How do I split a string in Java? - Stack Overflow
How to create and initialise in Java an array of String arrays in one line? - Stack Overflow
Java Strings: "String s = new String("silly");" - Stack Overflow
JEP 430: String Templates (Preview)
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Use the appropriately named method String#split().
String string = "004-034556";
String[] parts = string.split("-");
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556
Note that split's argument is assumed to be a regular expression, so remember to escape special characters if necessary.
there are 12 characters with special meanings: the backslash
\, the caret^, the dollar sign$, the period or dot., the vertical bar or pipe symbol|, the question mark?, the asterisk or star*, the plus sign+, the opening parenthesis(, the closing parenthesis), and the opening square bracket[, the opening curly brace{, These special characters are often called "metacharacters".
For instance, to split on a period/dot . (which means "any character" in regex), use either backslash \ to escape the individual special character like so split("\\."), or use character class [] to represent literal character(s) like so split("[.]"), or use Pattern#quote() to escape the entire string like so split(Pattern.quote(".")).
String[] parts = string.split(Pattern.quote(".")); // Split on the exact string.
To test beforehand if the string contains certain character(s), just use String#contains().
if (string.contains("-")) {
// Split it.
} else {
throw new IllegalArgumentException("String " + string + " does not contain -");
}
Note, this does not take a regular expression. For that, use String#matches() instead.
If you'd like to retain the split character in the resulting parts, then make use of positive lookaround. In case you want to have the split character to end up in left hand side, use positive lookbehind by prefixing ?<= group on the pattern.
String string = "004-034556";
String[] parts = string.split("(?<=-)");
String part1 = parts[0]; // 004-
String part2 = parts[1]; // 034556
In case you want to have the split character to end up in right hand side, use positive lookahead by prefixing ?= group on the pattern.
String string = "004-034556";
String[] parts = string.split("(?=-)");
String part1 = parts[0]; // 004
String part2 = parts[1]; // -034556
If you'd like to limit the number of resulting parts, then you can supply the desired number as 2nd argument of split() method.
String string = "004-034556-42";
String[] parts = string.split("-", 2);
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556-42
An alternative to processing the string directly would be to use a regular expression with capturing groups. This has the advantage that it makes it straightforward to imply more sophisticated constraints on the input. For example, the following splits the string into two parts, and ensures that both consist only of digits:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
class SplitExample
{
private static Pattern twopart = Pattern.compile("(\\d+)-(\\d+)");
public static void checkString(String s)
{
Matcher m = twopart.matcher(s);
if (m.matches()) {
System.out.println(s + " matches; first part is " + m.group(1) +
", second part is " + m.group(2) + ".");
} else {
System.out.println(s + " does not match.");
}
}
public static void main(String[] args) {
checkString("123-4567");
checkString("foo-bar");
checkString("123-");
checkString("-4567");
checkString("123-4567-890");
}
}
As the pattern is fixed in this instance, it can be compiled in advance and stored as a static member (initialised at class load time in the example). The regular expression is:
(\d+)-(\d+)
The parentheses denote the capturing groups; the string that matched that part of the regexp can be accessed by the Match.group() method, as shown. The \d matches and single decimal digit, and the + means "match one or more of the previous expression). The - has no special meaning, so just matches that character in the input. Note that you need to double-escape the backslashes when writing this as a Java string. Some other examples:
([A-Z]+)-([A-Z]+) // Each part consists of only capital letters
([^-]+)-([^-]+) // Each part consists of characters other than -
([A-Z]{2})-(\d+) // The first part is exactly two capital letters,
// the second consists of digits
You have forgotten to give the variable a name? Both of these work.
String[][] array = {emptyArr, filledArr, moreFilledArr};
String[][] array = new String[][] {emptyArr, filledArr, moreFilledArr};
I recommend you to go through the basic Java syntax specification and tutorials. Start with The Java Tutorials by Oracle Corp, free of cost.
these two are right too String[][] array = {emptyArr, filledArr, moreFilledArr};
String[][] array = new String[][] {emptyArr, filledArr, moreFilledArr};
please do check these too
Docs for Array in Java Have a Look for more clarification
tutorial of array with java docs
,
I had to add this as a comment but I couldn't, I want to extend with more resources mentioned above , Thanks
String is a special built-in class of the language. It is for the String class only in which you should avoid saying
String s = new String("Polish");
Because the literal "Polish" is already of type String, and you're creating an extra unnecessary object. For any other class, saying
CaseInsensitiveString cis = new CaseInsensitiveString("Polish");
is the correct (and only, in this case) thing to do.
I believe the main benefit of using the literal form (ie, "foo" rather than new String("foo")) is that all String literals are 'interned' by the VM. In other words it is added to a pool such that any other code that creates the same string will use the pooled String rather than creating a new instance.
To illustrate, the following code will print true for the first line, but false for the second:
System.out.println("foo" == "foo");
System.out.println(new String("bar") == new String("bar"));