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Why am I getting an error for actual and formal argument lists differ in length? [closed]

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Trying to make it so pasting a 60 character long url, will print only the first 56 characters. I have only ever used Java in high school, so I am very inexperienced.
import java.util.Scanner;
public class CopyOfCopyOfethanrun
{
public static void main()
{
Scanner kb = new Scanner(System.in);
link();
}
static void link(Scanner kb, String[] args)
{
String link;
System.out.print("\n\n Enter Link: ");
link = kb.nextLine();
String input = link;
String firstfiftysix = "";
if (input.length() > 56)
{firstfiftysix = input.substring(0, 56);}
else{firstfiftysix = input;
}
System.out.println(firstfiftysix);
}
}
[Here is an image showing the error I am experiencing][1]
[1]: https://i.stack.imgur.com/T6ux7.png

Your link method has 2 arguments: When invoking link, you need to provide [A] a Scanner instance, and [B] an instance of an array of Strings.
When you call link();, you provide neither of those. Presumably you want link(kb, args); there. Your main must look like:
public static void main(String[] args)
it currently doesn't. Once you fix that, voila, you have your args. But, you should just get rid of that, you don't use it anywhere in link.

How to iterate a for loop for a user input amount of times Java [closed]

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I want to get a user input number to print out "Color 1 Color 2..." etc. depending on what number the input is.
I want to do this but in java, and I'm not quite sure where to find it.
How to iterate a for loop for a user input in Java?

That will do the work, using only standard java classes (java.io.Console):
import java.io.Console;
public class Consoler {
public static void main(String[] args) {
final Console console = System.console();
console.printf("How may times?\n");
final String line = console.readLine();
try {
final int quantity = Integer.parseInt(line);
for (int i = 1; i <= quantity; i++) {
System.out.printf("Color %d ",i);
}
System.out.println();
} catch (final NumberFormatException e) {
System.err.println(line + " is not a number.");
}
}
}

java.util.Scanner is probably what you're looking for. As for the for loop, you should google Java for loop and read up on how they work. Then combine the two concepts.

Java getString method [closed]

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So I have the questions in the top part, but I want to have all the questions at the top, and when I need to ask the questions, I can just pull them down using a variable defined as the question. Right now however, the code is asking the questions from where the questions are, not using the variable "ask" and asking from System.out.print(ask). Any ideas on how to get it to do that?
import java.util.Scanner;
public class Greetings {
public static void main(String[] args) {
Scanner newscanner = new Scanner(System.in);
String ask = getString(newscanner, "Please enter your first name: ");
// String ask2 = getString(newscanner, "Please enter your last name: ");
// String ask3 = getString(newscanner, "Please enter your year of birth:
// ");
}
public static String getString(Scanner newscanner, String ask) {
System.out.print(ask);
String first = newscanner.next();
String firstletter = first.substring(0, 1).toUpperCase();
return firstletter;
}
}

Perhaps what you are looking to do is have the question be printed, and then the answer typed on the line below it? If so, what you need to do is change the first call in getString from System.out.print to System.out.println, which should add on a newline after the question, moving the input to the next line.
EDIT: This is what it might look like now:
Please enter your first name:John
And here's what it would change to:
Please enter your first name:
John

How do I check for an empty string in a Boolean statement? [closed]

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I have an assignment where I have to attach the letters "un" to any word that the user inputs (unless the inputted word already has "un" in front of it, in which case I just return the inputted word). I'm testing my method but I encountered one problem: my program keeps returning an error if I were to test for an empty input. Here is my code:
Scanner keyboard = new Scanner(System.in);
System.out.print("Enter: ");
String input = keyboard.nextLine();
if(input.substring(0,2).equalsIgnoreCase("un"))
{
System.out.println(input);
}
else if(input.equals(""))
{
System.out.println("un");
}
else
{
System.out.println("un" + input);
}
So I wanted to ask how I can test for an empty input/blank string since, evidently, the "" quotations do not work.

There's nothing wrong with checking input.equals("") per-se. The problem is that you have another test beforehand that throws an exception if input is shorter than 2 characters.
There are several ways to solve this, but I'd just simplify things and use startsWith. An empty string doesn't really need a special case of its own - just slap un before it, and you'll get un:
if (input.toLowerCase().startsWith("un")) {
System.out.println(input);
} else {
System.out.println("un" + input);
}

You are having this problem because you are trying to get the substring of string that doesnt have the required length. Put the empty string check first.
if(input.equals("")||input.length==1)
{
System.out.println("un");
}
else if(input.substring(0,2).equalsIgnoreCase("un"))
{
System.out.println(input);
}
else
{
System.out.println("un" + input);
}

If this weren't homework, and the library could be used for other things ( using it in this single purpose may be overkill ), you could use StringUtils.PrependIfMissing().
It does exactly this and handles nulls as well.

In java, how do I check if a string contains only 1 or 0 [closed]

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Also, how would I prompt the user to try again until they enter something with only 1 or 0 in it?
I realize I must use a for, or a while loop, but I'm not sure what to put as the condition.
I'm trying to have it so the user is prompted to enter something in binary, and if they don't enter something in binary, to be asked again to enter something in binary, and repeated until they do.
Thanks in advance!

You can do this by a simple regular expression matching:
if (inputString.matches("^[01]+$")) {
// accept this input
}

Simply use Integer.parseInt (str, 2);
it will throw a NumberFormatException if not binary

You can inspect every character of the String like so:
String s;//user input
boolean bad=false;//Starts false-will change to true if the input is bad
for(char c:s.toCharArray())
if(!(c=='0'||c=='1')){//if c isn't 0 or 1
bad=true;
break;//break out of loop because we've already found a problem
}

You may want to use the pattern below. The concept is to provide a "regular expression" that provides the rules for a conforming string, along with the message to prompt the user for input and the source to read the user's input from. The loop continues until a conforming string is found, or the user breaks out with "exit". The function would return the conforming string, or a null if the user wants to exit.
public String getConformingString(Scanner source, String message, String pattern) {
String result = null;
boolean isConformingString = false;
System.out.println(message);
String trialString = source.mextLine();
while (!isConformingString) {
if (trialString.matches(pattern) {
isConformingString = true;
result = isConformingString;
} else if ("exit".equalsIgnoreString(trialString)) {
isConformingString = true;
} else {
System.out.println(message);
trialString = source.nextline();
}
}
return result;
}