(amend) How to display input match with array?

Question

May someone tell me the error of this code? My question is display "Found" if the input is in the array, display "Not Found" if the input is not in the array. Why I only can display "Found" for whatever I key in?

String [] deptName = {"Accounting", "Human Resources","Sales"};

//input
String key = JOptionPane.showInputDialog("Enter a department: ");

for(int i=0; i<deptName.length; i++)
{
    if(deptName [i] == key);
}
System.out.println("Found");

Edit:

I have amend my code like this, how can I do not let it display 3 times?

String [] deptName = {"Accounting", "Human Resources","Sales"};

//input
String key = JOptionPane.showInputDialog("Enter a department: ");

for(int i=0; i<deptName.length; i++)
{
    if(deptName [i].equals(key))
        JOptionPane.showMessageDialog(null, "Found");
    else
        JOptionPane.showMessageDialog(null, "Not Found");
}

Hi, I've edited your question so that anyone else coming along later can see the whole conversation, not just the latest step. — Basic
– Basic, Commented Dec 8, 2011 at 15:41

Kent · Accepted Answer · 2011-12-08 10:12:43Z

3

"found" was always printed, because it is out of the for loop. and your check (which has problem too) does actually nothing

try

    your For loop {
         if(deptName [i].equals( key)){
          System.out.println("Found");
          break;
}
    }

answered Dec 8, 2011 at 10:12

Kent

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Comments

Thomas Johan Eggum · Accepted Answer · 2011-12-08 10:13:25Z

2

If you use == on a Object in java, you only check if the object is the exact same object. To compare the content of an object, you should use equals().

In case of the class String, equals() compares the actual String content.

deptName[i].equals(key);  //returns true if text is the same

answered Dec 8, 2011 at 10:13

Thomas Johan Eggum

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Comments

Naveed S · Accepted Answer · 2011-12-08 10:30:18Z

1

Do it like this

for(int i=0;i<deptName.length;i++){
    if(key.equals(deptName[i]){
       JOptionPane.showMessageDialog(null,"Found"); break;
}}
if(i==deptName.length)
   {JOptionPane.showMessageDialog(null,"Not Found");}

answered Dec 8, 2011 at 10:30

Naveed S

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Comments

stacker · Accepted Answer · 2011-12-08 10:13:32Z

0

You have several issues, first you need to call equals() and must not terminate the if with ';'

for(int i=0; i<deptName.length; i++)  {         
    if(deptName [i].equals( key )) {
       System.out.println("Found"); 
    }
}

would work as intented.

answered Dec 8, 2011 at 10:13

stacker

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Prince John Wesley · Accepted Answer · 2011-12-08 10:14:49Z

0

int i=0 ;   
for(; i<deptName.length; i++) { 
    if(key.equals(deptName[i])) break;    
} 
if( i < deptName.length) {
       // found
} else {
      // not found
}

answered Dec 8, 2011 at 10:14

Prince John Wesley

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2 Comments

Annie Tan Over a year ago

I have amend my code like this, how can I let it do not display 3 times?

Annie Tan Over a year ago

String [] deptName = {"Accounting", "Human Resources","Sales"}; //input String key = JOptionPane.showInputDialog("Enter a department: "); for(int i=0; i<deptName.length; i++) { if(deptName [i].equals(key)) JOptionPane.showMessageDialog(null, "Found"); else JOptionPane.showMessageDialog(null, "Not Found"); }

st0le · Accepted Answer · 2011-12-08 11:06:18Z

0

Screw the loop.

JOptionPane.showMessageDialog(null,Arrays.asList(deptName).contains(key) ? "Found" : "Not Found");

I understand this could be a programming exercise, so here you go...

for(int i=0;i<deptName.length;i++){
    if(key.equals(deptName[i])
       break;
}

if(i==deptName.length)
   JOptionPane.showMessageDialog(null,"Not Found");
else
   JOptionPane.showMessageDialog(null,"Found at " + i);

answered Dec 8, 2011 at 11:06

st0le

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