CODEWITHSUNDEEP
javastring
Bluefire
Bluefire

Reputation: 14099

Get string character by index

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.

Upvotes: 287

Views: 1182384

Answers (13)

Samiksha Jagtap
Samiksha Jagtap

Reputation: 603

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

Upvotes: 0

uzairshahid901
uzairshahid901

Reputation: 39

CharAt function not working

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

This code working fine

Upvotes: -1

Ababneh A
Ababneh A

Reputation: 1134

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

Upvotes: -1

Thiago Silva
Thiago Silva

Reputation: 796

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

Upvotes: 1

Ricardo Altamirano
Ricardo Altamirano

Reputation: 15198

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.

Upvotes: 406

silversk8terz
silversk8terz

Reputation: 91

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);
    }
}

Upvotes: 4

Sylvain Leroux
Sylvain Leroux

Reputation: 51990

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 π“ƒ₯ jumps over the lazy π“Šƒπ“Ώπ“…“π“ƒ‘";

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   π“ƒ₯   j u m p s   o v e r   t h e   l a z y   π“Šƒ 𓍿 π“…“ 𓃑 

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

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.

Upvotes: 12

Aqif Hamid
Aqif Hamid

Reputation: 3521

It is as simple as:

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

Upvotes: 5

Francisco Spaeth
Francisco Spaeth

Reputation: 23893

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.

Upvotes: 9

Kadar Kabouya Kalo
Kadar Kabouya Kalo

Reputation: 9

Like this:

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

Upvotes: -3

ametren
ametren

Reputation: 2246

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.

Upvotes: 56

fvu
fvu

Reputation: 32953

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.

Upvotes: 5

Thomas
Thomas

Reputation: 5094

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.

Upvotes: 8

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