how do i fix a line of code that prints before it gives a user a chance to give input

[amber kelp]

i am trying to get my user to input a string then an int. however the system.out.println code is printing before the user can respond to the first one.

[uneven dome]

<@&987246399047479336> please have a look, thanks.

[uneven dome]

While you are waiting for getting help, here are some tips to improve your experience:

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[vast widget]

post the code

[uneven dome]

Please use this format for posting code:

```java
// Example java program
int value = 5;
System.out.println(value);
```

Which results in:

// Example java program
int value = 5;
System.out.println(value);

For syntax highlighting, you have to add the name of the language after the three backticks, like ```java. Please make sure to use exactly this format, so no space between the backticks and the language name, and a newline before the code starts. If done right, the syntax highlighting will even be applied to your text as you type, before sending.

click • used /tag id: code · Mar 25, 2023, 5:43 AM

[amber kelp]

System.out.println();
        System.out.println("Jelly information");
        System.out.println("Enter the name of the jelly.");
        String jN1 = keyboard.nextLine();
        keyboard.nextLine();
        System.out.println("Enter the number of calories.");
        int jC1 = keyboard.nextInt();
        keyboard.nextLine();
        System.out.println("Enter the type of jelly. Must be \"Apple\", \"Blueberry\"," +
                " \"Grape\", \"Strawberry\", or \"Tomato\".");
        String jT1 = keyboard.nextLine();
        Jelly jelly1 = new Jelly(jN1, jC1, jT1);

[uneven dome]

Detected code, here are some useful tools:

Formatted code

System.out.println();
System.out.println("Jelly information");
System.out.println("Enter the name of the jelly.");
String jN1 = keyboard.nextLine();
keyboard.nextLine();
System.out.println("Enter the number of calories.");
int jC1 = keyboard.nextInt();
keyboard.nextLine();
System.out.println("Enter the type of jelly. Must be \"Apple\", \"Blueberry\"," + " \"Grape\", \"Strawberry\", or \"Tomato\".");
String jT1 = keyboard.nextLine();
Jelly jelly1 = new Jelly(jN1, jC1, jT1);

[amber kelp]

i fixed the problem i was having with th e keyboard.nextline after string jN1, but it is not storing the inputted string to jN1

[vast widget]

wdym?

It is storing it to jN1

Also prefer parsing rather then using nextInt() since it does not skip the new line

For example

int jC1 = Integer.parseInt(keyboard.nextLine());

[amber kelp]

System.out.println("-----Sandwich 1-----");
        System.out.println("PBJ Sandwich");
        System.out.println("Top Slice:");
        System.out.println(topSlice1.toString());
        System.out.println("Peanut Butter:");
        System.out.println(peanutButter1.toString());
        System.out.println("Jelly:");
        System.out.println(jelly1.toString());
        System.out.println("Bottom Slice:");
        System.out.println(bottomSlice1.toString());

[uneven dome]

Detected code, here are some useful tools:

Formatted code

System.out.println("-----Sandwich 1-----");
System.out.println("PBJ Sandwich");
System.out.println("Top Slice:");
System.out.println(topSlice1.toString());
System.out.println("Peanut Butter:");
System.out.println(peanutButter1.toString());
System.out.println("Jelly:");
System.out.println(jelly1.toString());
System.out.println("Bottom Slice:");
System.out.println(bottomSlice1.toString());

[amber kelp]

so im trying to print it out with this code and the name of the jelly is blank

[vast widget]

could you post your Jelly class

[amber kelp]

public class Jelly {
    private String name;
    private int calories; //Between 50-200, def is 50
    private String fruitType; //Apple, blackberry, GRAPE, blueberry, and tomato
    public Jelly() {
        name = "none";
        calories = 50;
        fruitType = "Grape";
    }
    public Jelly(String aN, int aC, String fT) {
        this.setName(aN);
        this.setCalories(aC);
        this.setFruitType(fT);
    }
    public String getName() {
        return this.name;
    }
    public int getCalories() {
        return this.calories;
    }
    public String getFruitType() {
        return this.fruitType;
    }
    public void setName(String aN) {
        if(aN != null)
            name = aN;
        else
            name = "none";
    }
    public void setCalories(int aC) {
        if(aC >= 50 && aC <= 200)
            calories = aC;
        else
            calories = 50;
    }
    public void setFruitType(String fT) {
        if(fT != null && (fT.equalsIgnoreCase("Apple") || fT.equalsIgnoreCase("Blackberry") || fT.equalsIgnoreCase("Grape") || fT.equalsIgnoreCase("Blueberry") || fT.equalsIgnoreCase("Tomato")))
            fruitType = fT;
        else
            fruitType = "Grape";
    }
    public boolean equals(Jelly j) {
        return j != null &&
                name.equals(j.getName()) &&
                calories == j.getCalories() &&
                fruitType.equals(j.getFruitType());
    }
    public String toString() {
        return "Jelly Name: "+getName()+" Calories: "+getCalories()+" Fruit Type: "+getFruitType();
    }
}

[uneven dome]

Detected code, here are some useful tools:

[vast widget]

In this code, why are you calling nextLine()

twice in each request

Jelly information
Enter the name of the jelly.
hello
Enter the number of calories.
500
Enter the type of jelly. Must be "Apple", "Blueberry", "Grape", "Strawberry", or "Tomato".
tomato
Jelly Name: hello Calories: 50 Fruit Type: tomato

besides it works for me

[amber kelp]

i didn't have it and this was happening

[vast widget]

you are getting the number of calories

and not inputting anything

and so it's trying to parse t into a number

[amber kelp]

yeah that's why i added the extra nextLine()

is that what's fucking up the storing of the name of the jelly

[vast widget]

no it should work just fine

        Scanner keyboard = new Scanner(System.in);
        System.out.println();
        System.out.println("Jelly information");
        System.out.println("Enter the name of the jelly.");
        String jN1 = keyboard.nextLine();
        System.out.println("Enter the number of calories.");
        int jC1 = Integer.parseInt(keyboard.nextLine());
        System.out.println("Enter the type of jelly. Must be \"Apple\", \"Blueberry\"," + " \"Grape\", \"Strawberry\", or \"Tomato\".");
        String jT1 = keyboard.nextLine();
        Jelly jelly1 = new Jelly(jN1, jC1, jT1);
        System.out.println(jelly1);

It works

Is this replit?

ah

[amber kelp]

i created a separate class and it works in that class but once I put the code back into my whole FE class it doesnt work

[vast widget]

Your entire code

Yeah I know the issue

You are using keyboard.nextBoolean()

always parse the result from nextLine()

don't use those methods

they are bad

When you do keyboard.nextBoolean()

it will read in a boolean but not the resulting new line - \n

so your String jn1 = keyboard.nextLine() is getting passed the new line character

because the cursor in the keyboard is all messed up

[uneven dome]

Mixing any nextXXX method with nextLine from the Scanner class for user input, will not ask you for input again but instead result in an empty line read by nextLine.

To prevent this, when reading user input, always only use nextLine. If you need an int, do

int value = Integer.parseInt(scanner.nextLine());

instead of using nextInt.

Assume the following:

Scanner scanner = new Scanner(System.in);

System.out.println("Enter your age:");
int age = scanner.nextInt();
System.out.println("Enter your name:");
String name = scanner.nextLine();

System.out.println("Hello " + name + ", you are " + age + " years old");

When executing this code, you will be asked to enter an age, suppose you enter 20.
However, the code will not ask you to actually input a name and the output will be:

Hello , you are 20 years old.

The reason why is that when you hit the enter button, your actual input is

20\n

and not just 20. A call to nextInt will now consume the 20 and leave the newline symbol \n in the internal input buffer of System.in. The call to nextLine will now not lead to a new input, since there is still unread input left in System.in. So it will read the \n, leading to an empty input.

So every user input is not only a number, but a full line. As such, it makes much more sense to also use nextLine(), even if reading just an age. The corrected code which works as intended is:

Scanner scanner = new Scanner(System.in);

System.out.println("Enter your age:");
// Now nextLine, not nextInt anymore
int age = Integer.parseInt(scanner.nextLine());
System.out.println("Enter your name:");
String name = scanner.nextLine();

System.out.println("Hello " + name + ", you are " + age + " years old");

The nextXXX methods, such as nextInt can be useful when reading multi-input from a single line. For example when you enter 20 John in a single line.

click • used /tag id: scanner · Mar 25, 2023, 6:35 AM

[amber kelp]

thank you so much i just switched it

my professor uses it so i didnt realize that it messed my code up

[vast widget]

Yeah it's a tricky thing, most schools teach to use nextInt() and the corresponding methods

It just doesn't work like you would expect, that's why the best practice for this is just to ALWAYS parse it

[amber kelp]

i didnt know boolean messed it up, i assumed it worked like nextLine() bc my prof always uses nextLine() as a fix-up when using nextInt()

[vast widget]

Yeah that's just an ugly way to fix the cursor problem, it works just not optimal