Develop Reference

Regex to split a string when there's nothing between two occurrences of the delimiter

Suppose I want to split this string a^b^c^d^e^^^f^g^h^^^ , I simply do string.split("\\^") which returns me an array of length 10 i.e [a, b, c, d, e, , , f, g, h] . However I want an array of length 13 which takes occurrences of the delimiter after h into consideration.
I can do something like this to achieve what I want
string = string.replace("^", "^ ");
String[] split = string.split("\\^");
for(String x : split){
System.out.println(x.trim());
}
but this seems like an overburden. Is there a regex to do this?

You can do this
String[] split = string.split("\\^", -1);
and it won't drop trailing separators.
If you really want to trim the last separator to get 12 values, you can do
String[] split = string.replaceAll("\\^$", "").split("\\^", -1);
or this will ensure you always have 12. It will either trim or expand the number of elements as required (adding null if expanding)
String[] split = Arrays.copyOf(string.split("\\^", -1), 12);

why split() produces extra , after sets limit -1

I want to split Area Code and preceding number from Telephone number without brackets so i did this.
String pattern = "[\\(?=\\)]";
String b = "(079)25894029".trim();
String c[] = b.split(pattern,-1);
for (int a = 0; a < c.length; a++)
System.out.println("c[" + a + "]::->" + c[a] + "\nLength::->"+ c[a].length());
Output:
c[0]::-> Length::->0
c[1]::->079 Length::->3
c[2]::->25894029 Length::->8
Expected Output:
c[0]::->079 Length::->3
c[1]::->25894029 Length::->8
So my question is why split() produces and extra blank at the start, e.g
[, 079, 25894029]. Is this its behavior, or I did something go wrong here?
How can I get my expected outcome?

First you have unnecessary escaping inside your character class. Your regex is same as:
String pattern = "[(?=)]";
Now, you are getting an empty result because ( is the very first character in the string and split at 0th position will indeed cause an empty string.
To avoid that result use this code:
String str = "(079)25894029";
toks = (Character.isDigit(str.charAt(0))? str:str.substring(1)).split( "[(?=)]" );
for (String tok: toks)
System.out.printf("<<%s>>%n", tok);
Output:
<<079>>
<<25894029>>

From the Java8 Oracle docs:
When there is a positive-width match at the beginning of this string
then an empty leading substring is included at the beginning of the
resulting array. A zero-width match at the beginning however never
produces such empty leading substring.
You can check that the first character is an empty string, if yes then trim that empty string character.

Your regex has problems, as does your approach - you can't solve it using your approach with any regex. The magic one-liner you seek is:
String[] c = b.replaceAll("^\\D+|\\D+$", "").split("\\D+");
This removes all leading/trailing non-digits, then splits on non-digits. This will handle many different formats and separators (try a few yourself).
See live demo of this:
String b = "(079)25894029".trim();
String[] c = b.replaceAll("^\\D+|\\D+$", "").split("\\D+");
System.out.println(Arrays.toString(c));
Producing this:
[079, 25894029]

Split string containing newline characters Java

Say I have a following string str:
GTM =0.2
Test =100
[DLM]
ABCDEF =5
(yes, it contains newline characters) That I am trying to split with [DLM] delimiter substring like this:
String[] strArr = str.split("[DLM]");
Why is it that when I do:
System.out.print(strArr[0]);
I get this output: GT
and when I do
System.out.print(strArr[1]);
I get =0.2
Does this make any sense at all?

str.split("[DLM]"); should be str.split("\\[DLM\\]");
Why?
[ and ] are special characters and String#split accepts regex.
A solution that I like more is using Pattern#quote:
str.split(Pattern.quote("[DLM]"));
quote returns a String representation of the given regex.

Yes, you're giving a regex which says "split with either D, or L, or M".
You should escape those boys like this: str.split("\[DLM\]");
It's being split at the first M.

Escape the brackets
("\\[DLM\\]")
When you use brackets inside the " ", it reads it as, each character inside of the brackets is a delimiter. So in your case, M was a delimiter

use
String[] strArr = str.split("\\[DLM]\\");
Instead of
String[] strArr = str.split("[DLM]");
Other wise it will split with either D, or L, or M.

Writing a regex in java. Using the string.split() method. I want it to stop splitting after the first occurrence of '('

I have strings with this format: "a,b, c,d" and this format: "a(b,c,d)"
I want to split on ',' or ', ' but I want to terminate splitting when I encounter the '(' in the second format.
This is what I had before I started hacking.
String [] stringArray = string.split(", |,");
The array of the first format would contain: 'a', 'b', 'c', 'd'
The array of the second format would conaint 'a(b,c,d)'
Example:
String string1 = "ab,cd, de";
String string2 = "ab(de,ef);
String [] array1 = string1.split(...);
String [] array2 = string2.split(...);
array1 result: ["ab" "cd" "de"]
array2 result: ["ab(de,ef)"]
The number of characters between the commas are not limited. I hope this is more clear.
Thanks.

If you know the parentheses are always properly balanced and they'll never be nested inside other parens, this will work:
String[] result = source.split(",\\s*(?![^()]*\\))");
If the lookahead finds a ) without seeing a ( first, it must be inside a pair of parens. Given this string:
"ab,cd, de,ef(gh,ij), kl,mn"
...result will be:
["ab", "cd", "de", "ef(gh,ij)", "kl", "mn"]

I think what you could need is a negative lookbehind; according to the doc, Java regex are like (more or less) Perl regex; but variable length lookbehind is not implemented in Perl, so that (?<!\(.*),\s* won't work (it would match comma followed by any number of spaces or no space, and not preceded by a ( followed by anything, i.e. would match comma only if not preceded by a ().
I believe the easiest thing is to split on the first ( occurrence (you can avoid regex to do so) and treat the two resulting segments differently, splitting the first on , and adding to the final array the second (prepended with the maybe lost ().
EDIT
since "a(b,d)" should give "a(b,d)", you must append whatever comes after ( (included) to the last splited string from the "first" segment. However, the concept is as written before.

Use the indexOf() method.
Initially, check if a string has a "(".
index = string.indexOf('(');
if(index ==-1) // it means there is no '('
{
string.split(...);
}
else
{
subString = string.subString(0,index); // the part of the string before the '('
// now do the following-
// 1. proceed with split on substring
// array1 = substring.split(...)
// 2. Create a new array, insert the elements of array1 in it,
// followed by the remaining part of the string
// array2 = combine(array1, string.subString(index+1)); // <-- you will need to write this method
}

Apply regex conditions after certain character

I have a string that i want to parse into an array.
The given string has the form P[AB, AC, AD] (A1, A2, A3).
I want to store it in an array like this so that all the data after the first ( will be filtered by the regex conditions:
P[AB, AC, AD]
A1
A2
A3
This is what i came up with:
String regex = "/^(([,()]+)$";
String[] numbers = stringIn.split(regex);
My problem is that it simply does not work because the regex won't filter out the pieces, everything is stored at numbers[0].

I think this is what you were trying for:
String regex = "[ ,()]+(?=[^\\]]*$)";
String stringIn = "P[AB, AC, AD] (A1, A2, A3)";
String[] numbers = stringIn.split(regex);
for (String n : numbers)
{
System.out.println(n);
}
output:
P[AB, AC, AD]
A1
A2
A3
[ ,()]+ part tries match spaces, commas and parentheses wherever they appear, but (?=[^\\]]*$) (a positive lookahead) filters out any match before the ]. I'm assuming there's only the one set of square brackets in the string.

Thought of a less elegant, but possibly easier to follow, way of doing this:
// Split the string into two parts.
// parts[0] == "P[AB, AC, AD]"
// parts[1] == "A1, A2, A3"
String[] parts = stringIn.split(" *\\(|\\)");
// Split the second part into its components.
String[] secondParts = stringInParts[1].split(", *");
// Combine the results.
String[] numbers = new String[secondParts.length + 1];
numbers[0] = parts[0];
System.arrayCopy(secondParts, 0, numbers, 1, secondParts.length);

You need to escape the parenthesis that you want to match literally, and you don't need back-slashes around the body, and you probably want to match instead of splitting.
Pattern regex = Pattern.compile("\\(([^)]+)");
Matcher m = regex.matcher(stringIn);
if (m.find()) {
String[] numbers = m.group(1).split("[,\\s]+");
}
Finally, unlike in JavaScript and C# and Python, $ in Java does not match the end of the input. You need to use \\z instead. $ in Java matches at the end of input or just before the last linebreak if there is a linebreak at the end of the input.