Java string array length error

Question

I'm a beginner in Java and practicing to code by myself. I'm using a book to teach me how to code and I'm stuck on a project where it includes creating a String array and determining its length. Here's my sample work. It's a simple baby name suggestion program.

public class BabyName {
    public static void main(String[] args) {
        String[] nameListOne = {"Alvin", "Elmer", "Angel", "Michael", "Tim", "Jude", "Gabriel", "Raphael", "John", "Smith", "Carl", "Mike"};
        String[] nameListTwo = {"", "Mark", "Posh", "Jake", "Cloud", "Star", "Ben", "Sam", "Kim", "Mark", "Simeon", "Louie", "Nat", "Matt", ""};

        int nameOneLength = nameListOne.Length;
        int nameTwoLength = nameListTwo.Length;

        int rand1 = (int) (Math.random() * nameOneLength);
        int rand2 = (int) (Math.random() * nameTwoLength);

        String babyname = nameListOne[rand1] + " " + nameListTwo[rand2];

        System.out.println("Your suggested baby name is " + babyname);
    }
}

I then get this errors when I tried to compile it:

BabyName.java:6: error: cannot find symbol
                int nameOneLength = nameListOne.Length;
                                               ^
  symbol:   variable Length
  location: variable nameListOne of type String[]
BabyName.java:7: error: cannot find symbol
                int nameTwoLength = nameListTwo.Length;
                                               ^
  symbol:   variable Length
  location: variable nameListTwo of type String[]
2 errors

I don't know what I did wrong here. I only followed what's written on the book.

Hope someone can help.

You want length, not Length.

Jon Skeet
– Jon Skeet

2015-07-06 07:44:37 +00:00
Commented Jul 6, 2015 at 7:44 — Jon Skeet
– Jon Skeet, Commented Jul 6, 2015 at 7:44
Upvote for @JonSkeet who technically answered it first!

Tim Biegeleisen
– Tim Biegeleisen

2015-07-06 07:45:52 +00:00
Commented Jul 6, 2015 at 7:45 — Tim Biegeleisen
– Tim Biegeleisen, Commented Jul 6, 2015 at 7:45
... but voted to close as a typo, basically.

Jon Skeet
– Jon Skeet

2015-07-06 07:46:16 +00:00
Commented Jul 6, 2015 at 7:46 — Jon Skeet
– Jon Skeet, Commented Jul 6, 2015 at 7:46

codeaholicguy · Accepted Answer · 2015-07-06 07:44:36Z

2

nameListOne.length length should be lowercase

answered Jul 6, 2015 at 7:44

codeaholicguy

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Erwin Bolwidt · Accepted Answer · 2015-07-06 07:45:20Z

2

Java keywords are case sensitive. Use lowercase 'l' for 'length': nameListOne.length

answered Jul 6, 2015 at 7:45

Erwin Bolwidt

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1 Comment

Alvin Over a year ago

Yes. Really need to get used with these case sensitive thing. Thanks.

Igor Lopes · Accepted Answer · 2015-07-06 07:47:09Z

1

It should be:

int nameOneLength = nameListOne.length;
int nameTwoLength = nameListTwo.length;

answered Jul 6, 2015 at 7:47

Igor Lopes

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Mohit Kanwar · Accepted Answer · 2015-07-06 08:00:17Z

0

Since you are a beginner, this would be helpful for you: Following are the naming conventions in Java ( a standard that should be (but not a must) followed)

All the variable names begin with a lower case e.g. length and are in CamelCasing
All Constants are all caps e.g. CONSTANT_LENGTH
All Class names begin with Capital letter and are in CamelCasing, e.g. LengthType

So, with this information in mind, you can detect such issues earlier (by following conventions)

The issue with your code is "Length" is not a field, where as "length" is.

Remember; Java is Case-Sensitive.

answered Jul 6, 2015 at 8:00

Mohit Kanwar

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1 Comment

Alvin Over a year ago

Thank you for the additional info. Will take note of that. Thanks again.

Amit · Accepted Answer · 2015-07-06 07:48:23Z

0

Use nameListOne.length Becoz Java is case-Sensetive

answered Jul 6, 2015 at 7:48

Amit

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