Java Array matching input with array

Question

I'm attempting to write a method in a class that will prompt the user to enter the name of a student. Then search the list of already existing names for a match. But I cant seem to figure out how to proceed in coding it on how to search for a valid match.

 public void modifyExam(String [] names)
 {
 String name;
 Scanner scanner = new Scanner(System.in);
 System.out.println("Please enter the name of the student whose grade you would like to modify: ");
 name = scanner.nextLine();
 boolean nameMatch = true;
      for (int i=0; i<names.length; i++) 
      {
       // ....
      }

I have no idea how to code it to search for a valid match and was hoping to get some insight. — Leoinu
– Leoinu, Commented Apr 20, 2014 at 20:35

Ayman · Accepted Answer · 2014-04-20 20:35:25Z

2

You should use .equals() to compare strings in Java. Example:

public void modifyExam(String [] names) {
    String name;
    Scanner scanner = new Scanner(System.in);
    System.out.println("Please enter the name of the student whose grade you would like to modify: ");
    name = scanner.nextLine();
    boolean nameMatch = false;
    for (int i=0; i<names.length; i++) {
        if( names[i].equals(name) ) {
            // do your logic here ...
        }
    }
}

answered Apr 20, 2014 at 20:35

Ayman

1,84216 silver badges17 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Jeff Ward · Accepted Answer · 2014-04-20 20:47:15Z

1

I would recommend that you store your students in a Map with the name as the key. Then you wouldn't have to iterate (assuming your names are unique).

Map<String, Student> students = new HashMap<String, Student>();
Student s = new Student("Joe Smoe");
students.put(s.getName(), s);

Then you can lookup the student to update like this:

Student studentToUpdate = students.get(name);
if (studentToUpdate != null) {
   // logic here...
}

Note Returns null if this map contains no mapping for the key. So you would want to add a null check before using the return value of the get call and deal with it accordingly.

edited Apr 20, 2014 at 20:47

answered Apr 20, 2014 at 20:36

Jeff Ward

1,1196 silver badges17 bronze badges

Comments

zgc7009 · Accepted Answer · 2014-04-20 20:37:18Z

0

Equalities.

if(name.equals(name[i]))
    System.out.println("match at " i);

answered Apr 20, 2014 at 20:37

zgc7009

3,3995 gold badges24 silver badges34 bronze badges

Comments

JamesB · Accepted Answer · 2014-04-20 20:38:25Z

0

boolean nameMatch = false;
for (int i=0; i<names.length; i++) { 
    if(names[i].equals(name)) {
        namesMatch = true;
        break;
     }
}

The break means you don't carry on searching the array as a match has been found.

answered Apr 20, 2014 at 20:38

JamesB

7,9242 gold badges24 silver badges21 bronze badges

Comments

Yudhishthir Singh · Accepted Answer · 2014-04-20 21:01:07Z

I would break this function into two functions. One called findName() or even more generic findString() and the other called modifyExam. Let findString() return an index, use the index in modify exam.
Here is what findString() should do

    int findString(String [] names, String name) {
        for ( int i = 0; i < names.length; i++ ) {
                if names[i].equals(name) return // either a boolean or an index or something.
        }
        return -1 // or null
   }

You could also use binary search if the search array is large and already sorted. Using binarySearch() the findString method will be something like :-

int findString(String[] names, String name, int startIndex, int stopIndex) {
    if ( startIndex > stopIndex) return;
    int mid = (stopIndex - startIndex)/2
    if ( name < names[mid] ) return findString(names, name, startIndex, mid-1); 
    else if ( names[mid].equals(name) ) return mid;
    else return findString(names, name, mid+1, stopIndex);
}

Collectives™ on Stack Overflow

Java Array matching input with array

5 Answers 5

Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related