Check if a String is in an ArrayList of Strings

Question

How can I check if a String is there in the List?

I want to assign 1 to temp if there is a result, 2 otherwise.

My current code is:

Integer temp = 0;
List<String> bankAccNos = new ArrayList<String>();//assume list contains values
String bankAccNo = "abc";
for(String no : bankAccNos)
    if(no.equals(bankAccNo))
        temp = 1;

Why not just set temp to be 2 at the begining? And why use object Integer instead of int? Solution looks good enough if you change what i said — tskulbru
– tskulbru, Commented Apr 18, 2012 at 11:49
What you have written already is a working solution, except that you are missing the else statement. Also notice that comparing strings is case sensitive which might be an issue if you didn't get it to work properly. Also, using contains like suggested is another valid approach — John Snow
– John Snow, Commented Apr 18, 2012 at 12:01
@JimmyGustafsson : yes man i was doin that only but i got confused... anyway thanks — abhi
– abhi, Commented Apr 19, 2012 at 4:09

jazzytomato · Accepted Answer · 2012-04-18 12:38:50Z

101

temp = bankAccNos.contains(no) ? 1 : 2;

edited Apr 18, 2012 at 12:38

answered Apr 18, 2012 at 11:50

jazzytomato

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2 Comments

Michal Over a year ago

The List.contains() method checks for identity, not equality. It works well for integers but not strings (which is what OP asked for). Iterating over the list, or better, using Java8 List stream is the way to go for List<String>

jazzytomato Over a year ago

@Michal the documentation states otherwise docs.oracle.com/javase/8/docs/api/java/util/… It checks for equality, not identity: " returns true if and only if this list contains at least one element e such that (o==null ? e==null : o.equals(e))"

Angelo Fuchs · Accepted Answer · 2017-05-28 18:00:26Z

16

The List interface already has this solved.

int temp = 2;
if(bankAccNos.contains(bakAccNo)) temp=1;

More can be found in the documentation about List.

edited May 28, 2017 at 18:00

answered Apr 18, 2012 at 11:49

Angelo Fuchs

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3 Comments

Michal Over a year ago

The List.contains() method checks for identity, not equality. It works well for integers but not strings (which is what OP asked for). Iterating over the list, or better, using Java8 List stream is the way to go for List<String>

Angelo Fuchs Over a year ago

@Michal Not true as the documentation says otherwise (and I've just tested). Hit the link in my answer and read for yourself.

Michael Laffargue Over a year ago

More formally, returns true if and only if this list contains at least one element e such that (o==null ? e==null : o.equals(e)).

So it's equality not identity.

Alberto Trindade Tavares · Accepted Answer · 2017-07-08 03:01:31Z

7

    List list1 = new ArrayList();
    list1.add("one");
    list1.add("three");
    list1.add("four");

    List list2 = new ArrayList();
    list2.add("one");
    list2.add("two");
    list2.add("three");
    list2.add("four");
    list2.add("five");


    list2.stream().filter( x -> !list1.contains(x) ).forEach(x -> System.out.println(x));

The output is:

two
five

edited Jul 8, 2017 at 3:01

Alberto Trindade Tavares

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answered Jul 7, 2017 at 22:02

Abhi

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Comments

Adam reuben · Accepted Answer · 2021-10-19 22:11:59Z

*public class Demo {
   public static void main(String[] args) {
      List aList = new ArrayList();
      aList.add("A");
      aList.add("B");
      aList.add("C");
      aList.add("D");
      aList.add("E");
      if(aList.contains("C"))
         System.out.println("The element C is available in the ArrayList");
      else
         System.out.println("The element C is not available in the ArrayList");
      if(aList.contains("H"))
         System.out.println("The element H is available in the ArrayList");
      else
         System.out.println("The element H is not available in the ArrayList");
   }
}*

The Output will be:

The element C is available in the ArrayList The element H is not available in the ArrayList

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