CodeGym/Help with Java Tasks/Can anyone

Level 4

sialkot

08.11.2018
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15comments

Can anyone

Question about the task

Come on, lucky seven!

Java Syntax, Level 4, Lesson 4

Under discussion

What Is wrong in this code?

Write the checkInterval method. The method should check whether an integer is in the range from 50 to 100 and display the result on the screen as follows: "The number a is not in the interval." or "The number a is in the interval.", where a is the method argument.

Example for 112:
The number 112 is not in the interval.

Example for 60:
The number 60 is in the interval.

Requirements:

The program should display text on the screen.
The main method should not call System.out.println or System.out.print().
The main method must call the checkInterval method.
The checkInterval method must be static void.
The checkInterval method should display text on the screen in accordance with the task conditions.

package com.codegym.task.task04.task0410; /* Come on, lucky seven! */ public class Solution { public static void main(String[] args) { checkInterval(60); checkInterval(112); checkInterval(10); } public static void checkInterval(int a) { if ( a >= 50 && a <= 100){ System.out.println("The number" + a + "is not in the interval."); } else { System.out.println("The number" + a + "is not in the interval."); } } }

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J H U X

Level 7 , Alkmaar, Netherlands

15 January 2021, 15:11

You wrote not twice and you forgot to input spaces. Good luck!

Kelly Reed

Level 4 , Wichita, United States

26 October 2020, 14:38

best possible solution- use the helpful hints code blocks to keep from making grammatical mistakes.... as you have it written, wht you will display on the console is This number100is not in the interval. (notice the lack of spacing before or after the number) This is due to a lack of spaces within your system.out.println("blah blah "+a+" blah blah") notice the spacing within the code block around the quotations surrounding the addition signs surrounding your passed in variable.... hope this helps. thank you everyone for your help on my speed bumps.

Asen Popov

Level 9 , Miziya, Bulgaria

10 June 2020, 11:47

The text should be printed exactly as required :)

Asen Popov

Level 9 , Miziya, Bulgaria

10 June 2020, 11:46

No intervals before and after the number

Sanjay Chauhan

Level 28 , Delhi, India

28 January 2020, 03:15

You are missing space in the message. Try this: if(a >=50 && a <=100) System.out.println("The number "+ a +" is in the interval."); else System.out.println("The number "+a +" is not in the interval.");

Sushant

Level 4 , goa, india

13 August 2019, 09:56

try this this might help you. if(a >= 50 && a <= 100){ System.out.println("The number "+ a +" is in the interval."); } else System.out.println("The number "+ a +" is not in the interval.");

Okanlawon Oluwatobi Damilare

Level 9 , Banjul, Gambia

18 March 2019, 17:10

try if(a >= 50 && a <= 100){ System.out.println("The number "+ a +" is in the interval."); }else System.out.println("The number "+ a +" is in not the interval.");

-2

Balen Asangbeh

Level 23 , Douala, Cameroon

22 January 2019, 08:10

on lines 17 and 20, give a space after The number and before is. i.e ("The number " + a + " is not in the interval.");

Jason Ahn

Level 4 , Ottawa, Canada

22 January 2019, 01:21

look at the first statement. It should be "The number __ is in the interval."

Level 8 , Wietzendorf, Germany

20 January 2019, 00:39

You wrote " The number .. is not in the interval" two times. Also you are missing some space in between. You wrote, f.e. : "The number60is not in the interval".