如何在字符串数组中找到给定String的索引？

Question

Surely the solution to this is the following:

public long myFunc(String name) throws Exception {
    for(int i=0;i<amount;i++){ 
       if(this.otherString[i].equals(name)) 
           return longArray[i]; 
    } 
    throw new Exception("Not found"); 
} 

However, this does not seem to be the case.

Answer 1

You can use Guava, then your code can looks like this

String[] stringArray = {"s1", "s2", "s3"};

int index = Iterators.indexOf(Iterators.forArray(stringArray), new Predicate<String>() {
    @Override
    public boolean apply(String input) {
        return input.equals("s2");
    }
});

or simpler

int index = Arrays.asList(stringArray).indexOf("s2");

Your code can also look like this

public class Finder {

    private String[] stringArray = {"s1", "s2", "s3"};

    public int findIndex(String name) {
        for (int i = 0; i < stringArray.length; i++) {
            if (stringArray[i].equals(name))
                return i;
        }

        throw new RuntimeException("Not found");
    }

    public static void main(String... s) {
        int index = new Finder().findIndex("s1");

        System.out.println(index);
    }
}

Answer 2

You can either run your code through a debugger and find out why it doesn't work, or add some println traces to your original code and you'll see what the problem is:

public long myFunc(String name) throws Exception {
    System.out.println("Looking for: " + name);
    for (int i = 0; i < amount; i++){
        if(this.otherString[i].equals(name))
            return longArray[i];
        System.out.printf("%4d: \"%s\": No match.%n", i, this.otherString[i]);
    }
    for (int i = amount; i < this.otherString.length; i++)
        System.out.printf("%4d: \"%s\": Not checked.%n", i, this.otherString[i]);
    throw new Exception("Not found");
}

By the way, make sure you are correctly interpreting the method's behaviour. Could it be that it's finding it but throwing an ArrayIndexOutOfBoundsException because i is greater than longArray.length, and you misinterpret that as the exception you are explicitly throwing?

Answer 3

Turns out '\' was at the end of the string that was meant to be equal. This does not show in printing. Next time I will be more ruthless in the inspection of the debugging. Thank you for all the suggestions though, and sorry for wasting your time.

Answer 4

This might be completely wrong, but isn't the problem just missing curly brackets in the if statement? I am new to the java language and the example is probably messy, but it uses the same structure from your question and works just fine:

public class random_class {

public static void main(String[] args) {
    // TODO Auto-generated method stub
    String[] a = new String[] {"hi!", "hello world!", "ho ho ho world!"};       
    int b = getHWIndex(a);
    System.out.println(b);
}

public static int getHWIndex(String[] stringArray){
    int i;
    Boolean test = false;
    for(i=0;i<stringArray.length;i++){
        if(stringArray[i].equals("hello world!")){
            test = true;  
            break;
        }
    }
    if(test == true){
    return i;}else{
        return i = 0; // this is not a good answer... 
//but as you return an int I could not think on a quick way to fix   the return when there is no match.
        }
    }
}