For my project I need to input a first and last name with no numbers but I simply can't find anything online. If you could help, that would be terrific.
Also if you have the time, I need to flip the first and last name with a comma when the user inputs them.
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
public class PhoneListr {
public static void main (String[] args) throws IOException {
String firstName;
String lastName;
System.out.println ("Please enter your first name:");
firstName = PhoneList ("");
System.out.println ("Please enter your last name:");
lastName = PhoneList ("");
System.out.println ("Your full name is: " + lastName + "," + firstName);
}
public static String PhoneList (String input) throws IOException {
boolean continueInput = false;
while (continueInput == false) {
BufferedReader bufferedReader = new BufferedReader (new InputStreamReader (System.in));
input = bufferedReader.readLine();
continueInput = true;
if(input.matches("[a-zA-Z]+")) {
continueInput = true;
}
else {
System.out.println ("Error, you may only use the alphabet");
continueInput = false;
}
}
return input;
}
}
use String.matches(regex):
if(input.matches("[a-zA-Z]+")) {
System.out.println("your input contains no numerics");
}
else {
System.out.println("only alphabets allowed");
}
the above regex checks a through z, or A through Z, inclusive (range).
For this type of string matching you need to use Regular Expressions, or "RegEx"es. It is a very big topic to cover but here's an introduction. RegExes are tools used to test whether strings match certain criteria, and/or to pull certain patterns out of that string or replace certain parts that match those patterns with something else.
Here is an example using a RegEx to test whether your input contains a digit:
if(input.matches("\\d")) {
// matched a digit, do something
}
Based on OP problem and clarification here are few suggestions.
Chaitanya solution handles the check for only alphabets perfectly.
About the neglected areas of problem :
i would advice you to make two variable firstName and lastName inside the main()
String firstName;
String lastName;
Change retun type of method phoneList() to String
return the entered name input insted of number inside the method phoneList ( dont actually see why you are returning number) and store it inside the firstName and lastName
System.out.println ("Please enter your first name:");
firstName = PhoneList (0);
System.out.println ("Please input your last name:");
lastNamr =PhoneList (0);
now to print it in the "comma format" use
System.out.println("full name is: " +lastName+ "," +firstName);
As i read your program again , its a mess!!
About the method phoneList()
Use regex condition to set continueInput to true/flase and not exploit execptions.
P.s. I would appreciate "editing" to post , if any fellow member find any mistakes above, using a mobile, not sure about formatting etc. Thanks. :-) (y)
You can also use
if(input.matches("[^0-9]"))
{
System.out.println("Don't input numbers!");
continueInput = false;}
Can't understand what 2nd question asks. For answer what I get from your 2nd question is like this.
In main function change the code like this
String first_name = null;
String last_name = null;
System.out.println ("Please enter your first name:");
first_name = PhoneList();
System.out.println ("Please input your last name:");
second_name = PhoneList();
System.out.println (second_name+","+first_name);
then in PhoneList function last line should be changed to
return input;
Please check for a link! for more info
You can add a check by passing the argument to method StringUtils.isNumeric
This returns true if the String entered is numeric
Related
This question already has answers here:
Loop user input until conditions met
(3 answers)
Closed 7 years ago.
I'm currently working my way through a Udemy Java course and am practicing what i have learnt thus far.
I have the following simple program which i am planning on using to get the user to input his name.
import java.util.Scanner;
public class Adventure {
public static final int menuStars = 65;
private static Scanner input = new Scanner(System.in);
public static void main(String[] args) {
String firstName = "";
String lastName = "";
boolean validName = false;
while(!validName){
//Entering first name
System.out.println("Please enter your first name.");
try {
firstName = input.nextLine();
if(firstName.length() == 0){
throw new Exception("Please enter a first name of at least 1 character.");
}else{
//Entering last name
System.out.println("Please enter your last name.");
lastName = input.nextLine();
if(lastName.length() == 0){
throw new Exception("Please enter a last name of at least 1 character");
}else{
System.out.println("You have entered " + firstName +" " + lastName);
}
}
} catch (Exception e) {
System.out.println(e.getMessage());
continue;
}
//Used to terminate loop when both first & last names are valid
validName = true;
}
}
}
I want to make the program repeat the error message when the user inputs a blank name instead of restarting the entire program from the beginning.
E.g When the user enters a blank first name, i want the program to keep repeating "Please enter a first name of at least 1 character" and when the user enters a blank last name, for it to keep repeating "Please enter a last name of at least 1 character" until the user enters a valid name.
However, currently when the user enters a blank first name or last name, my program will repeat itself from the very beginning instead of repeating just the error message.
How would i go about making the program repeat just the error message?
Use a boolean variable that stores true when "Please enter your first name." is printed. Check before printing this string each time if this variable is false or not. Also, initialize it to false before the loop. Same idea goes for last name.
if(!printed)
{
System.out.println("Please enter your first name.");
printed=true;
}
havent tested that but i am guessing it can be like that, with out try/catch though, it just makes no sense to me using it in the way you have it on your code
String firstName = "";
String lastName = "";
System.out.println("Please enter your first name.");
firstName = input.nextLine();
while(firstName.length<1){
System.out.println("Please enter a first name of at least 1 character.");
firstName = input.nextLine();
}
lastName=input.nextLine();
while(firstName.length<1){
System.out.println("Please enter a last name of at least 1 character.");
lastName = input.nextLine();
}
System.out.println("You have entered " + firstName +" " + lastName);
Edit, some basic info about exceptions
try catch is used when something unexpected happens and you try to find a way round it. for example if an array of 10 positions is expected at some point and a smaller array (lets say 4 positions) is being used. Then this would cause an exception causing the program to terminate with no further information.
With try catch you can check what the problem is, and try to either inform the user to do something(if they can) or close the program in a better way, using System.exit() for example and saving all the work that was done till that point
An other example is that if you ask for 2 numbers to do an addition. if the user enters letters instead of number the int sum=numbA+numbB; would throw and exception. This of course could be handled using an if. but even better would be something like this
A whitespace is actually considered a character, so the check of (length == 0) doesn't work for your purposes.
Although the following code below is incomplete (ex: handles the potentially undesirable case of firstname=" foo", (see function .contains()), it does what the original post asks - when the user enters a blank first/last name, it keeps repeating "Please enter a first/last name of at least 1 character" until the user enters a valid first/last name.
import java.util.Scanner;
public class Adventure {
public static final int menuStars = 65;
private static Scanner input = new Scanner(System.in);
public static void main(String[] args) {
String firstName = "";
String lastName = "";
boolean firstNameLegal = false;
boolean lastNameLegal = false;
// Entering first name
while (!firstNameLegal) {
System.out.println("Please enter your first name.");
firstName = input.nextLine();
if (!firstName.equals(" "))
firstNameLegal = true;
else
System.out.println("Please enter a first name of at least 1 character.");
}
// Entering last name
while(!lastNameLegal){
System.out.println("Please enter your last name.");
lastName = input.nextLine();
if(!lastName.equals(" "))
lastNameLegal = true;
else
System.out.println("Please enter a last name of at least 1 character.");
}
System.out.println("You have entered " + firstName +" " + lastName);
}
}
For this program it asks for the user to input their full name. It then sorts out the first name and last name by separating them at the space the put between the first and last name. However, indexOf() is not recognizing the space and only returns -1. Why is that? Thanks.
Here is the prompt off of PracticeIt:
Write a method called processName that accepts a Scanner for the console as a parameter and that prompts the user to enter his or her full name, then prints the name in reverse order (i.e., last name, first name). You may assume that only a first and last name will be given. You should read the entire line of input at once with the Scanner and then break it apart as necessary. Here is a sample dialogue with the user:
Please enter your full name: Sammy Jankis
Your name in reverse order is Jankis, Sammy
import java.util.*;
public class Exercise15 {
public static void main(String[] args) {
Scanner inputScanner = new Scanner(System.in);
processName(inputScanner);
}
public static void processName(Scanner inputScanner) {
System.out.print("Please enter your full name: ");
String fullName = inputScanner.next();
int space = fullName.indexOf(" "); // always return -1 for spaces
int length = fullName.length();
String lastName = fullName.substring(space+1,length+1);
String firstname = fullName.substring(0, space);
System.out.print("Your name in reverse order is " + lastName + ", " + firstname);
}
}
As next will return the next token use nextLine not next to get the whole line
see http://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#next()
When you do String fullName = inputScanner.next() you only read till the next whitespace so obviously there is no whitespace in fullName since it is only the first name.
If you want to read the whole line use String fullName = inputScanner.nextLine();
for the last few hours i've been trying to get this code to only allow character input, as its for accepting names..
I have been following the same type of method i used to allow only integers, but with no luck.
any advice would be great,
Here is my code
String name;
System.out.println("Enter student's name: ");
name = in.nextLine();
while (!in.nextLine().matches("[a-zA-Z]"));
{
System.out.println("Please enter a valid name!: ");
in.next();
}
name = in.nextLine();
Your regex checks if the line matches exactly one [a-zA-Z] character. You probably want to check at least one: [a-zA-Z]+
Bear in mind that the way you are checking, you won't get what you have checked in the variable name, as you are consuming it in your checking. Can I propose an alternative:
System.out.println("Enter student's name: ");
String name = in.nextLine();
while(!name.matches("[a-zA-Z]+")){
System.out.println("Please enter a valid name!");
name = in.nextLine();
}
I'm not too familiar with regex, personally, so I would just go with a method that takes a string and determines whether it is only comprised of the alphabets.
[EDIT] Shame on me, the method for testing if a character is a letter is static. Thanks to dansalmo for reminding me.
public static boolean isAlphabetic(String s) {
for(int i = 0; i < s.length(); i++) {
if (!Character.isLetter(s.charAt(i))) return false;
}
return true;
}
Did you check the value of the variable name?
While you assign the input to name , you seem to read the next line for regex check.
Remove the in.next too and it should work fine
Similar to Bucco's solution but iterating over a char array:
public static boolean isAlphabetic(String s) {
for(char c : s.toCharArray())
if (!Character.isLetter(c)) return false;
return true;
}
I have just started the java programming and at the moment I am doing the basic things. I came across a problem that I can't solve and didn't found any answers around so I thought you might give me a hand. I want to write a program to prompt the user to enter their full name (first name, second name and surname) and output their initials.
Assuming that the user always types three names and does not include any unnecessary spaces. So the input data will always look like this : Name Middlename Surname
Some of my code that I have done and stuck in there as I get number of the letter that is in the code instead of letter itself.
import java.util.*;
public class Initials
{
public static void main (String[] args)
{
//create Scanner to read in data
Scanner myKeyboard = new Scanner(System.in);
//prompt user for input – use print to leave cursor on line
System.out.print("Please enter Your full Name , Middle name And Surname: ");
String name = myKeyboard.nextLine();
String initials1 = name.substring(0, 1);
int initials2 = name.
//output Initials
System.out.println ("Initials Are " + initials1 + initials2 + initials3);
}
}
Users will enter a string like
"first middle last"
so therefore you need to get each word from the string.
Loot at split.
After you get each word of the user-entered data, you need to use a loop to get the first letter of each part of the name.
First, the nextLine Function will return the full name. First, you need to .split() the string name on a space, perhaps. This requires a correctly formatted string from the user, but I wouldn't worry about that yet.
Once you split the string, it returns an array of strings. If the user put them in correectly, you can do a for loop on the array.
StringBuilder builder = new StringBuilder(3);
for(int i = 0; i < splitStringArray.length; i++)
{
builder.append(splitStringArray[i].substring(0,1));
}
System.out.println("Initials Are " + builder.toString());
Use the String split() method. This allows you to split a String using a certain regex (for example, spliting a String by the space character). The returned value is an array holding each of the split values. See the documentation for the method.
Scanner myKeyboard = new Scanner(System.in);
System.out.print("Please enter Your full Name , Middle name And Surname: ");
String name = myKeyboard.nextLine();
String[] nameParts = name.split(" ");
char firstInitial = nameParts[0].charAt(0);
char middleInitial = nameParts[1].charAt(0);
char lastInitial = nameParts[2].charAt(0);
System.out.println ("Initials Are " + firstInitial + middleInitial + lastInitial);
Note that the above assumes the user has entered the right number of names. You'll need to do some catching or checking if you need to safeguard against the users doing "weird" things.
This question already has an answer here:
How to use java.util.Scanner to correctly read user input from System.in and act on it?
(1 answer)
Closed 5 years ago.
Basically, my situation requires me to check to see if the String that is defined by user input from the keyboard is only alphabetical characters in one case and only digits in another case. This is written in Java.
my current code:
switch (studentMenu) {
case 1: // Change all four fields
System.out.println("Please enter in a first name: ");
String firstNameIntermediate = scan.next();
firstName = firstNameIntermediate.substring(0,1).toUpperCase() + firstNameIntermediate.substring(1);
System.out.println("Please enter in a middle name");
middleName = scan.next();
System.out.println("Please enter in a last name");
lastName = scan.next();
System.out.println("Please enter in an eight digit student ID number");
changeID();
break;
case 2: // Change first name
System.out.println("Please enter in a first name: ");
firstName = scan.next();
break;
case 3: // Change middle name
System.out.println("Please enter in a middle name");
middleName = scan.next();
break;
case 4: // Change last name
System.out.println("Please enter in a last name");
lastName = scan.next();
case 5: // Change student ID:
changeID();
break;
case 6: // Exit to main menu
menuExit = true;
default:
System.out.println("Please enter a number from 1 to 6");
break;
}
}
}
public void changeID() {
studentID = scan.next();
}
I need to make sure the StudentID is only numerical and each of the name segments are alphabetical.
java.util.Scanner can already check if the next token is of a given pattern/type with the hasNextXXX methods.
Here's an example of using boolean hasNext(String pattern) to validate that the next token consists of only letters, using the regular expression [A-Za-z]+:
Scanner sc = new Scanner(System.in);
System.out.println("Please enter letters:");
while (!sc.hasNext("[A-Za-z]+")) {
System.out.println("Nope, that's not it!");
sc.next();
}
String word = sc.next();
System.out.println("Thank you! Got " + word);
Here's an example session:
Please enter letters:
&###$
Nope, that's not it!
123
Nope, that's not it!
james bond
Thank you! Got james
To validate that the next token is a number that you can convert to int, use hasNextInt() and then nextInt().
Related questions
Validating input using java.util.Scanner - has many examples!
It's probably easiest to do this with a regular expression. Here's some sample code:
import java.util.regex.*;
public class Test
{
public static void main(String[] args) throws Exception
{
System.out.println(isNumeric("123"));
System.out.println(isNumeric("abc"));
System.out.println(isNumeric("abc123"));
System.out.println(isAlpha("123"));
System.out.println(isAlpha("abc"));
System.out.println(isAlpha("abc123"));
}
private static final Pattern NUMBERS = Pattern.compile("\\d+");
private static final Pattern LETTERS = Pattern.compile("\\p{Alpha}+");
public static final boolean isNumeric(String text)
{
return NUMBERS.matcher(text).matches();
}
public static final boolean isAlpha(String text)
{
return LETTERS.matcher(text).matches();
}
}
You should probably write methods of "getAlphaInput" and "getNumericInput" which perform the appropriate loop of prompt/fetch/check until the input is correct. Or possibly just getInput(Pattern) to avoid writing similar code for different patterns.
You should also work out requirements around what counts as a "letter" - the above only does a-z and A-Z... if you need to cope with accents etc as well, you should look more closely at the Pattern docs and adapt appropriately.
Note that you can use a regex to validate things like the length of the string as well. They're very flexible.
Im not sure this is the best way to do, but you could use Character.isDigit() and Character.IsLiteral() mabybe like this:
for( char c : myString.toCharArray() ) {
if( !Character.isLiteral(c) ) {
//
}
}
try regexp: \d+ -- numerical, [A-Za-z]+ -- alphabetical
I don't think you can prevent the users from entering invalid values, but you have the option of validating the data you receive. I'm a fan of regular expressions. Real quick, something like this maybe (all values initialized to empty Strings):
while (!firstName.matches("^[a-zA-Z]+$")) {
System.out.println("Please enter in a first name");
firstName = scan.next();
}
...
while (!studentID.matches("^\\d{8}$")) {
System.out.println("Please enter in an eight digit student ID number");
changeID();
}
If you go this route, you might as well categorize the different cases you need to validate and create a few helper methods to deal with each.
"Regex" tends to seem overwhelming in the beginning, but learning it has great value and there's no shortage of tutorials for it.
That is the code
public class InputLetters {
String InputWords;
Scanner reader;
boolean [] TF;
boolean FT;
public InputLetters() {
FT=false;
while(!FT){
System.out.println("Enter that you want to: ");
reader = new Scanner(System.in);
InputWords = reader.nextLine();
Control(InputWords);
}
}
public void Control(String s){
String [] b = s.split(" ");
TF = new boolean[b.length];
for(int i =0;i<b.length;i++){
if(b[i].matches("^[a-zA-Z]+$")){
TF[i]=true;
}else
{
TF[i]=false;
}
}
for(int j=0;j<TF.length;j++){
if(!TF[j]){
FT=false;
System.out.println("Enter only English Characters!");
break;
}else{
FT=true;
}
}
}