Get string character by index
JavaStringJava Problem Overview
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.
Java Solutions
Solution 1 - Java
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.
Solution 2 - Java
You want .charAt()
"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.
Solution 3 - Java
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₯ jumps over the lazy ππΏπ
π‘";
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₯ j u m p s o v e r t h e l a z y π πΏ π
π‘
Using Java 8 stream (works as expected)
T h e q u i c k b r o w n π₯ j u m p s o v e r t h e l a z y π πΏ π
π‘
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.
Solution 4 - Java
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.
Solution 5 - Java
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.
Solution 6 - Java
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.
Solution 7 - Java
It is as simple as:
String charIs = string.charAt(index) + "";
Solution 8 - Java
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);
}
}
Solution 9 - Java
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
Solution 10 - Java
CodePointAt instead of charAt is safer to use. charAt may break when there are emojis in the strtng.
Solution 11 - Java
CharAt function not working
Edittext.setText(YourString.toCharArray(),0,1);
This code working fine
Solution 12 - Java
Like this:
String a ="hh1hhhhhhhh";
char s = a.charAt(3);