String Choices(String value1, String value2)
{
Scanner x = new Scanner(System.in);
String option=x.next();
if(option!=null)
{
if(option.equals(value1)) return value1;
else if (option.equals(value2)) return value2;
}
else return option;
}
If you want to compare Strings just use equals no need to use Scanner here.
Related
I'm trying to create a simple method. Basically, I want this method (called "returnIndex") to return the word at the ArrayList index number the user types in.
Example:
If the user types in "1", is should return whatever String is at index 1 in the ArrayList.
This is what I have so far:
public void returnIndex ()
{
Scanner in = new Scanner (System.in)
while (in.hasNextLine())
{
if (in.equals(1))
{
//return item at that index
}
}
}
I'm just not sure how to say "return the item at that index" in Java. Of course, I'll have to make the code work with any other number, not just '1'. But for now, I'm focusing on '1'. Not even sure if the in.equals(1) part is even 100% right.
My apologies if this question seems a little elementary. I'm still working on my Java. Just hints please, no complete answers. Thank you very much.
public String returnIndex(Scanner in, List<String> list) {
return list.get(in.nextInt());
}
Don't create new Scanners as it can cause subtle problems. Instead, create only one and keep using it. That means you should pass it into this function.
There's no need to use ArrayList when List will do (as it will here).
You need to make the function return String, not void, if you want it to return a String.
public static void main(String[] args) {
List<String> values = new ArrayList<String>();
values.add("One");
values.add("Two");
values.add("Three");
String result = getStringAtIndex(values);
System.out.println("The result:" + result);
}
public static String getStringAtIndex(List<String> list) {
Scanner scanner = new Scanner(System.in);
int index = 0;
index = scanner.nextInt();
return list.get(index-1);
}
I created a JAVA code, and I don't have any errors, but when I run the code, the output does this:
Enter a word: Thank you for entering a word! And it does not let me enter anything, when I intend for the code to let me enter a word, then it checks if it is a word, and gives the answer if it is a word, or none if it isn't. (It is my first time asking on this site) Here's the code:
package files;
import java.util.Scanner;
public class Testprinter {
static boolean myBoolean = false;
static Scanner userInput = new Scanner(System.in);
public static void main(String[] args){
String usersInput;
while(myBoolean != true)
{
System.out.print("Enter a word: ");
usersInput = userInput.toString();
myBoolean = checkInput(usersInput);
}
checkifComplete();
}
public static boolean checkInput(String usersInput){
if(usersInput == (String)usersInput)
{
return true;
} else { return false; }
}
public static void checkifComplete(){
if(myBoolean = true){
System.out.print("Thank you for entering a word!");
}
}
}
This line is wrong:
if (usersInput == (String)usersInput)
It should be:
if (usersInput.equals(usersInput))
In Java, strings (and in general: all objects, that is all types that are non-primitive) must me compared using the equals() method, which tests for equality. The == operator is fine for testing equality between primitive types, but for objects it tests for identity - a different concept, and 99% of the time, not what you want.
And besides, you're comparing a string with itself! it'll always return true, I'm quite sure that's not what you want to do… notice that the parameter must have a different name, currently it's called just like the attribute. Perhaps this is what you meant?
public static boolean checkInput(String input) {
return usersInput.equals(input);
}
You forgot scanner.nextLine(); thats reason its not asking you enter anything.
Instead of usersInput = userInput.toString();
Use:
String usersInputStr = scanner.nextLine();
Follow this link - for how to use scanner: How can I read input from the console using the Scanner class in Java?
Your issue is using userinput.toString(), when you should be using usersInput = userInput.next();. You are currently retrieving the string representation of the scanner, not getting a word.
Corrected main:
public static void main(String[] args){
String usersInput;
while(myBoolean != true)
{
System.out.print("Enter a word: ");
usersInput = userInput.next();
myBoolean = checkInput(usersInput);
}
checkifComplete();
}
I am confused about returning the needed value. Here is a part of my code:
public class StrNum {
public static int getInt(String input) {
String str = new String(input);
int result;
if (str.startsWith("b")) {
str = str.substring(1);
result = Integer.parseInt(str, 2);
}
else if (str.startsWith("x")) {
str = str.substring(1);
result = Integer.parseInt(str, 16);
}
Now, what I need to return is result. When I write return result;, it asks me to initialize the variable (and I am aware that it hasn't been initialized). When I use return result inside of the if statements, Eclipse tells me that I have to return a value.
Where am I being stupid here? I would appreciate a good explanation.
What will the method getInt() return if Str doesn't start with neither "b" or "x"? That would be an error because result is not being initialized. You could solve this by intializing result with a value that you would like to return in that case:
int result = -1; // for example
Edit:
Since you want to use input to determine if the number will be parsed as binary or hexadecimal, I would recommend you to add an else statement to parse the number in base 10 as default:
if (...)
// ...
else if (...)
// ...
else
result = Integer.parseInt(Str);
Note:
Try to follow Java naming conventions. Use names like someVar for variables/methods and use names like SomeClass for classes.
It's not necessary to create a new string instance Str, unless you are going to use the original input later in the same method.
To create a String with the same content you can simply do
String str = input;
Try
int result = 0;
You must initialize variables before returning them or in other words do result = . The variable must always be initialized no matter what code path your application takes.
Your can either return the result from if and also from else, provided you are not doing additional calculations after the else block.
or simply initialize result = 0, it will change anyways before you return.
I am simply trying to see if the inputted value matches a value that is already in the array and if it does return "Valid". I realize this is very simple but I cannot get this to work:
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
String[] accountNums = { "5658845", "8080152", "1005231", "4520125", "4562555",
"6545231", "7895122", "5552012", "3852085", "8777541",
"5050552", "7576651", "8451277", "7881200", "1302850",
"1250255", "4581002" };
String newAccount;
String test = "Invalid";
newAccount = keyboard.next();
for (int i = 0; i < accountNums.length; i++)
{
if(newAccount == accountNums[i])
{
test = "Valid";
}
}
System.out.println(test);
}
}
thank you for any assistance (and patience)
Use equals method. Check here why to.
if (newAccount.equals(accountNums[i]))
Jayamohan's answer is correct and working but I suggest working with integers rather than Strings. It is a more effective approach as CPUs handle numbers (integers) with a lot more ease than they handle Strings.
What has to be done in this case is change newAccount and accountNums to ints instead of Strings and also remove all the quotation marks from the accountNums initialization. Instead of calling keyboard.next() you can call keyboard.nextInt(), which returns an integer. The if-statement is fine as it is.
Why are you using an array?
List<String> accountNums = Arrays.asList( "5658845", "8080152", "1005231", "4520125", "4562555",
"6545231", "7895122", "5552012", "3852085", "8777541",
"5050552", "7576651", "8451277", "7881200", "1302850",
"1250255", "4581002" );
String test = "Invalid";
Then you just need this (no loop):
if (accountNums.contains(newAccount)) {
test = "Valid";
}
Plus, it's easier to read and understand.
You cannot compare strings with ==
You must use .equals()
There are some situation that I need to convert string to float or some other numerical data-type but there is a probability of getting some nonconvertible values such as "-" or "/" and I can't verify all the values beforehand to remove them.
and I want to avoid using try/catch for this matter , is there any other way of doing a proper conversion in java? something similar to C# TryParse?
The simplest thing I can think of is java.util.Scanner . However this approach requires a new Scanner instance for each String.
String data = ...;
Scanner n = new Scanner(data);
if(n.hasNextInt()){//check if the next chars are integer
int i = n.nextInt();
}else{
}
Next you could write a regex pattern that you use to check the String (complex to fail too big values) and then call Integer.parseInt() after checking the string against it.
Pattern p = Pattern.compile("insert regex to test string here");
String data = ...;
Matcher m = p.matcher(data);
//warning depending on regex used this may
//only check part of the string
if(m.matches()){
int i = Integer.parseInt(data);
}
However both of these parse the string twice, once to test the string and a second time to get the value. Depending on how often you get invalid strings catching an exception may be faster.
Unfortunately, there is no such method in Java. There is no out parameter in Java, so writing such a method would need to return a null Float to signal an error, or to pass a FloatHolder object which could be modified by the method:
public class FloatHolder {
private float value;
public void setValue(float value) {
this.value = value;
}
public float getValue() {
return this.value;
}
}
public static boolean tryParseFloat(String s, FloatHolder holder) {
try {
float value = Float.parseFloat(s);
holder.setValue(value);
}
catch (NumberFormatException e) {
return false;
}
}
This is an old question, but since all the answers fail to mention this (and I wasn't aware of it myself until seeing it in a merge request written by a colleague), I want to point potential readers to the Guava Floats and Ints classes:
With the help of these classes, you can write code like this:
Integer i = Ints.tryParse("10");
Integer j = Ints.tryParse("invalid");
Float f = Floats.tryParse("10.1");
Float g = Floats.tryParse("invalid.value");
The result will be null if the value is an invalid int or float, and you can then handle it in any way you like. (Be careful to not just cast it to an int/float, since this will trigger a NullPointerException if the value is an invalid integer/floating point value.)
Note that these methods are marked as "beta", but they are quite useful anyway and we use them in production.
For reference, here are the Javadocs for these classes:
https://google.github.io/guava/releases/snapshot-jre/api/docs/com/google/common/primitives/Ints.html
https://google.github.io/guava/releases/snapshot-jre/api/docs/com/google/common/primitives/Floats.html
Java does not provide some built in tryParse type of methods, on of the solutions you can try is to create your own tryParse Method and put try/catch code in this method and then you can easily use this method across your application very easily and without using try/catch at all the places you use the method.
One of the sample functions can have following code
public static Long parseLong(String value) {
if(isNullOrEmpty(value)) {
return null;
}
try {
return Long.valueOf(value);
}
catch (NumberFormatException e) {
}
return null;
}
Regular expressions helped me solve this issue. Here is how:
Get the string input.
Use the expression that matches one or more digits.
Parse if it is a match.
String s = "1111";
int i = s.matches("^[0-9]+$") ? Integer.parseInt(s) : -1;
if(i != -1)
System.out.println("Integer");
else
System.out.println("Not an integer");