255
votes

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.

12
It is? The answer is pretty straightforward.ametren
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
@user845279 Character.toString fulfills all the necessary requirements and isn't messy at all.Ricardo Altamirano
@pythonscript I agree, but it isn't much different from using substring() directly.user845279
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

12 Answers

373
votes

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.

47
votes

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.

10
votes

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 𓃥 jumps over the lazy 𓊃𓍿𓅓𓃡";

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   𓃥   j u m p s   o v e r   t h e   l a z y   𓊃 𓍿 𓅓 𓃡 

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

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.

8
votes

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.

8
votes

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.
5
votes

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.

5
votes

It is as simple as:

String charIs = string.charAt(index) + "";
4
votes

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);
    }
}
0
votes

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

0
votes

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

0
votes

CharAt function not working

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

This code working fine

-3
votes

Like this:

String a ="hh1hhhhhhhh";
char s = a.charAt(3);