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String.split() not working as intended

I'm trying to split a string, however, I'm not getting the expected output.
String one = "hello 0xA0xAgoodbye";
String two[] = one.split(" |0xA");
System.out.println(Arrays.toString(two));
Expected output: [hello, goodbye]
What I got: [hello, , , goodbye]
Why is this happening and how can I fix it?
Thanks in advance! ^-^

If you'd like to treat consecutive delimiters as one, you could modify your regex as follows:
"( |0xA)+"
This means "a space or the string "0xA", repeated one or more times".

(\\s|0xA)+ This will match one or more number of space or 0xA in the text and split them

This result is caused by multiple consecutive matches in the string. You may wrap the pattern with a grouping construct and apply a + quantifier to it to match multiple matches:
String one = "hello 0xA0xAgoodbye";
String two[] = one.split("(?:\\s|0xA)+");
System.out.println(Arrays.toString(two));
A (?:\s|0xA)+ regex matches 1 or more whitespace symbols or 0XA literal character sequences.
See the Java online demo.
However, you will still get an empty value as the first item in the resulting array if the 0xA or whitespaces appear at the start of the string. Then, you will have to remove them first:
String two[] = one.replaceFirst("^(?:\\s|0xA)+", "").split("(?:\\s+|0xA)+");
See another Java demo.

How can we remove a ':' characters from a string?

I have strings like
#lle #mme: #crazy #upallnight:
I would like to remove the words which starts with either # or #. It works perfectly fine if those words doesn't contain the ':' character. However, that ':' character is left whenever I delete the words. Therefore I decided to replace those ':' characters before I delete the words using a string.replace() function. However, they are still not removed.
String example = "#lle #mme: #crazy #upallnight:";
example.replace(':',' ');
The result : #lle #mme: #crazy #upallnight:
I am pretty stuck here, anyhelp would be appreciated.

You can do this:
example = example.replaceAll(" +[##][^ ]+", "");
What this will do is replace any substrings in your string that match the regex pattern [##][^ ]+ with the empty string. Since that pattern matches the words you want to dump, it'll do what you want.
Demo of the pattern on Regex101

From Java docs:
String s = "Abc: abc#:";
String result = s.replace(':',' ');
Output in variable result= Abc abc#
I think you forgot to store the returned result of replace() method in some other String variable.

Split a string in Java containing (

I am trying to split a string in Java.
For example
Hello (1234)
The part after ( must not be included in the string.
I am expecting the following:
Hello
How would you do it?

Just split according to
zero or more spaces.
And the following ( character.
Then get the value from index 0 of splitted parts.
"Hello (1234)".split("\\s*\\(")[0];
or
"Hello (1234)".split("\\s+")[0];

You can replace the contents in the parenthesis by nothing.
String str = "Hello(1234)";
String result = str.replaceAll("\\(.*\\)", "");
System.out.println(result);
You mention split operation in the question, but you say
The part after ( must not be included in the string. I am expecting
the following:
So I'm assuming you are discarding the (1234) ? If you need to save it, consider the other answer (using split)

You may try the following regex
String[] split = "Hello (1234)".split("\\s*\\([\\d]*\\)*");
System.out.println(split[0]);
\\s* space may be ignored
\\( ( or left parenthesis has special meaning in regex so \\( means only (
Same is the case with ) or right parenthesis \\)
\\d* digits may be present
You may use + instead of * if that character is present at-least once

Remove unwanted characters from string by regex in Java

I have a string here:
javax.swing.JLabel[,380,30,150x25,alignmentX=0.0,alignmentY=0.0]: Hello
I want to remove everything before the ":", including the ":" itself. This would leave only "Hello". I read about regex, but no combination I tried worked. Can someone tell me how to do it. Thanks in advance!

You need to use replaceAll method or replaceFirst.
string.replaceFirst(".*:\\s*", "");
or
string.replaceAll(".*:\\s*", "");
This would give you only Hello. If you remove \\s* pattern,then it would give you <space>Hello string.
.* Matches any character zero or more times, greedily.
: Upto the colon.
\\s* Matches zero or more space characters.

You could also just split the string by : and take the second string. Like this
String sample = "javax.swing.JLabel[,380,30,150x25,alignmentX=0.0,alignmentY=0.0]: Hello";
System.out.println(sample.split(":", -1)[1]);
This will output
<space>Hello
If you want to get rid of that leading space just trim it off like
System.out.println(sample.split(":", -1)[1].trim());

Remove repeating set of characters in a string

I want to remove the sequesnce "-~-~-" if it repeats in a string, but only if they are together.
I have tried to create a regex based on the removing of multiple white spaces regex:
test.replaceAll("\\s+", " ");
Unfortunately I was unsuccessful. Can someone please help me write the correct regex? thanks.
Example:
string test = "hello-~-~--~-~--~-~-"
output:
hello-~-~-
Another example
string test = "-~-~--~-~--~-~-hello-~-~--~-~--~-~-"
output:
-~-~-hello-~-~-

The regex is:
test.replaceAll("(-~-~-){2,}", "-~-~-")
replaceAll replaces all occurrences matched by the regex (the first parameter) with the second parameter.
the () groups the expression -~-~- together, {2,} means two or more occurrences.
EDIT
Like #anubhava said, instead of using -~-~- for the replacement string, you could also use $1 which backreferences the first capturing group (i.e. the expression in the regex surrounded by ()).

test.replaceAll("(-~-~-)+", "-~-~-");

This is the regex you need:
(-~-~-){2}