280

I know how to work out the index of a certain character or number in a string, but is there any predefined method I can use to give me the character at the nth position? So in the string "foo", if I asked for the character with index 0 it would return "f".

Note - in the above question, by "character" I don't mean the char data type, but a letter or number in a string. The important thing here is that I don't receive a char when the method is invoked, but a string (of length 1). And I know about the substring() method, but I was wondering if there was a neater way.

Mark Rotteveel
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Bluefire
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    It is? The answer is pretty straightforward. – ametren Jun 27 '12 at 15:41
  • Did you notice he doesn't want a `char` value? And he knows how to do `substring()` but just wants a "neater" way. FYI, I can say that `substring()` is the neatest way. – user845279 Jun 27 '12 at 15:43
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    @user845279 `Character.toString` fulfills all the necessary requirements and isn't messy at all. – Ricardo Altamirano Jun 27 '12 at 15:46
  • @pythonscript I agree, but it isn't much different from using `substring()` directly. – user845279 Jun 27 '12 at 15:48
  • @user845279 - substring has the syntax `substring(index, index + 1)`. I'd like to avoid the "index + 1" bit. – Bluefire Jun 27 '12 at 15:59
  • @Bluefire If that's what you're trying to avoid, my answer should solve that. `substring` isn't ideal because if you want the last letter in the string, you can't use `index + 1` because it will throw an exception. – Ricardo Altamirano Jun 27 '12 at 16:01
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    I'm late to this party, but @RicardoAltamirano is a bit mistaken. The `endIndex` (second parameter) of `String.substring(int, int)` is an **exclusive** index, and it _won't_ throw an exception for `index + 1` as long as `index < length()` -- which is true even for the last character in the string. – William Price May 14 '15 at 02:48

13 Answers13

396

The method you're looking for is charAt. Here's an example:

String text = "foo";
char charAtZero = text.charAt(0);
System.out.println(charAtZero); // Prints f

For more information, see the Java documentation on String.charAt. If you want another simple tutorial, this one or this one.

If you don't want the result as a char data type, but rather as a string, you would use the Character.toString method:

String text = "foo";
String letter = Character.toString(text.charAt(0));
System.out.println(letter); // Prints f

If you want more information on the Character class and the toString method, I pulled my info from the documentation on Character.toString.

Matthias Braun
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Ricardo Altamirano
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    "The important thing here is that I don't receive a char when the method is invoked, but a string", but thanks anyway (upvote) :D – Bluefire Jun 27 '12 at 15:45
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    I think Sylvain Leroux's answer is better.[doc about Character](https://docs.oracle.com/javase/7/docs/api/java/lang/Character.html#unicode) – lin Feb 08 '18 at 07:08
  • I agree with @ChaojunZhong [this](https://stackoverflow.com/a/33588633/3555586) is a more suitable answer since it is not advisable to use charAt() cause you would be having problems when you have characters that needs 2 code units. – bpunzalan Jun 19 '19 at 13:53
53

You want .charAt()

Here's a tutorial

"mystring".charAt(2)

returns s

If you're hellbent on having a string there are a couple of ways to convert a char to a string:

String mychar = Character.toString("mystring".charAt(2));

Or

String mychar = ""+"mystring".charAt(2);

Or even

String mychar = String.valueOf("mystring".charAt(2));

For example.

ametren
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  • @ametren Is string string concatenation preferable to `Character.toString`? – Ricardo Altamirano Jun 27 '12 at 15:47
  • I think that may come down to a matter of personal preference. You could also do `String mychar = String.valueOf("mystring".charAt(2));` – ametren Jun 27 '12 at 15:48
  • To pile on, my personal preference in this case would be `String mychar = ""+"mystring".charAt(2);` because it's the most concise. Others will differ in their opinion on this. – ametren Jun 27 '12 at 15:54
10

None of the proposed answers works for surrogate pairs used to encode characters outside of the Unicode Basic Multiligual Plane.

Here is an example using three different techniques to iterate over the "characters" of a string (incl. using Java 8 stream API). Please notice this example includes characters of the Unicode Supplementary Multilingual Plane (SMP). You need a proper font to display this example and the result correctly.

// String containing characters of the Unicode 
// Supplementary Multilingual Plane (SMP)
// In that particular case, hieroglyphs.
String str = "The quick brown  jumps over the lazy ";

Iterate of chars

The first solution is a simple loop over all char of the string:

/* 1 */
System.out.println(
        "\n\nUsing char iterator (do not work for surrogate pairs !)");
for (int pos = 0; pos < str.length(); ++pos) {
    char c = str.charAt(pos);
    System.out.printf("%s ", Character.toString(c));
    //                       ^^^^^^^^^^^^^^^^^^^^^
    //                   Convert to String as per OP request
}

Iterate of code points

The second solution uses an explicit loop too, but accessing individual code points with codePointAt and incrementing the loop index accordingly to charCount:

/* 2 */
System.out.println(
        "\n\nUsing Java 1.5 codePointAt(works as expected)");
for (int pos = 0; pos < str.length();) {
    int cp = str.codePointAt(pos);

    char    chars[] = Character.toChars(cp);
    //                ^^^^^^^^^^^^^^^^^^^^^
    //               Convert to a `char[]`
    //               as code points outside the Unicode BMP
    //               will map to more than one Java `char`
    System.out.printf("%s ", new String(chars));
    //                       ^^^^^^^^^^^^^^^^^
    //               Convert to String as per OP request

    pos += Character.charCount(cp);
    //     ^^^^^^^^^^^^^^^^^^^^^^^
    //    Increment pos by 1 of more depending
    //    the number of Java `char` required to
    //    encode that particular codepoint.
}

Iterate over code points using the Stream API

The third solution is basically the same as the second, but using the Java 8 Stream API:

/* 3 */
System.out.println(
        "\n\nUsing Java 8 stream (works as expected)");
str.codePoints().forEach(
    cp -> {
        char    chars[] = Character.toChars(cp);
        //                ^^^^^^^^^^^^^^^^^^^^^
        //               Convert to a `char[]`
        //               as code points outside the Unicode BMP
        //               will map to more than one Java `char`
        System.out.printf("%s ", new String(chars));
        //                       ^^^^^^^^^^^^^^^^^
        //               Convert to String as per OP request
    });

Results

When you run that test program, you obtain:

Using char iterator (do not work for surrogate pairs !)
T h e   q u i c k   b r o w n   ? ?   j u m p s   o v e r   t h e   l a z y   ? ? ? ? ? ? ? ? 

Using Java 1.5 codePointAt(works as expected)
T h e   q u i c k   b r o w n      j u m p s   o v e r   t h e   l a z y       

Using Java 8 stream (works as expected)
T h e   q u i c k   b r o w n      j u m p s   o v e r   t h e   l a z y       

As you can see (if you're able to display hieroglyphs properly), the first solution does not handle properly characters outside of the Unicode BMP. On the other hand, the other two solutions deal well with surrogate pairs.

Peter Mortensen
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Sylvain Leroux
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9

You could use the String.charAt(int index) method result as the parameter for String.valueOf(char c).

String.valueOf(myString.charAt(3)) // This will return a string of the character on the 3rd position.
Peter Mortensen
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Francisco Spaeth
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8

You're pretty stuck with substring(), given your requirements. The standard way would be charAt(), but you said you won't accept a char data type.

Thomas
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  • Fair enough. But, since char is a primitive type, I assume `toString()` won't work on it, and `valueOf()` is only for numbers (I think, I may be wrong), so how do I convert a char to a string? – Bluefire Jun 27 '12 at 15:43
  • "in the above question, by "character" I don't mean the char data type" -- I don't read this as "I won't accept a `char`" – ametren Jun 27 '12 at 15:43
  • @Bluefire See my answer. `Character.toString` should work (it's a static method from the [`Character`](http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/Character.html) class. – Ricardo Altamirano Jun 27 '12 at 15:45
5

A hybrid approach combining charAt with your requirement of not getting char could be

newstring = String.valueOf("foo".charAt(0));

But that's not really "neater" than substring() to be honest.

fvu
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5

It is as simple as:

String charIs = string.charAt(index) + "";
Peter Mortensen
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Aqif Hamid
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4

Here's the correct code. If you're using zybooks this will answer all the problems.

for (int i = 0; i<passCode.length(); i++)
{
    char letter = passCode.charAt(i);
    if (letter == ' ' )
    {
        System.out.println("Space at " + i);
    }
}
Peter Mortensen
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silversk8terz
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1

if someone is strugling with kotlin, the code is:

var oldStr: String = "kotlin"
var firstChar: String = oldStr.elementAt(0).toString()
Log.d("firstChar", firstChar.toString())

this will return the char in position 1, in this case k remember, the index starts in position 0, so in this sample: kotlin would be k=position 0, o=position 1, t=position 2, l=position 3, i=position 4 and n=position 5

Thiago Silva
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0

I come across this question yeasterday and I am aware this has accepted answer, just want to add one more solution to this in javascript scenario which can help people like me who are looking for this -

let name = 'Test'
console.log(name[2])

// Here at index 2 we have 's' value and this will simply give the expected output
Samiksha Jagtap
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-1

CodePointAt instead of charAt is safer to use. charAt may break when there are emojis in the strtng.

Ababneh A
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-1

CharAt function not working

Edittext.setText(YourString.toCharArray(),0,1);

This code working fine

Hkachhia
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-3

Like this:

String a ="hh1hhhhhhhh";
char s = a.charAt(3);
Leistungsabfall
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  • OP stated that a `String` of length 1 is desired, **not** a `char`. – William Price May 13 '15 at 22:17
  • The 6 other answers, including the accepted one, suggested `charAt()` as a possible solution. What does this answer add? – Dan Getz May 14 '15 at 00:26
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    Also, it looks like you're hinting that `charAt()` uses 1-based indices, by having the only different character in `a` in the third position. If that were true, then it would be better for you to say or explain that than to hint at it. In reality that's *not* true: `charAt()` uses 0-based indices, so `s` will be `'h'`. – Dan Getz May 14 '15 at 12:47