Develop Reference

Java String Regex replacement

Sample Input:
a:b
a.in:b
asds.sdsd:b
a:b___a.sds:bc___ab:bd
Sample Output:
a:replaced
a.in:replaced
asds.sdsd:replaced
a:replaced___a.sds:replaced___ab:replaced
String which comes after : should be replaced with custom function.
I have done the same without Regex. I feel it can be replaced with regex as we are trying to extract string out of specific pattern.
For first three cases, it's simple enough to extract String after :, but I couldn't find a way to deal with third case, unless I split the string ___ and apply the approach for first type of pattern and again concatenate them.

Just replace only the letters with exists next to : with the string replaced.
string.replaceAll("(?<=:)[A-Za-z]+", "replaced");
DEMO
or
If you also want to deal with digits, then add \d inside the char class.
string.replaceAll("(?<=:)[A-Za-z\\d]+", "replaced");

(:)[a-zA-Z]+
You can simply do this with string.replaceAll.Replace by $1replaced.See demo.
https://regex101.com/r/fX3oF6/18

How check that string is NOT blank by java regular expression?

There is regular expression for finding blank string and I want only negation. I also see this question but it does not work for java (see examples). Solution also not work for me (see 3-rd line in example).
For example
Pattern.compile("/^$|\\s+/").matcher(" ").matches() - false
Pattern.compile("/^$|\\s+/").matcher(" a").matches()- false
Pattern.compile("^(?=\\s*\\S).*$").matcher("\t\n a").matches() - false
return false in both cases.
P.S. If something is not clear ask me questions.
UPDATED
I want to use this regular expression in #Pattern annotation without creating custom annotation and programmatic validator for it. That's why I want a "plain" regexp solution without using find function.

It's not clear what you mean by negation.
If you mean "a string that contains at least one non-blank character," then you can use this:
Pattern.compile("\\S").matcher(str).find()
If it's really necessary to use matches, then you can do it with this.
Pattern.compile("\\A\\s*\\S.*\\Z").matcher(str).matches()
This just matches 0 or more spaces followed by a non-space followed by any characters at all up to the end of the string.
If you mean "a string that is all non-blank with at least one such character," then you can use this:
Pattern.compile("\\A\\S+\\Z").matcher(str).matches()
You need to study the Java regex syntax. In Java, regular expressions are compiled from strings, so there's no need for special delimiters like /.../ or %r{...} as you'll see in other languages.

How about this:
if(!string.trim().isEmpty()) {
// do something
}

Use regex \s : A whitespace character: \t\n\x0B\f\r.
Pattern.compile("\\s")

if then condition using regex in java

I have a pattern which goes like this
String1 :"String2",
i have to validate this pattern. here if u see there are two cases, the somestring1 can contain special characters if it is given within double quotes.
eg: "xxxx-xxx" :"yyyyyyyy",--------> is valid
but xxxx-xxx :"yyyyyyyy",--------> is not valid
"xxxx-xxx :"yyyyyyyy",--------> is not valid
So i need to create a regex which will check whether the double quotes is closed properly if it is present in String1.

Short answer: Regex doesn't work like that.
What you can do however, is to use two separate patterns to validate:
\"[^\"]+?\" :.*
To check the one that can contain special characters, and:
[a-zA-Z]+? :.*
To check the one that can't
EDIT:
Thinking some more about it, you could combine the two patterns above like so:
^(\"[^\"]+?\"|[a-zA-Z]+?) :.*$
Which will match something :"something" and "some-thing" :"something" but not "some-thing : "something" or some-thing : "something". Assuming that the string only contains the given text.

If I understand your question right, this simple regex should work
\"string1\" :\"string2\"

Maybe something like this?
(?<normalString>^[a-zA-Z]+$)|(?<specialString>^".*?"$)
This will capture only a-z characters and put them in the "normalString" group, or if there's an string within quotation marks, capture that and put it in the "specialString" group.

Necessary to escape a java regular expression in matches()?

I'm currently doing a test on an HTTP Origin to determine if it came from SSL:
(HttpHeaders.Names.ORIGIN).matches("/^https:\\/\\//")
But I'm finding it's not working. Do I need to escape matches() strings like a regular expression or can I leave it like https://? Is there any way to do a simple string match?
Seems like it would be a simple question, but surprisingly I'm not getting anywhere even after using a RegEx tester http://www.regexplanet.com/advanced/java/index.html. Thanks.

Java's regex doesn't need delimiters. Simply do:
.matches("https://.*")
Note that matches validates the entire input string, hence the .* at the end. And if the input contains line break chars (which . will not match), enable DOT-ALL:
.matches("(?s)https://.*")
Of couse, you could also simply do:
.startsWith("https://")
which takes a plain string (no regex pattern).

How about this Regex:
"^(https:)\/\/.*"
It works in your tester

How do I write a regular expression to find the following pattern?

I am trying to write a regular expression to do a find and replace operation. Assume Java regex syntax. Below are examples of what I am trying to find:
12341+1
12241+1R1
100001+1R2
So, I am searching for a string beginning with one or more digits, followed by a "1+1" substring, followed by 0 or more characters. I have the following regex:
^(\d+)(1\\+1).*
This regex will successfully find the examples above, however, my goal is to replace the strings with everything before "1+1". So, 12341+1 would become 1234, and 12241+1R1 would become 1224. If I use the first grouped expression $1 to replace the pattern, I get the wrong result as follows:
12341+1 becomes 12341
12241+1R1 becomes 12241
100001+1R2 becomes 100001
Any ideas?

Your existing regex works fine, just that you are missing a \ before \d
String str = "100001+1R2";
str = str.replaceAll("^(\\d+)(1\\+1).*","$1");
Working link

IMHO, the regex is correct.
Perhaps you wrote it wrong in the code. If you want to code the regex ^(\d+)(1\+1).* in a string, you have to write something like String regex = "^(\\d+)(1\\+1).*".
Your output is the result of ^(\d+)(1+1).* replacement, as you miss some backslash in the string (e.g. "^(\\d+)(1\+1).*").

Your regex looks fine to me - I don't have access to java but in JavaScript the code..
"12341+1".replace(/(\d+)(1\+1)/g, "$1");
Returns 1234 as you'd expect. This works on a string with many 'codes' in too e.g.
"12341+1 54321+1".replace(/(\d+)(1\+1)/g, "$1");
gives 1234 5432.

Personally, I wouldn't use a Regex at all (it'd be like using a hammer on a thumbtack), I'd just create a substring from (Pseudocode)
stringName.substring(0, stringName.indexOf("1+1"))
But it looks like other posters have already mentioned the non-greedy operator.
In most Regex Syntaxes you can add a '?' after a '+' or '*' to indicate that you want it to match as little as possible before moving on in the pattern. (Thus: ^(\d+?)(1+1) matches any number of digits until it finds "1+1" and then, NOT INCLUDING the "1+1" it continues matching, whereas your original would see the 1 and match it as well).