How to accept user input in Java ⌨️【8 minutes】

Java user input scanner
#Java #input #scanner
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("What is your name? ");
String name = scanner.nextLine();
System.out.println("How old are you? ");
int age = scanner.nextInt();
scanner.nextLine();
System.out.println("What is your favorite food?");
String food = scanner.nextLine();
System.out.println("Hello "+name);
System.out.println("You are "+age+" years old");
System.out.println("You like "+food);
}
}

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[@doronfeldman5152]

Agree

[@BroCodez]

// It's good practice to close your scanner when you are done using it. scanner.close(); // I forgot to. So I didn't import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); System.out.println("What is your name? "); String name = scanner.nextLine(); System.out.println("How old are you? "); int age = scanner.nextInt(); scanner.nextLine(); System.out.println("What is your favorite food?"); String food = scanner.nextLine(); System.out.println("Hello "+name); System.out.println("You are "+age+" years old"); System.out.println("You like "+food); scanner.close(); } }

[@nairaac]

thanks for giving it in the comments! also mine kept on saying error on the first line.

[@CS_CHILD]

Umm....I'm having a problem here please It says error..system cannot be resolved to a variable

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Bro you have no idea how long it took me to wonder why I couldn't enter any input with the nextLine after nextInt. You are awesome!!

[@JuliHoffman]

That broke down the WHY of scanners for me, rather than "because that's just the way it is." Thank you!

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[@mailiesathare]

I like how you tackled the newline character stuck in the buffer integrated concept (I would still call it an error). I was taught that the buffer needed to be cleared, and I still do not know how to visualize a buffer.

[@1jokerpro780]

for strings , when using scanner , its better to use .next(); instead of .nextLine(); to avoid that error , that worked for me at least . and i have to say , great video , thanks for the efforts . this is by far the most beginner friendly , right to the point , including tricks and errors tutorial. i sincerely thank you bro hahahaha

[@stephaniehernandez3641]

Thank you!!! I took a note of this :)

[@paia6447]

Next() keyword won't be able to print the whole text, for like name and tittle at the same time

[@ayansiddiqui7383]

Thanks , it helped a lot

[@cubebrick7019]

helped me thanks

[@sairos4057]

If you use ".next();" instead of ".nextLine();", It will be not possible to type more than one word Example: Joker - works fine Joker pro - error

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This is the best Java tutorial for beginners, so you can learn Java and English in one hit. Please keep going! I vote for Java advance tutorial. Thanks a lot Bro

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Thanks, Bro!! /n explanation was what I was looking for throughout the youtube

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I hope you will get more than 1M views each time you uploaded a new video after one day, in the future.

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Man! You saved my java project! Thank you!

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Love your style bro!

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Finally, get this tutorial that explained this error clearly

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your videos are so useful, thank you

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Thank you for giving us such a great learning material !:)

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Great tutorial! Thanks!

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Nice and smoothly explained

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[@user-rs2ks1pb1s]

Great explanation of the mysterious behavior of the Scanner class.

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For the first time my code executes without error!! # Thank you so much sir, I love your pattern of writing codes so simple ...

[@medjl6083]

Very important video of Java for input data by user with keyboard.

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this is a comprehensive explanation

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Very easy to understand. Thank you :)

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The shia Labeouf "Just do it" icon made me subscribe. Thanks bro

[@joshbenjamin1100]

Great video! One question. I made a program that has an IF/ELSE statement in it, so depending on the user's input of age, it goes to one or the other. The program continues to ask questions and receive input if the user's input leads to the ELSE statement, but if it leads to the IF (i.e., the statement before the ELSE) statement, it stops taking input. Any idea why?

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lovin' it

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Another lesson that i understood

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Sometimes it does not work or maybe i just missing something in my code after i implement the nextLine() method to eat some extra white space; but yeah its very helpful, Thanks bro

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