We Keep Coding

Regular expression to not to split if it's inside a particular pattern

I have a custom string which is a streamed version of multiple item. Item can be Integer, String and List of String where each item
represented by I,S,L respectively.
For eg: String d = "I:123;S:345;L:{S:45;S:67;S:789};I:23";
I am trying to write a single regular expression which could split this string and should produce 4 items
1. I:123, 2. S:345, 3.L:{s:45;s:67;s:789}; ,4:I:23
If I just split based on ;then it will split items List of also that should not happen.Then I tried to write some complex regular expression but nothing worked.
Could some one please give some pointer?

You may use this lookahead based regex for splitting:
String[] arr = str.split(";(?![^{}]*})");
RegEx Demo
Details:
;: Match literal ;
(?![^{}]*}): Negative lookahead to make sure we don'r have a } after non { and } characters

You should use the lazy quantifiers:
(.+?);(.+?);(.*)
Links:
https://regex101.com/r/3nrRxP/1
https://www.regular-expressions.info/repeat.html

String split method returning first element as empty using regex

I'm trying to get the digits from the expression [1..1], using Java's split method. I'm using the regex expression ^\\[|\\.{2}|\\]$ inside split. But the split method returning me String array with first value as empty, and then "1" inside index 1 and 2 respectively. Could anyone please tell me what's wrong I'm doing in this regex expression, so that I only get the digits in the returned String array from split method?

You should use matching. Change your expression to:
`^\[(.*?)\.\.(.*)\]$`
And get your results from the two captured groups.
As for why split acts this way, it's simple: you asked it to split on the [ character, but there's still an "empty string" between the start of the string and the first [ character.

Your regex is matching [ and .. and ]. Thus it will split at this occurrences.
You should not use a split but match each number in your string using regex.

You've set it up such that [, ] and .. are delimiters. Split will return an empty first index because the first character in your string [1..1] is a delimiter. I would strip delimiters from the front and end of your string, as suggested here.
So, something like
input.replaceFirst("^[", "").split("^\\[|\\.{2}|\\]$");
Or, use regex and regex groups (such as the other answers in this question) more directly rather than through split.

Why not use a regex to capture the numbers? This will be more effective less error prone. In that case the regex looks like:
^\[(\d+)\.{2}(\d+)\]$
And you can capture them with:
Pattern pat = Pattern.compile("^\\[(\\d+)\\.{2}(\\d+)\\]$");
Matcher matcher = pattern.matcher(text);
if(matcher.find()) { //we've found a match
int range_from = Integer.parseInt(matcher.group(1));
int range_to = Integer.parseInt(matcher.group(2));
}
with range_from and range_to the integers you can no work with.
The advantage is that the pattern will fail on strings that make not much sense like ..3[4, etc.

java Regex - split but ignore text inside quotes?

using only regular expression methods, the method String.replaceAll and ArrayList
how can i split a String into tokens, but ignore delimiters that exist inside quotes?
the delimiter is any character that is not alphanumeric or quoted text
for example:
The string :
hello^world'this*has two tokens'
should output:
hello
worldthis*has two tokens

I know there is a damn good and accepted answer already present but I would like to add another regex based (and may I say simpler) approach to split the given text using any non-alphanumeric delimiter which not inside the single quotes using
Regex:
/(?=(([^']+'){2})*[^']*$)[^a-zA-Z\\d]+/
Which basically means match a non-alphanumeric text if it is followed by even number of single quotes in other words match a non-alphanumeric text if it is outside single quotes.
Code:
String string = "hello^world'this*has two tokens'#2ndToken";
System.out.println(Arrays.toString(
string.split("(?=(([^']+'){2})*[^']*$)[^a-zA-Z\\d]+"))
);
Output:
[hello, world'this*has two tokens', 2ndToken]
Demo:
Here is a live working Demo of the above code.

Use a Matcher to identify the parts you want to keep, rather than the parts you want to split on:
String s = "hello^world'this*has two tokens'";
Pattern pattern = Pattern.compile("([a-zA-Z0-9]+|'[^']*')+");
Matcher matcher = pattern.matcher(s);
while (matcher.find()) {
System.out.println(matcher.group(0));
}
See it working online: ideone

You cannot in any reasonable way. You are posing a problem that regular expressions aren't good at.

Do not use a regular expression for this. It won't work. Use / write a parser instead.
You should use the right tool for the right task.

Split a string with regular expression

How can I split string for 3 character? (I don't want to do loop for this, maybe some regular expression will be help)
I give example:
String str = "111222333444";
String[] result = str.split("help?"); // get "111", "222", "333"

Using guava-library
Iterable<String> strNums = Splitter.fixedLength(3).split("111222333444")
Readable than using regex. You can then use Ints.tryParse(...) to get Integer version if you want.

Using .split will match regular expressions in the string which, in the underlying implementation, involves traversing the entire string anyway. Writing a simple loop to just create a token from every 3 characters would probably be more efficient.

Frankly, I don't think you can do it for a string of undefined length, without a loop.

You can not use split because the arg of split is the separator, not the resulting sub-strings.
So, your separator regex would be nothing !?
Sorry, you heve write loop. BTW, the regex engine for splitis full of loops.

Problem replacing words using [^a-zA-Z] regex

Just could not get this one and googling did not help much either..
First something that I know: Given a string and a regex, how to replace all the occurrences of strings that matches this regular expression by a replacement string ? Use the replaceAll() method in the String class.
Now something that I am unable to do. The regex I have in my code now is [^a-zA-Z] and I know for sure that this regex is definitely going to have a range. Only some more characters might be added to the list. What I need as output in the code below is Worksheet+blah but what I get using replaceAll() is Worksheet++++blah
String homeworkTitle = "Worksheet%#5_blah";
String unwantedCharactersRegex = "[^a-zA-Z]";
String replacementString = "+";
homeworkTitle = homeworkTitle.replaceAll(unwantedCharactersRegex,replacementString);
System.out.println(homeworkTitle);
What is the way to achieve the output that I wish for? Are there any Java methods that I am missing here?

[^a-zA-Z]+
Will do it nicely.
You just need a greedy quantifier in order to match as many non-alphabetical characters you can, and replace the all match by one '+' (a - by default - greedy quantifier)
Note: [^a-zA-Z]+? would make the '+' quantifier lazy, and would have give you the same result than [^a-zA-Z], since it would only have matched only one non-alphabetical character at a time.

String unwantedCharactersRegex = "[^a-zA-Z]"
This matches a single non-letter. So each single non-letter is replaced by a +. You need to say "one or more", so try
String unwantedCharactersRegex = "[^a-zA-Z]+"